Waves

Given,

• Mass of the string, M = 2.50 kg
• Tension in the string, T = 200 N
• Length of the stretched string, l = 20.0 m

The mass per unit length of the string is

$\text m = \frac{\text M}{\text l} = \frac{2.50\ \text{kg}}{20.0\ \text m} = 0.125\ \text{kg m}^{-1}$

The speed of a transverse wave in a stretched string is given by

$\text v = \sqrt{\frac{\text T}{\text m}}$

Substituting the values,

$\text v = \sqrt{\frac{200\ \text N}{0.125\ \text{kg m}^{-1}}} = \sqrt{1600} = 40\ \text{m s}^{-1}$

The time taken by the transverse jerk to travel the whole length of the string is

$\text t = \frac{\text l}{\text v} = \frac{20.0\ \text m}{40\ \text{m s}^{-1}}$

$\text t = 0.5\ \text s$

Hence, the disturbance takes 0.5 s to reach the other end of the string.

Given,

• Height of the tower, h = 300 m
• Speed of sound in air, v = 340 m s^-1
• Acceleration due to gravity, g = 9.8 m s^-2

The splash is heard at the top after two separate intervals — the time t₁ taken by the stone to fall to the water surface, and the time t₂ taken by the sound of the splash to travel back up to the top.

Time of fall of the stone : The stone is dropped, so its initial velocity u = 0. From

$\text h = \text u\text t + \frac{1}{2}\text g\text t^2$

with u = 0,

$\text t_1 = \sqrt{\frac{2\text h}{\text g}} = \sqrt{\frac{2 \times 300}{9.8}} = \sqrt{61.22}$

$\text t_1 = 7.82\ \text s$

Time taken by the sound to travel up :

$\text t_2 = \frac{\text h}{\text v} = \frac{300\ \text m}{340\ \text{m s}^{-1}} = 0.88\ \text s$

Therefore, the total time after which the splash is heard at the top is

$\text t = \text t_1 + \text t_2 = 7.82 + 0.88$

$\text t = 8.70\ \text s$

Hence, the splash is heard at the top of the tower 8.70 s after the stone is dropped.

Given,

• Length of the steel wire, l = 12.0 m
• Mass of the steel wire, M = 2.10 kg
• Required speed of the transverse wave, v = 343 m s^-1

The mass per unit length of the wire is

$\text m = \frac{\text M}{\text l} = \frac{2.10\ \text{kg}}{12.0\ \text m} = 0.175\ \text{kg m}^{-1}$

The speed of a transverse wave in a stretched wire is given by

$\text v = \sqrt{\frac{\text T}{\text m}} \qquad \Rightarrow \qquad \text T = \text v^2 \text m$

Substituting the values,

$$\begin{aligned}\text T = (343\ \text{m s}^{-1})^2 \times 0.175\ \text{kg m}^{-1} \\[1em] = 117649 \times 0.175\end{aligned}$$

$\text T = 2.06 \times 10^4\ \text N$

Hence, the tension in the wire should be 2.06 × 10⁴ N.

The speed of sound in a gas is given by Laplace's formula,

$\text v = \sqrt{\frac{\gamma \text P}{\rho}} \qquad \dots(\text i)$

(a) Independent of pressure : Let one gram-molecule (mole) of the gas have pressure P and volume V. According to the gas equation,

$\text{PV} = \text{RT}$

If M is the molecular-weight of the gas and ρ its density, then $\text V = \frac{\text M}{\rho}$, so that

$\frac{\text P \text M}{\rho} = \text{RT} \qquad \Rightarrow \qquad \frac{\text P}{\rho} = \frac{\text{RT}}{\text M} \qquad \dots(\text{ii})$

At a constant temperature the ratio $\frac{\text P}{\rho}$ is a constant. That is, if P changes, the density ρ also changes in such a way that the ratio remains constant. Hence, if the temperature of the gas remains constant, then there is no effect of the pressure-change on the speed of sound.

(b) Increases with temperature : Substituting the value of $\frac{\text P}{\rho}$ from equation (ii) in equation (i),

$\text v = \sqrt{\frac{\gamma \text{RT}}{\text M}} \qquad \Rightarrow \qquad \text v \propto \sqrt{\text T}$

Here there is no term of pressure, and the speed of sound is directly proportional to the square-root of the absolute temperature of the gas. Hence the speed of sound increases with rise in temperature.

(c) Increases with humidity : The density of moist air, that is, air mixed with water-vapour, is less than the density of dry air. Assuming the value of γ for moist air to be the same as that for dry air, it is clear from equation (i) that at a constant pressure

$\text v \propto \frac{1}{\sqrt{\rho}}$

Hence, with increase in humidity the density decreases and the speed of sound in moist air is slightly greater than in dry air.

No, the converse is not true.

Every function of the combination (x − vt) or (x + vt) does represent a travelling wave, but the mere fact that x and t occur in this combination is not by itself sufficient. An acceptable function for a travelling wave must also be finite everywhere and at all times, since the displacement of a particle of the medium can never become infinitely large.

Examining the three functions on this basis :

(a) (x − vt)² — as x or t increases without limit, the value of the function increases without limit. It is not finite everywhere, and so it cannot represent a travelling wave.

(b) $\log \left(\frac{\text x + \text{vt}}{\text x_0}\right)$ — when (x + vt) tends to zero, the logarithm tends to − ∞, and the function is not finite. Hence it cannot represent a travelling wave.

(c) $\frac{1}{\text x + \text{vt}}$ — this function is finite at all points and at all times, and it also contains x and t in the combination (x + vt). Hence it can represent a travelling wave.

Hence, out of the three, only the function in part (c) can possibly represent a travelling wave.

Given,

• Frequency of the ultrasonic sound, n = 1000 kHz = 1000 × 10³ Hz
• Speed of sound in air, v_a = 340 m s^-1
• Speed of sound in water, v_w = 1486 m s^-1

The frequency of a wave is decided by the source and remains unchanged whether the wave travels in air or in water. Hence the frequency of both the reflected and the transmitted sound is 1000 kHz.

(a) Reflected sound : The reflected sound travels back in air, so

$\lambda_\text a = \frac{\text v_\text a}{\text n} = \frac{340\ \text{m s}^{-1}}{1000 \times 10^3\ \text{s}^{-1}}$

$\lambda_\text a = 0.34 \times 10^{-3}\ \text m$

(b) Transmitted sound : The transmitted sound travels in water, so

$\lambda_\text w = \frac{\text v_\text w}{\text n} = \frac{1486\ \text{m s}^{-1}}{1000 \times 10^3\ \text{s}^{-1}}$

$\lambda_\text w = 1.486 \times 10^{-3}\ \text m$

Hence, the wavelength of the reflected sound is 0.34 × 10^-3 m and that of the transmitted sound is 1.486 × 10^-3 m.

Given,

• Speed of the ultrasonic sound in the tissue, v = 1.7 km s^-1 = 1.7 × 10³ m s^-1
• Operating frequency of the scanner, n = 4.2 MHz = 4.2 × 10⁶ Hz

From the relation between speed, frequency and wavelength,

$\text v = \text n \lambda \qquad \Rightarrow \qquad \lambda = \frac{\text v}{\text n}$

Substituting the values,

$\lambda = \frac{1.7 \times 10^3\ \text{m s}^{-1}}{4.2 \times 10^6\ \text{s}^{-1}}$

$\lambda = 4.0 \times 10^{-4}\ \text m$

Hence, the wavelength of sound in the tissue is 4.0 × 10^-4 m.

Given, the wave is

$\text y \text{ (x, t)} = 3.0 \sin\left(36 \text t + 0.018 \text x + \frac{\pi}{4}\right)$

where x, y are in cm and t in s.

The standard equation of a plane progressive simple harmonic wave travelling along the − X direction, that is, from right to left, is

$\text y = \text a \sin (\omega \text t + \text k \text x + \phi) \qquad \dots(\text i)$

where a is the amplitude, ω the angular frequency, k the propagation constant and φ the initial phase at the origin.

(a) The argument of the sine contains x and t in the combination (ωt + kx), so the given equation is identical in form with equation (i). Hence it represents a travelling wave moving from right to left, that is, along the negative direction of the X-axis. Comparing, ω = 36 s^-1 and k = 0.018 cm^-1, so the speed is

$\text v = \frac{\omega}{\text k} = \frac{36}{0.018} = 2000\ \text{cm s}^{-1} = 20\ \text{m s}^{-1}$

(b) Comparing with equation (i), the amplitude is

$\text a = 3.0\ \text{cm}$

and the frequency is

$\text n = \frac{\omega}{2\pi} = \frac{36}{2 \times 3.14} = 5.7\ \text{Hz}$

(c) The initial phase at the origin is

$\phi = \frac{\pi}{4}$

(d) The least distance between two successive crests is the wavelength λ. Since $\text k = \frac{2\pi}{\lambda}$,

$\lambda = \frac{2\pi}{\text k} = \frac{2 \times 3.14}{0.018} = 348.9\ \text{cm}$

$\lambda = 3.5\ \text m$

Hence, it is a travelling wave moving from right to left with a speed of 20 m s^-1, its amplitude is 3.0 cm, its frequency is 5.7 Hz, its initial phase at the origin is $\frac{\pi}{4}$ and the least distance between two successive crests is 3.5 m.

The wave of Question 8 is

$\text y \text{ (x, t)} = 3.0 \sin\left(36 \text t + 0.018 \text x + \frac{\pi}{4}\right)$

At x = 0 :

$\text y = 3.0 \sin\left(36 \text t + \frac{\pi}{4}\right)$

Here

$\omega = \frac{2\pi}{\text T} = 36 \qquad \Rightarrow \qquad \text T = \frac{2\pi}{36}$

Taking values of t at intervals of $\frac{\text T}{8}$, the corresponding values of y are

Plotting these values gives the following graph.

At x = 2 cm :

$\text y = 3.0 \sin\left(36 \text t + 0.036 + \frac{\pi}{4}\right)$

At x = 4 cm :

$\text y = 3.0 \sin\left(36 \text t + 0.072 + \frac{\pi}{4}\right)$

Shapes of the graphs : In each of the three cases the graph so obtained is sinusoidal, the three curves being identical in shape.

Aspect in which the motion differs : The amplitude of every curve is 3.0 cm and the angular frequency of every curve is 36 s^-1, so the amplitude and the frequency are the same at all the three points. Only the constant term inside the sine, that is, the initial phase, changes from 0 to 0.036 and then to 0.072.

Hence, the oscillatory motion of a travelling wave differs from one point to another only in phase, and not in amplitude or frequency.

Given, the travelling harmonic wave is

y (x, t) = 2.0 cos 2π (10 t − 0.0080 x + 0.35)

where x and y are in cm and t in s.

Re-writing the equation,

$\text y = 2.0 \cos \left[2\pi(10 \text t - 0.0080 \text x) + 2\pi (0.35)\right]$

$= 2.0 \cos \left[\frac{2\pi}{\left(\frac{1}{0.0080}\right)}\left(\frac{10}{0.0080}\text t - \text x\right) + 2\pi(0.35)\right]$

Comparing this with the general equation of a travelling wave,

$\text y = \text a \cos \left[\frac{2\pi}{\lambda}(\text{vt} - \text x) + \phi_0\right]$

we get

$\lambda = \frac{1}{0.0080\ \text{cm}^{-1}} = 125\ \text{cm} = 1.25\ \text m$

The phase difference between two points separated by a path difference Δx is

$\Delta \phi = \frac{2\pi}{\lambda} \times \Delta \text x$

(a) For Δx = 4 m :

$\Delta \phi = \frac{2\pi}{1.25\ \text m} \times 4\ \text m = 6.4\ \pi\ \text{rad}$

(b) For Δx = 0.5 m :

$\Delta \phi = \frac{2\pi}{1.25\ \text m} \times 0.5\ \text m = 0.8\ \pi\ \text{rad}$

(c) For Δx = λ/2 :

$\Delta \phi = \frac{2\pi}{\lambda} \times \frac{\lambda}{2} = \pi\ \text{rad}$

(d) For Δx = 3λ/4 :

$\Delta \phi = \frac{2\pi}{\lambda} \times \frac{3\lambda}{4} = \frac{3}{2}\pi\ \text{rad}$

Hence, the phase differences are 6.4π rad, 0.8π rad, π rad and $\frac{3\pi}{2}$ rad respectively.

Given,

• Displacement, $\text y \text{ (x, t)} = 0.06 \sin \left(\frac{2\pi \text x}{3}\right) \cos (120 \pi \text t)$, with x and y in metre and t in second
• Length of the string, l = 1.5 m
• Mass of the string, M = 3.0 × 10^-2 kg

(a) In the argument of the trigonometric functions x and t do not occur in the combination (ωt ± kx), which is the characteristic of a travelling wave. The equation is instead a product of a harmonic function of x and a harmonic function of t taken separately. Hence the function represents a stationary wave.

(b) Let a wave travelling on the string in the positive direction of the X-axis be

$\text y_1 = \text a \sin (\omega \text t - \text k\text x)$

The string is clamped at its ends, so the wave reflected at the clamped end suffers a phase change of π and travels in the negative direction of the X-axis,

$\text y_2 = - \text a \sin (\omega \text t + \text k\text x)$

By the principle of superposition, the stationary wave formed is

$$\begin{aligned}\text y = \text y_1 + \text y_2 = \text a\left[\sin (\omega \text t - \text k\text x) - \sin (\omega \text t + \text k\text x)\right] \\[1em] = - 2\text a \cos \omega \text t \sin \text k\text x \qquad \dots(\text i)\end{aligned}$$

Comparing equation (i) with the given equation,

$\text k = \frac{2\pi}{\lambda} = \frac{2\pi}{3} \qquad \Rightarrow \qquad \lambda = 3\ \text m$

$\omega = 2\pi \text n = 120\pi \qquad \Rightarrow \qquad \text n = 60\ \text{Hz}$

The speed of propagation of each wave is

$\text v = \text n \lambda = 60 \times 3 = 180\ \text{m s}^{-1}$

(c) The mass per unit length of the string is

$\text m = \frac{\text M}{\text l} = \frac{3.0 \times 10^{-2}\ \text{kg}}{1.5\ \text m} = 2.0 \times 10^{-2}\ \text{kg m}^{-1}$

The speed of a transverse wave in a stretched string is $\text v = \sqrt{\frac{\text T}{\text m}}$, so that

$\text T = \text v^2 \text m = (180)^2 \times (2.0 \times 10^{-2})$

$\text T = 648\ \text N$

Hence, it is a stationary wave; each component wave has a wavelength of 3 m, a frequency of 60 Hz and a speed of 180 m s^-1, and the tension in the string is 648 N.

The stationary wave on the string is

$\text y \text{ (x, t)} = 0.06 \sin \left(\frac{2\pi \text x}{3}\right) \cos (120 \pi \text t)$

(i) (a) Frequency — Yes.

Every point of the string carries the common time factor cos (120πt) with the same angular frequency ω = 120π rad s^-1. Hence all the points oscillate with the same frequency,

$\text n = \frac{\omega}{2\pi} = \frac{120\pi}{2\pi} = 60\ \text{Hz}$

(i) (b) Phase — Not the same for all points.

The time dependence cos (120πt) is the same for every x, but the spatial factor $\sin \left(\frac{2\pi \text x}{3}\right)$ may be positive or negative. Where this factor is positive the point oscillates as + |A(x)| cos ωt, and where it is negative it oscillates as |A(x)| cos (ωt + π). Hence all the points lying in the same segment between two successive nodes vibrate in the same phase, while the points on the two sides of a node are opposite in phase.

(i) (c) Amplitude — No.

The amplitude of a point at a distance x is

$\text A(\text x) = 0.06 \left|\sin \left(\frac{2\pi \text x}{3}\right)\right|\ \text m$

which depends on x. It is zero at the nodes and maximum, equal to 0.06 m, at the antinodes. Hence different points have different amplitudes.

(ii) Amplitude at x = 0.375 m :

Here $\text k = \frac{2\pi}{3}$, so

$\text k\text x = \frac{2\pi}{3} \times 0.375 = \frac{\pi}{4}$

$\sin (\text k\text x) = \sin \frac{\pi}{4} = \frac{\sqrt{2}}{2} = 0.707$

Therefore,

$\text A(0.375) = 0.06 \times 0.707$

$\text A(0.375) = 0.042\ \text m$

Hence, all the points oscillate with the same frequency of 60 Hz but not with the same phase or amplitude, and the amplitude of a point 0.375 m from one end is 0.042 m.

A function represents a travelling wave only if x and t occur in it in the combination (ωt ± kx), and it represents a stationary wave only if it contains harmonic functions of x and of t separately. Applying this test :

(a) y = 2 cos (3 x) sin (10 t)

The equation contains harmonic functions of x and of t separately. Hence it represents a stationary wave.

(b) $\text y = 2\sqrt{\text x - \text v \text t}$

Although x and t occur in the combination (x − vt), the function is not a harmonic function. Hence it represents none of the two — it cannot represent any type of wave.

(c) y = 3 sin (5 x − 0.5 t) + 4 cos (5 x − 0.5 t)

Both the terms are harmonic functions in which x and t occur in the combination (5x − 0.5t). Their sum is therefore a single harmonic function of the same combination, so it represents a travelling wave.

(d) y = cos x sin t + cos 2 x sin 2 t

Each of the two terms separately contains a harmonic function of x and a harmonic function of t. Hence the equation is a sum of two stationary waves, that is, it represents a superposition of two stationary waves.

Given,

• Frequency of the fundamental mode, n = 45 Hz
• Mass of the wire, M = 3.5 × 10^-2 kg
• Linear mass density of the wire, m = 4.0 × 10^-2 kg m^-1

The length of the wire is

$\text l = \frac{\text M}{\text m} = \frac{3.5 \times 10^{-2}\ \text{kg}}{4.0 \times 10^{-2}\ \text{kg m}^{-1}} = 0.875\ \text m$

(a) Speed of the transverse wave : The wire is stretched between two rigid supports, so in the fundamental mode there is a node at each end and an antinode in the middle. The frequency of the fundamental mode is

$\text n = \frac{\text v}{2\text l} \qquad \Rightarrow \qquad \text v = \text n \times 2\text l$

Substituting the values,

$\text v = 45 \times 2 \times 0.875$

$\text v = 78.75\ \text{m s}^{-1}$

(b) Tension in the wire : The speed of a transverse wave in a stretched wire is $\text v = \sqrt{\frac{\text T}{\text m}}$, so that

$$\begin{aligned}\text T = \text v^2 \text m = (78.75)^2 \times (4.0 \times 10^{-2}) \\[1em] = 6201.56 \times 4.0 \times 10^{-2}\end{aligned}$$

$\text T = 248\ \text N$

Hence, the speed of the transverse wave in the wire is 78.75 m s^-1 and the tension in the wire is 248 N.

Given,

• Frequency of the tuning fork, n = 340 Hz
• First resonating length of the air column, l₁ = 25.5 cm
• Second resonating length of the air column, l₂ = 79.3 cm

The tube is open at one end and closed by the movable piston at the other, so it behaves as a closed pipe. In a closed pipe the successive resonating lengths differ by half a wavelength,

$\text l_2 - \text l_1 = \frac{\lambda}{2} \qquad \Rightarrow \qquad \lambda = 2(\text l_2 - \text l_1)$

Therefore the speed of sound in the air column is

$\text v = \text n \lambda = 2\text n(\text l_2 - \text l_1)$

Substituting the values,

$$\begin{aligned}\text v = 2 \times 340 \times (79.3 - 25.5)\ \text{cm s}^{-1} \\[1em] = 2 \times 340 \times 53.8 = 36584\ \text{cm s}^{-1}\end{aligned}$$

$\text v = 365.84\ \text{m s}^{-1}$

Hence, the speed of sound in air at the temperature of the experiment is 365.84 m s^-1.

Given,

• Length of the steel rod, l = 100 cm
• Fundamental frequency of longitudinal vibrations, n = 2.53 kHz = 2.53 × 10³ Hz

The rod is clamped at its middle point, so the middle point is necessarily a node (N). In the fundamental mode of longitudinal vibration the two end-points of the rod are free to vibrate, so each end is an antinode (A).

The distance between two consecutive antinodes is $\frac{\lambda}{2}$, where λ is the wavelength of the stationary wave set up in the rod. Hence

$\text l = \frac{\lambda}{2} \qquad \Rightarrow \qquad \lambda = 2\text l$

$\lambda = 2 \times 100 = 200\ \text{cm}$

The speed of sound in the rod is

$$\begin{aligned}\text v = \text n \lambda = (2.53 \times 10^3\ \text{s}^{-1}) \times (200\ \text{cm}) \\[1em] = 5.06 \times 10^5\ \text{cm s}^{-1}\end{aligned}$$

$\text v = 5.06 \times 10^3\ \text{m s}^{-1}$

Hence, the speed of sound in steel is 5.06 × 10³ m s^-1.

Given,

• Length of the pipe, l = 20 cm = 0.2 m
• Frequency of the source, n = 430 Hz
• Speed of sound in air, v = 340 m s^-1

Pipe closed at one end : A closed organ pipe produces only odd harmonics. The frequency of the mth mode of vibration is

$\text n = \frac{(2\text m - 1)\text v}{4\text l}, \qquad \text m = 1, 2, 3, \dots$

Substituting the values,

$430 = \frac{(2\text m - 1) \times 340}{4 \times 0.2}$

$(2\text m - 1) = \frac{430 \times 4 \times 0.2}{340} = \frac{344}{340} \approx 1$

$\text m = 1$

Hence the fundamental mode, that is, the first harmonic, of the air column in the closed pipe is resonantly excited.

Pipe open at both ends : An open organ pipe produces all the harmonics, and the frequency of the mth mode is

$\text n = \frac{\text m \text v}{2\text l}$

Substituting the values,

$430 = \frac{\text m \times 340}{2 \times 0.2} \qquad \Rightarrow \qquad \text m = \frac{430 \times 2 \times 0.2}{340} = 0.5$

But m must be an integer, and 0.5 is not an integer.

Hence, the 430 Hz source excites the fundamental mode (first harmonic) of the closed pipe, and the same source cannot be in resonance with the pipe if both of its ends are open.

Given,

• Original frequency of the string A, n_A = 324 Hz
• Beat frequency before the change, = 6 Hz
• Beat frequency after the tension in A is reduced, = 3 Hz

The number of beats per second is equal to the difference in the frequencies of the two sources. Hence the possible frequencies of the string B are

$\text n_\text B = 324 \pm 6 = 330\ \text{Hz} \quad \text{or} \quad 318\ \text{Hz}$

The frequency of a stretched string is proportional to the square-root of the tension in it,

$\text n = \frac{1}{2\text l}\sqrt{\frac{\text T}{\text m}} \qquad \Rightarrow \qquad \text n \propto \sqrt{\text T}$

When the tension in A is slightly reduced, the frequency of A becomes slightly lower than 324 Hz.

• If the frequency of B were 330 Hz, then on lowering the frequency of A below 324 Hz the difference (n_B − n_A) would become more than 6 Hz, so the beat frequency would increase.
• If the frequency of B were 318 Hz, then on lowering the frequency of A below 324 Hz the difference (n_A − n_B) would become less than 6 Hz, so the beat frequency would decrease.

It is given that the beat frequency is reduced to 3 Hz. This is possible only in the second case.

Hence, the frequency of the string B is 318 Hz.

(a) In a sound wave, a displacement node is a pressure antinode and vice-versa.

A displacement node is a point where the amplitude of vibration of the particles of the medium is the minimum, that is, zero. At such a point the particles on the two sides move towards it or away from it together, so the crowding together and thinning out of the particles is the greatest there. Hence the variation in pressure is maximum, which makes it a pressure antinode. At a displacement antinode, on the other hand, the amplitude of vibration is maximum but the particles on the two sides move in the same direction by the same amount, so there is no change in pressure. Hence a displacement antinode is a pressure node.

(b) Bats can ascertain distances, directions, nature and sizes of the obstacles without any 'eyes'.

Bats emit ultrasonic sound waves of very high frequency. These waves are reflected back towards the bat by the obstacles in its path. From the reflected waves the bat estimates the distance, the direction, the size and the nature of the obstacle with the help of its brain senses.

(c) A violin note and a sitar note may have the same frequency, yet we can distinguish between the two notes.

Along with the fundamental tone, a musical instrument also produces a number of overtones. The overtones produced by a violin and those produced by a sitar are of different strengths, so the quality or timbre of the sound produced by them is different. Hence the two notes of the same frequency can be distinguished from each other.

(d) Solids can support both longitudinal and transverse waves, but only longitudinal waves can propagate in gases.

Solids have a shear modulus, so they can sustain shearing stress. The propagation of a transverse wave is such that it produces shearing stress in the medium, so its propagation is possible only in solids. Gases do not have any definite shape and so they yield to shearing stress. Hence a transverse wave cannot propagate in a gas, and only longitudinal waves, for which the medium is alternately compressed and rarefied, can travel in it.

(e) The shape of a pulse gets distorted during propagation in a dispersive medium.

A pulse is actually a combination of waves having different wavelengths. In a dispersive medium the speed of a wave depends upon its wavelength, so these component waves travel with different velocities. As the pulse advances, the component waves get separated from one another and this results in the distortion of the shape of the pulse.

Given,

• y₁ = 4 sin (2x − 6t), so a₁ = 4
• $\text y_2 = 3 \sin \left(2\text x - 6\text t - \frac{\pi}{2}\right)$, so a₂ = 3
• Phase difference between the two waves, $\phi = \frac{\pi}{2}$

The two waves have the same frequency and travel along the same path in the same direction, so by the principle of superposition the amplitude of the resultant wave is

$\text A = \sqrt{\text a_1^2 + \text a_2^2 + 2\text a_1 \text a_2 \cos \phi}$

Substituting the values,

$$\begin{aligned}\text A = \sqrt{(4)^2 + (3)^2 + 2 \times 4 \times 3 \times \cos 90^\circ} \\[1em] = \sqrt{16 + 9 + 0} = \sqrt{25}\end{aligned}$$

$\text A = 5$

Hence, the amplitude of the resultant wave is 5 units.

π /3

Given, the two particles execute simple harmonic motion of the same amplitude a and the same angular frequency ω along the X-axis, their mean positions being separated by a distance x₀, and the maximum separation between them is (x₀ + a).

Let the displacements of the two particles from their own mean positions be

$\text x_1 = \text a \sin \omega \text t, \qquad \text x_2 = \text x_0 + \text a \sin (\omega \text t + \phi)$

where φ is the phase difference between their motions.

The separation between the two particles at any instant is

$\text x_2 - \text x_1 = \text x_0 + \text a \sin (\omega \text t + \phi) - \text a \sin \omega \text t$

Using $\sin \text C - \sin \text D = 2 \cos \frac{\text C + \text D}{2} \sin \frac{\text C - \text D}{2}$,

$\text x_2 - \text x_1 = \text x_0 + 2\text a \sin \frac{\phi}{2}\cos \left(\omega \text t + \frac{\phi}{2}\right)$

The cosine term has a maximum value of 1, so the maximum separation is

$(\text x_2 - \text x_1)_{max} = \text x_0 + 2\text a \sin \frac{\phi}{2}$

It is given that this maximum separation is (x₀ + a). Therefore,

$\text x_0 + 2\text a \sin \frac{\phi}{2} = \text x_0 + \text a$

$\sin \frac{\phi}{2} = \frac{1}{2} \qquad \Rightarrow \qquad \frac{\phi}{2} = \frac{\pi}{6}$

$\phi = \frac{\pi}{3}$

Hence, the phase difference between their motions is $\frac{\pi}{3}$.

$4\sqrt{2}$

Given, the displacement of the particle is

x = 4 (cos π t + sin πt)

To find the amplitude, the expression is written as a single harmonic function. Multiplying and dividing by $\sqrt{2}$,

$\text x = 4\sqrt{2}\left[\frac{1}{\sqrt{2}}\cos \pi \text t + \frac{1}{\sqrt{2}}\sin \pi \text t\right]$

Since $\frac{1}{\sqrt{2}} = \sin \frac{\pi}{4} = \cos \frac{\pi}{4}$,

$\text x = 4\sqrt{2}\left[\sin \frac{\pi}{4}\cos \pi \text t + \cos \frac{\pi}{4}\sin \pi \text t\right]$

Using sin A cos B + cos A sin B = sin (A + B),

$\text x = 4\sqrt{2}\sin \left(\pi \text t + \frac{\pi}{4}\right)$

Comparing this with the standard equation x = A sin (ωt + φ), the amplitude is

$\text A = 4\sqrt{2}$

Hence, the amplitude of the particle is $4\sqrt{2}$.

n = 5, f₂ = (5 / 4) f₁

Given, an open pipe resonates in its second harmonic with frequency f₁, and after closing one end the same pipe resonates in its nth harmonic with frequency f₂.

Let l be the length of the pipe and v the speed of sound in air.

Open pipe : The frequency of the second harmonic of an open pipe is

$\text f_1 = 2 \times \frac{\text v}{2\text l} = \frac{\text v}{\text l}$

Closed pipe : A closed pipe produces only odd harmonics, and the frequency of its nth harmonic is

$\text f_2 = \text n \times \frac{\text v}{4\text l} = \frac{\text n}{4}\left(\frac{\text v}{\text l}\right) = \frac{\text n}{4}\text f_1$

where n is odd, that is, n = 1, 3, 5, ...

It is given that the frequency is slowly raised until resonance occurs, so f₂ must be greater than f₁. That is,

\dfrac{\text n}{4} \gt 1 \qquad \Rightarrow \qquad \text n &ampgt 4

The smallest odd value of n greater than 4 is n = 5. Hence the first resonance on raising the frequency occurs at

$\text f_2 = \frac{5}{4}\text f_1$

Hence, n = 5 and $\text f_2 = \frac{5}{4}\text f_1$.

$\frac{4\text L}{3}\sqrt{\frac{\rho_1}{\rho_2}}$

Given,

• Length of the closed organ pipe = L, gas density = ρ₁
• Length of the open organ pipe = L' (say), gas density = ρ₂
• The compressibility of the gases is the same in both the pipes
• Both pipes vibrate in their first overtone with the same frequency

Let v₁ and v₂ be the speeds of sound in the two gases. Since the compressibility, and hence the elasticity, is the same for both,

$\text v = \sqrt{\frac{\gamma \text P}{\rho}} \qquad \Rightarrow \qquad \frac{\text v_1}{\text v_2} = \sqrt{\frac{\rho_2}{\rho_1}}$

First overtone of the closed pipe : For a closed pipe the first overtone is the third harmonic, so

$\text n_2 = 3\text n_1 = 3\left(\frac{\text v_1}{4\text L}\right)$

First overtone of the open pipe : For an open pipe the first overtone is the second harmonic, so

$\text n_2' = 2\text n_1' = 2\left(\frac{\text v_2}{2\text L'}\right)$

The two frequencies are equal, that is, n₂ = n₂'. Therefore,

$3\left(\frac{\text v_1}{4\text L}\right) = 2\left(\frac{\text v_2}{2\text L'}\right)$

$\text L' = \frac{4}{3}\text L\left(\frac{\text v_2}{\text v_1}\right)$

Substituting $\frac{\text v_2}{\text v_1} = \sqrt{\frac{\rho_1}{\rho_2}}$,

$\text L' = \frac{4\text L}{3}\sqrt{\frac{\rho_1}{\rho_2}}$

Hence, the length of the open organ pipe is $\frac{4\text L}{3}\sqrt{\frac{\rho_1}{\rho_2}}$.

argon $\left(\text M = 36, \sqrt{\frac{10}{36}} = \frac{17}{32}\right)$

Given,

• Frequency of the tuning fork, n = 244 s^-1
• Minimum height at which resonance occurs, l = (0.350 ± 0.005) m

The resonance column tube is closed at the lower end by the gas column, so at the minimum height the air column vibrates in its fundamental mode,

$\text l = \frac{\lambda}{4} \qquad \Rightarrow \qquad \lambda = 4\text l$

The speed of sound in the gas is therefore

$\text v = \text n \lambda = 4\text n\text l = 4 \times 244 \times 0.35 = 341.6\ \text{m s}^{-1}$

Maximum error in the speed : Since n is fixed, $\frac{\Delta \text v}{\text v} = \frac{\Delta \text l}{\text l}$, so

$\Delta \text v = \text v \times \frac{\Delta \text l}{\text l} = 341.6 \times \frac{0.005}{0.350} = 4.88\ \text{m s}^{-1} \approx 5\ \text{m s}^{-1}$

Hence the permissible range of the speed is v = (342 ± 5) m s^-1.

Speed of sound in each gas : Using $\text v = \sqrt{\frac{\gamma \text{RT}}{\text M \times 10^{-3}}}$, where M is the molar mass in gram,

$\text v = \sqrt{100\gamma \text{RT}} \times \sqrt{\frac{10}{\text M}}$

(a) Neon is monoatomic, γ = 1.67 and M = 20,

$\text v = \sqrt{167\text{RT}} \times \sqrt{\frac{10}{20}} = 640 \times \frac{7}{10} = 448\ \text{m s}^{-1}$

(b) Nitrogen is diatomic, γ = 1.4 and M = 28,

$\text v = \sqrt{140\text{RT}} \times \sqrt{\frac{10}{28}} = 590 \times \frac{3}{5} = 354\ \text{m s}^{-1}$

(c) Oxygen is diatomic, γ = 1.4 and M = 32,

$\text v = \sqrt{140\text{RT}} \times \sqrt{\frac{10}{32}} = 590 \times \frac{9}{16} = 331.8\ \text{m s}^{-1}$

(d) Argon is monoatomic, γ = 1.67 and M = 36,

$\text v = \sqrt{167\text{RT}} \times \sqrt{\frac{10}{36}} = 640 \times \frac{17}{32} = 340\ \text{m s}^{-1}$

Only the value 340 m s^-1 lies within the permissible range (342 ± 5) m s^-1.

Hence, the gas in the tube is argon.

6

Given,

• Length of the pipe, l = 85 cm = 0.85 m
• Speed of sound in air, v = 340 m/s
• Upper limit of frequency = 1250 Hz

The pipe is closed at one end, so it produces only odd harmonics. The frequency of the mode corresponding to m = 0, 1, 2, ... is

$\text n = \frac{(2\text m + 1)\text v}{4\text l}$

The natural oscillations are required to have frequencies below 1250 Hz, so

$\frac{(2\text m + 1) \times 340}{4 \times 0.85} \le 1250$

$(2\text m + 1) \le \frac{1250 \times 4 \times 0.85}{340} = \frac{4250}{340} = 12.5$

$2\text m \le 11.5 \qquad \Rightarrow \qquad \text m \le 5.75$

Since m must be a whole number, the permissible values are

$\text m = 0, 1, 2, 3, 4, 5$

Hence, there are 6 possible natural oscillations of the air column whose frequencies lie below 1250 Hz.

0.025 m

Given,

• Length of the air column resonating in the fundamental mode, l₁ = 0.1 m
• Length of the air column resonating with the first overtone, l₂ = 0.35 m

Let e be the end correction. In a resonance column, which behaves as a closed pipe,

Fundamental mode :

$\text l_1 + \text e = \frac{\lambda}{4} \qquad \dots(\text i)$

First overtone, which is the third harmonic :

$\text l_2 + \text e = \frac{3\lambda}{4} \qquad \dots(\text{ii})$

Multiplying equation (i) by 3,

$3\text l_1 + 3\text e = \frac{3\lambda}{4} \qquad \dots(\text{iii})$

Comparing equations (ii) and (iii),

$\text l_2 + \text e = 3\text l_1 + 3\text e$

$\text e = \frac{\text l_2 - 3\text l_1}{2}$

Substituting the values,

$\text e = \frac{0.35 - 3(0.1)}{2} = \frac{0.35 - 0.30}{2} = \frac{0.05}{2}$

$\text e = 0.025\ \text m$

Hence, the end correction is 0.025 m.

15.2 cm

Given,

• Diameter of the column tube, d = 4 cm
• Frequency of the tuning fork, n = 512 Hz
• Speed of sound in air, v = 336 m/s = 33600 cm/s
• The zero of the metre scale coincides with the top of the resonance column tube

At the first resonance the air column vibrates in its fundamental mode, so the distance between the antinode at the open end and the node at the water surface is

$\frac{\lambda}{4} = \text l_1 + \text e$

where l₁ is the length of the air column and e is the end correction. Therefore,

$\text l_1 = \frac{\lambda}{4} - \text e$

The end correction for a tube of diameter d is e = 0.3 d. Also $\lambda = \frac{\text v}{\text n}$, so

$\text l_1 = \frac{\text v}{4\text n} - 0.3\text d$

Substituting the values,

$$\begin{aligned}\text l_1 = \frac{336 \times 100}{4 \times 512}\ \text{cm} - 0.3 \times 4\ \text{cm} \\[1em] = 16.4\ \text{cm} - 1.2\ \text{cm}\end{aligned}$$

$\text l_1 = 15.2\ \text{cm}$

Since the zero of the scale coincides with the top of the tube, the reading of the water level is equal to the length of the air column.

Hence, the reading of the water level in the column is 15.2 cm.

a low-pressure pulse starts travelling up the pipe, if the other end of the pipe is open

a high-pressure pulse starts travelling up the pipe, if the other end of the pipe is closed.

For sound waves, which are pressure waves, an optically rarer medium behaves as the denser medium, so the behaviour of sound waves on reflection is just the opposite of that of light waves.

If the other end is open : The open end behaves as a free boundary for the pressure wave. A compression, that is, a high-pressure pulse, is reflected back as a rarefaction, that is, a low-pressure pulse. Hence a low-pressure pulse starts travelling up the pipe.

If the other end is closed : The closed end behaves as a rigid boundary. A compression is reflected back as a compression, so a high-pressure pulse starts travelling up the pipe.

Hence, a low-pressure pulse starts travelling up the pipe if the other end of the pipe is open, and a high-pressure pulse starts travelling up the pipe if the other end is closed.

Note: Both the second and the fourth options are correct statements of what happens, and the printed answer key marks both of them.

$\text y(\text t) = \text A \sin \frac{\pi \text x}{6} \cos \frac{50 \pi \text t}{3}$, $\text y(\text t) = \text A \sin \frac{5\pi \text x}{6} \cos \frac{250 \pi \text t}{3}$ and $\text y(\text t) = \text A \sin \frac{5\pi \text x}{2} \cos 250 \pi \text t$

Given,

• Length of the taut string, l = 3 m, fixed at x = 0
• Speed of the waves in the string, v = 100 m s^-1
• The other end is vibrated so that stationary waves are set up

Since the end x = 0 is fixed, it must be a node, and every option already has the factor sin kx, which is zero at x = 0. The end x = 3 m is being vibrated, so it must be an antinode. Hence at x = 3, sin kx must be ± 1.

Also, the speed of the wave in the string must come out to be 100 m s^-1 in every acceptable waveform, that is, $\text v = \frac{\omega}{\text k} = 100$.

Testing each option :

(1) $\sin \frac{\pi \text x}{6}$ at x = 3 is $\sin \frac{3\pi}{6} = 1$, so x = 3 is an antinode. Here $\omega = \frac{50\pi}{3}$ and $\text k = \frac{\pi}{6}$, so $\text v = \frac{50\pi/3}{\pi/6} = 100$ m s^-1. Acceptable.

(2) $\sin \frac{\pi \text x}{3}$ at x = 3 is $\sin \pi = 0$, so x = 3 is a node, not an antinode. Not acceptable.

(3) $\sin \frac{5\pi \text x}{6}$ at x = 3 is $\sin \frac{15\pi}{6} = 1$, so x = 3 is an antinode. Here $\omega = \frac{250\pi}{3}$ and $\text k = \frac{5\pi}{6}$, so $\text v = \frac{250\pi/3}{5\pi/6} = 100$ m s^-1. Acceptable.

(4) $\sin \frac{5\pi \text x}{2}$ at x = 3 is $\sin \frac{15\pi}{2} = -1$, so x = 3 is an antinode. Here ω = 250π and $\text k = \frac{5\pi}{2}$, so $\text v = \frac{250\pi}{5\pi/2} = 100$ m s^-1. Acceptable.

Hence, the possible waveforms are those given in the first, the third and the fourth options.

the intensity of the sound heard at the first resonance was more than that at the second resonance, and the length of the air-column at the first resonance was some what shorter than 1/4th of the wavelength of the sound in air

Given, a resonance air-column experiment in which the first resonance corresponds to the shorter air-column and the second resonance to the longer air-column.

First option — correct. As the length of the air column increases, a part of the energy supplied by the fork is dissipated in the longer column. Hence the sound heard at the first resonance is louder than that heard at the second resonance.

Second option — incorrect. The prongs of the tuning fork are held in a vertical plane above the mouth of the resonance tube, so that the vibrations are communicated along the length of the air column.

Third option — incorrect. The amplitude of vibration of the ends of the prongs of a tuning fork is of the order of a fraction of a millimetre, and never as large as 1 cm.

Fourth option — correct. On account of the end correction e, the antinode is formed slightly outside the open end. Hence

$\text l_1 + \text e = \frac{\lambda}{4} \qquad \Rightarrow \qquad \text l_1 = \frac{\lambda}{4} - \text e$

that is, the length of the air column at the first resonance is somewhat shorter than $\frac{\lambda}{4}$.

Hence, the first and the fourth alternatives are correct.

5

Given,

• Length of the string, l = 20 cm = 20 × 10^-2 m
• Mass of the string, M = 1.0 g = 1.0 × 10^-3 kg
• Tension in the string, T = 0.5 N
• Frequency of the external vibrator, n = 100 Hz

The mass per unit length of the string is

$\text m = \frac{\text M}{\text l} = \frac{1.0 \times 10^{-3}\ \text{kg}}{20 \times 10^{-2}\ \text m} = \frac{1}{200}\ \text{kg m}^{-1}$

The speed of the transverse wave in the string is

$\text v = \sqrt{\frac{\text T}{\text m}} = \sqrt{\frac{0.5}{1/200}} = \sqrt{100} = 10\ \text{m s}^{-1}$

The wavelength of the wave set up in the string is

$\lambda = \frac{\text v}{\text n} = \frac{10\ \text{m s}^{-1}}{100\ \text{s}^{-1}} = 0.1\ \text m = 10\ \text{cm}$

In a stationary wave the separation between two successive nodes is $\frac{\lambda}{2}$. Therefore,

$\text{Separation} = \frac{\lambda}{2} = \frac{10}{2}$

$\text{Separation} = 5\ \text{cm}$

Hence, the separation between the successive nodes on the string is 5 cm.

(i) a^-2

The particle is restricted between x = 0 and x = a, with nodes at both the ends. If n be the number of loops in the standing wave, then since the distance between two consecutive nodes is $\frac{\lambda}{2}$,

$\text n\frac{\lambda}{2} = \text a \qquad \Rightarrow \qquad \lambda = \frac{2\text a}{\text n}$

By the de Broglie relation, the linear momentum of the particle is

$\lambda = \frac{\text h}{\text p} \qquad \Rightarrow \qquad \text p = \frac{\text h}{\lambda}$

The energy of the particle is

$\text E = \frac{\text p^2}{2\text m} = \frac{\text h^2}{\lambda^2 \times 2\text m}$

Substituting the value of λ,

$\text E = \frac{\text h^2}{\left(\frac{2\text a}{\text n}\right)^2 \times 2\text m} = \frac{\text h^2 \text n^2}{4\text a^2 \times 2\text m} = \frac{\text h^2 \text n^2}{8\text a^2 \text m}$

For a particular value of n, all the other quantities being constant,

$\text E \propto \text a^{-2}$

Hence, the allowed energy for a particular value of n is proportional to a^-2.

(ii) 8 meV

Given, m = 1.0 × 10^-30 kg, a = 6.6 nm = 6.6 × 10^-9 m, h = 6.6 × 10^-34 J-s and e = 1.6 × 10^-19 C.

For the ground state n = 1, so

$\text E_1 = \frac{\text h^2}{8\text a^2 \times \text m}$

Substituting the values,

$$\begin{aligned}\text E_1 = \frac{(6.6 \times 10^{-34})^2}{8 \times (6.6 \times 10^{-9})^2 \times 1.0 \times 10^{-30}} \\[1em] = \frac{10^{-20}}{8}\ \text J\end{aligned}$$

Converting into electron-volt,

$$\begin{aligned}\text E_1 = \frac{10^{-20}}{8 \times 1.6 \times 10^{-19}}\ \text{eV} = \frac{1}{8 \times 16}\ \text{eV} \\[1em] = 7.8 \times 10^{-3}\ \text{eV} \approx 8 \times 10^{-3}\ \text{eV}\end{aligned}$$

$\text E_1 = 8\ \text{meV}$

Hence, the energy of the particle in its ground state is closest to 8 meV.

(iii) n

For the nth state,

$\text a = \text n\frac{\lambda}{2} \qquad \Rightarrow \qquad \lambda = \frac{2\text a}{\text n}$

The linear momentum of the particle is

$\text p = \frac{\text h}{\lambda} = \frac{\text{nh}}{2\text a}$

Since a and h are constants, p ∝ n. But p = mv, and the mass m is fixed, so

$\text v \propto \text n$

Hence, the speed of the particle is proportional to n.

6.25 N

Given,

• Linear mass density of the string, m = 0.04 kg m^-1
• $\text y = 0.02 \sin \left[2\pi\left(\frac{\text t}{0.04} - \frac{\text x}{0.50}\right)\right]$

Comparing the given equation with the standard equation of a progressive wave y = a sin (ωt − kx),

$\omega = \frac{2\pi}{0.04} \qquad \text{and} \qquad \text k = \frac{2\pi}{0.50}$

The speed of the wave in the string is

$\text v = \frac{\omega}{\text k} = \frac{2\pi/0.04}{2\pi/0.50} = \frac{0.50}{0.04} = 12.5\ \text{m s}^{-1}$

The speed of a transverse wave in a stretched string is $\text v = \sqrt{\frac{\text T}{\text m}}$, so that

$\text T = \text v^2 \text m$

Substituting the values,

$\text T = (12.5)^2 \times 0.04 = 156.25 \times 0.04$

$\text T = 6.25\ \text N$

Hence, the tension in the string is 6.25 N.

10 gram

Given,

• Length of the hollow pipe closed at one end, l₁ = 0.8 m
• Length of the uniform string, l₂ = 0.5 m
• Tension in the wire, T = 50 N
• Speed of sound in air, v = 320 m s^-1

Fundamental frequency of the pipe closed at one end :

$\text n_1 = \frac{\text v}{4\text l_1} = \frac{320}{4 \times 0.8} = 100\ \text{Hz}$

Frequency of the second harmonic of the string :

$\text n_2 = \frac{2}{2\text l_2}\sqrt{\frac{\text T}{\text m}}$

where m is the mass per unit length of the string.

For resonance, n₁ = n₂. Therefore,

$\frac{\text v}{4\text l_1} = \frac{2}{2\text l_2}\sqrt{\frac{\text T}{\text m}}$

Rearranging,

$$\begin{aligned}\sqrt{\text m} = \frac{4\text l_1}{\text l_2 \text v}\sqrt{\text T} = \frac{4 \times 0.8}{0.5 \times 320}\sqrt{50} = \frac{3.2}{160}\sqrt{50} \\[1em] = 0.02 \times 7.071 = 0.1414\end{aligned}$$

$\text m = (0.1414)^2 = \frac{1}{50}\ \text{kg m}^{-1}$

The mass of the string is

$\text M = \text m \times \text l_2 = \frac{1}{50} \times 0.5 = 0.01\ \text{kg}$

$\text M = 10\ \text{gram}$

Hence, the mass of the string is 10 gram.

178.2 Hz

Given,

• Length of the sonometer wire, l = 1.5 m
• Elastic strain produced in the wire = 1% = $\frac{1}{100}$
• Density of steel, ρ = 7.7 × 10³ kg/m³
• Young's modulus of steel, Y = 2.2 × 10¹¹ N/m²

Stress in the wire : By the definition of Young's modulus,

$\text{Stress} = \text Y \times \text{strain} = 2.2 \times 10^{11} \times \frac{1}{100}$

$\frac{\text T}{\text A} = 2.2 \times 10^9\ \text{N m}^{-2}$

Fundamental frequency : The fundamental frequency of the vibrating sonometer wire is

$\text n = \frac{1}{2\text l}\sqrt{\frac{\text T}{\text m}}$

The mass per unit length of the wire is

$\text m = \frac{\text M}{\text l} = \frac{\text V \times \rho}{\text l} = \frac{\text A \times \text l \times \rho}{\text l} = \text A \rho$

Substituting this value,

$\text n = \frac{1}{2\text l}\sqrt{\frac{\text T}{\text A \rho}} = \frac{1}{2\text l}\sqrt{\frac{(\text T/\text A)}{\rho}}$

Substituting the values,

$$\begin{aligned}\text n = \frac{1}{2 \times 1.5}\sqrt{\frac{2.2 \times 10^9}{7.7 \times 10^3}} = \frac{1}{3}\sqrt{2.857 \times 10^5} \\[1em] = \frac{534.5}{3}\end{aligned}$$

$\text n = 178.2\ \text{Hz}$

Hence, the fundamental frequency of the steel wire is 178.2 Hz.

The length of the string is 0.25 m, and the maximum displacement of the mid-point of the string, from its equilibrium position is 0.01 m

Given, the string is vibrating in its fifth harmonic according to

y(x, t) = 0.01 m sin [(62.8 m^-1)x] cos [(628 s^-1)t]

Comparing with the standard equation of a stationary wave,

$\text y = 2\text a \cos \frac{2\pi \text{vt}}{\lambda}\sin \frac{2\pi \text x}{\lambda}$

Number of nodes : In the fifth harmonic the string vibrates in five segments, so the number of nodes is 5 + 1 = 6, not 5. Hence the first statement is wrong.

Length of the string :

$\frac{2\pi}{\lambda} = 62.8 \qquad \Rightarrow \qquad \lambda = \frac{2\pi}{62.8} = \frac{2 \times 3.14}{62.8} = 0.1\ \text m$

For the fifth harmonic,

$\text L = \frac{5\lambda}{2} = \frac{5 \times 0.1}{2} = 0.25\ \text m$

Hence the second statement is correct.

Maximum displacement of the mid-point : The maximum displacement at an antinode is 2a, and from the given equation 2a = 0.01 m. In the fifth harmonic the mid-point of the string is an antinode, so its maximum displacement is 0.01 m. Hence the third statement is correct.

Fundamental frequency :

$\frac{2\pi \text v}{\lambda} = 628 \qquad \Rightarrow \qquad \text v = \frac{628\lambda}{2\pi} = \frac{628 \times 0.1}{2 \times 3.14} = 10\ \text{m s}^{-1}$

$\text n_1 = \frac{\text v}{2\text L} = \frac{10}{2 \times 0.25} = 20\ \text{Hz}$

Hence the fourth statement is wrong.

Hence, the second and the third statements are correct.

Given, the travelling harmonic wave is

y (x, t) = 7.5 sin (0.0050 x + 12 t + π/4) ...(i)

(i) Displacement and velocity at x = 1 cm and t = 1 s :

From equation (i), the displacement is

$$\begin{aligned}\text y(1, 1) = 7.5 \sin\left(0.0050 \times 1 + 12 \times 1 + \frac{\pi}{4}\right) \\[1em] = 7.5 \sin (12.79\ \text{rad})\end{aligned}$$

Since 1 rad = 57.3°,

$$\begin{aligned}\text y(1, 1) = 7.5 \sin (12.79 \times 57.3^\circ) = 7.5 \sin (733^\circ) \\[1em] = 7.5 \sin (720^\circ + 13^\circ) = 7.5 \sin 13^\circ \\[1em] = 7.5 \times 0.225\end{aligned}$$

$\text y(1, 1) = 1.687\ \text{cm}$

The velocity of oscillation of the particle is obtained by differentiating equation (i) with respect to t, keeping x constant,

$$\begin{aligned}\text u = \frac{\text{dy}}{\text{dt}} = \frac{\text d}{\text{dt}}\left[7.5 \sin\left(0.0050\text x + 12\text t + \frac{\pi}{4}\right)\right] \\[1em] = 7.5 \cos\left(0.0050\text x + 12\text t + \frac{\pi}{4}\right) \times 12 \\[1em] = 90 \cos\left(0.0050\text x + 12\text t + \frac{\pi}{4}\right) \qquad \dots(\text{ii})\end{aligned}$$

At x = 1 cm and t = 1 s,

$$\begin{aligned}\text u(1, 1) = 90 \cos (12.79\ \text{rad}) = 90 \cos 13^\circ \\[1em] = 90 \times 0.9744 = 87.7\ \text{cm s}^{-1}\end{aligned}$$

$\text u(1, 1) = 0.877\ \text{m s}^{-1}$

Speed of wave propagation : The wave travels along the negative direction of the X-axis, and is described by

$\text y (\text x, \text t) = \text a \sin \frac{2\pi}{\lambda}(\text{vt} + \text x) \qquad \dots(\text{iii})$

Comparing equation (i) with equation (iii),

$\frac{2\pi \text v}{\lambda} = 12 \qquad \text{and} \qquad \frac{2\pi}{\lambda} = 0.0050$

Dividing,

$\text v = \frac{12}{0.0050} = 2400\ \text{cm s}^{-1} = 24\ \text{m s}^{-1}$

Hence, the velocity of oscillation of the particle, 0.877 m s^-1, is not equal to the velocity of wave propagation, which is 24 m s^-1. The wave velocity is the same at all points of the wave, whereas the particle velocity changes from point to point.

(ii) Points having the same displacement and velocity :

Also from equation (iii),

$\lambda = \frac{2\pi}{0.0050} = \frac{2 \times 3.14}{0.0050} = 1256\ \text{cm} = 12.56\ \text m$

All the points that are separated from the point x = 1 cm by a whole number of wavelengths have the same transverse displacement and the same velocity, and these values do not change with the instant of time chosen. Hence the required points lie at distances

$\pm \lambda,\ \pm 2\lambda,\ \pm 3\lambda, \dots$

from the point x = 1 cm.

Hence, the required points are at distances of ± 12.56 m, ± 25.12 m, ± 37.68 m, ... from the point x = 1 cm, and these are the same for t = 2 s, 5 s and 11 s.

Given,

• Speed of the S wave, v_S = 4.0 km s^-1
• Speed of the P wave, v_P = 8.0 km s^-1
• The first P wave arrives 4 min = 240 s before the first S wave

Let d be the distance of the earthquake from the seismograph. Both the waves travel in a straight line and cover the same distance d.

The time taken by the S wave to reach the seismograph is

$\text t_\text S = \frac{\text d}{4.0\ \text{km s}^{-1}}$

and that taken by the P wave is

$\text t_\text P = \frac{\text d}{8.0\ \text{km s}^{-1}}$

It is given that

$\text t_\text S - \text t_\text P = 240\ \text s$

Substituting the values,

$\frac{\text d}{4.0} - \frac{\text d}{8.0} = 240$

$\frac{2\text d - \text d}{8.0} = 240 \qquad \Rightarrow \qquad \frac{\text d}{8.0} = 240$

$\text d = 8.0\ \text{km s}^{-1} \times 240\ \text s$

$\text d = 1920\ \text{km}$

Hence, the earthquake occurs at a distance of 1920 km from the seismograph.

Given,

• Linear mass density of the string, m = 8.0 × 10^-3 kg m^-1
• Frequency of the tuning fork, n = 256 Hz
• Mass in the pan, M = 90 kg
• Amplitude of the wave, a = 5.0 cm = 0.05 m
• At t = 0, the end x = 0 has y = 0 and is moving along the positive y-direction

Tension in the string : The string passes over a pulley and is tied to the pan, so

$\text T = \text{Mg} = 90\ \text{kg} \times 9.8\ \text{m s}^{-2}$

Speed of the wave :

$\text v = \sqrt{\frac{\text T}{\text m}} = \sqrt{\frac{90 \times 9.8}{8.0 \times 10^{-3}}} = \sqrt{110250}$

$\text v = 332\ \text{m s}^{-1}$

Angular frequency :

$\omega = 2\pi \text n = 2 \times 3.14 \times 256 = 1.6 \times 10^3\ \text{rad s}^{-1}$

Wavelength and propagation constant :

$\lambda = \frac{\text v}{\text n} = \frac{332\ \text{m s}^{-1}}{256\ \text{s}^{-1}} = 1.3\ \text m$

$\text k = \frac{2\pi}{\lambda} = \frac{2 \times 3.14}{1.3} = 4.83\ \text{m}^{-1}$

Equation of the wave : The pulley end absorbs all the incoming energy, so there is no reflected wave and only a progressive wave travels along the positive direction of the X-axis,

$\text y = \text a \sin (\omega \text t - \text k\text x + \phi)$

At t = 0 and x = 0 it is given that y = 0 and the end is moving along the positive y-direction. Hence sin φ = 0 with cos φ positive, which gives φ = 0. Therefore,

$\text y = \text a \sin (\omega \text t - \text k\text x)$

Substituting the values of a, ω and k,

$\text y = 0.05 \sin (1.6 \times 10^3\text t - 4.83\text x)$

Hence, the transverse displacement of the string is y = 0.05 sin (1.6 × 10³ t − 4.83 x), where y and x are in metre and t in second.

the temperature of the air

Reason — The speed of sound in a gas is given by $\text v = \sqrt{\frac{\gamma \text{RT}}{\text M}}$, so that $\text v \propto \sqrt{\text T}$. As the temperature of the air rises, the molecules move faster and the speed of sound increases. The amplitude, the frequency and the wavelength of the sound are not properties of the medium, so they do not decide the speed of sound in air.

the medium through which it travels

Reason — The speed of a mechanical wave depends only upon the properties of the medium through which it travels, that is, upon its elasticity and its inertia (density). For example, the speed of a transverse wave in a stretched string is $\text v = \sqrt{\frac{\text T}{\text m}}$ and that of sound in a gas is $\text v = \sqrt{\frac{\gamma \text P}{\text d}}$. Neither expression contains the frequency, the wavelength or the amplitude of the wave.

the difference between the two frequencies

Reason — When two waves of nearly equal frequencies n₁ and n₂ are sounded together, the intensity of the resultant sound rises and falls (n₁ − n₂) times in one second. Hence the number of beats per second, that is, the beat frequency, is equal to the difference in the frequencies of the two sound-sources.

antinodes

Reason — In a stationary wave, the points at which the displacement of the medium particles is always maximum are called antinodes, and the points at which the displacement remains permanently zero are called nodes. At an antinode constructive interference takes place, so the amplitude there is maximum.

angular frequency

Reason — In the equation of a progressive wave y = a sin (ωt − kx), the quantity ω is the angular frequency of the oscillating particles, and is related to the frequency n and the time-period T by

$\omega = 2\pi \text n = \frac{2\pi}{\text T}$

Here a is the amplitude and $\text k = \frac{2\pi}{\lambda}$ is the propagation constant, or wave number.

the length of the pipe

Reason — In a closed organ pipe a node is formed at the closed end and an antinode at the open end, so that $\text l = \frac{\lambda}{4}$. Hence the fundamental frequency is

$\text n = \frac{\text v}{4\text l}$

which is decided by the length l of the pipe. The frequency of the source only selects which mode is excited; it does not decide the natural frequency of the pipe.

2π rad

Reason — The phase difference between two points whose path difference is Δx is

$\Delta \phi = \frac{2\pi}{\lambda} \times \Delta \text x$

Putting Δx = λ,

$\Delta \phi = \frac{2\pi}{\lambda} \times \lambda = 2\pi\ \text{radian}$

Thus, two particles separated by one wavelength are in the same phase of oscillation.

mass per unit length of the string

Reason — In the formula $\text v = \sqrt{\frac{\text T}{\text m}}$, T is the tension in the string and m is the mass per unit length, that is, the linear mass density of the string. It is not the mass of the whole string. If r be the radius of the string and d the density of its material, then m = πr²d.

have slightly different frequencies

Reason — Beats are produced by the superposition of two waves of nearly equal frequencies travelling in the same direction. If the frequencies were exactly equal, the phase difference between the waves at any point would remain steady and the intensity of sound at that point would remain constant, so no beats would be heard.

half a wavelength

Reason — In a stationary wave the nodes are situated at x = 0, $\frac{\lambda}{2}$, $\frac{2\lambda}{2}$, $\frac{3\lambda}{2}$, ... Hence the distance between two consecutive nodes, and also between two consecutive antinodes, is $\frac{\lambda}{2}$, while the distance between a node and its neighbouring antinode is $\frac{\lambda}{4}$.

inversely proportional to the length of the pipe

Reason — In an open organ pipe an antinode is formed at each end, so that $\text l = \frac{\lambda}{2}$. Hence the fundamental frequency is

$\text n = \frac{\text v}{2\text l} \qquad \Rightarrow \qquad \text n \propto \frac{1}{\text l}$

that is, the fundamental frequency is inversely proportional to the length of the pipe.

nodes

Reason — In a stationary wave certain points of the bounded medium, situated at equal distances, always remain in the position of rest, that is, their displacement remains zero. These points are called nodes, and destructive interference takes place there.

the properties of the medium

Reason — The speed of propagation of a wave is decided by the elasticity and the density of the medium alone. For example, $\text v = \sqrt{\frac{\text E}{\text d}}$ for a longitudinal wave. Changing the frequency of the source changes the wavelength in such a way that the product nλ, that is, the speed, remains the same.

its frequency and the speed of sound in air

Reason — The wavelength of a sound wave in air is the ratio of the speed of sound in air to its frequency,

$\lambda = \frac{\text v}{\text n}$

so it is decided by the frequency of the wave together with the speed of sound in air.

Note: The book's answer key prints option 3, the temperature of the air, but its own explanation states that the wavelength is the ratio of the speed of sound to its frequency, which is option 2. The temperature affects the wavelength only indirectly, through its effect on the speed of sound.

the third harmonic

Reason — A closed organ pipe produces only odd harmonics, whose frequencies are in the ratio n₁ : n₂ : n₃ ... = 1 : 3 : 5 ... The tone next above the fundamental is therefore three times the fundamental frequency. This tone is called the first overtone or the third harmonic of the closed pipe.

0.9 m

Reason — Given, n = 300 Hz and v = 330 m/s. From v = nλ,

$\lambda = \frac{\text v}{\text n} = \frac{330\ \text{m/s}}{300\ \text{Hz}} = 1.1\ \text m$

Note: The book's answer key prints option (b), 0.9 m, but its own explanation carries out exactly this calculation and obtains 1.1 m. Hence the correct answer should be 1.1 m.

Three

Reason — In an open organ pipe an antinode is formed at each of the two ends. In the second harmonic the length of the pipe is l = λ, so there are antinodes at x = 0, $\frac{\text l}{2}$ and l, that is, three antinodes, and nodes at $\frac{\text l}{4}$ and $\frac{3\text l}{4}$, that is, two nodes.

Note: The book's answer key prints option (c), Three, and its explanation states that the second harmonic of an open pipe has two antinodes and three nodes. In fact the counts are the other way round — three antinodes and two nodes — so the number of nodes is two and the correct answer should be Two.

half the wavelength

Reason — In a stationary wave the antinodes are situated at x = $\frac{\lambda}{4}$, $\frac{3\lambda}{4}$, $\frac{5\lambda}{4}$, ... Hence the distance between two consecutive antinodes is $\frac{\lambda}{2}$, which is the same as the distance between two consecutive nodes.

beats

Reason — When two sound waves of nearly equal frequencies travelling in the same direction superpose, the intensity of the resultant sound rises and falls alternately with time. This periodic rise and fall in the intensity of sound is called the phenomenon of beats.

5 Hz

Reason — The beat frequency is the difference between the frequencies of the two sound waves,

$\text n_1 - \text n_2 = 1005\ \text{Hz} - 1000\ \text{Hz} = 5\ \text{Hz}$

That is, five beats are heard in one second.

diffraction

Reason — The bending of waves around obstacles, or their spreading out on passing through narrow openings, is called diffraction. It is the confirmative evidence of the wave nature of a disturbance.

twice the length of the pipe

Reason — In an organ pipe open at both ends there is an antinode at each end and a node in the middle. Hence for the fundamental mode,

$\text l = \frac{\lambda}{2} \qquad \Rightarrow \qquad \lambda = 2\text l$

that is, the wavelength of the fundamental mode is twice the length of the pipe.

the temperature of the air increases

Reason — The speed of sound in air is directly proportional to the square-root of its absolute temperature, $\text v \propto \sqrt{\text T}$. Numerically,

$\text v_\text t = (332 + 0.61\text t)\ \text{m s}^{-1}$

so the speed increases by about 0.61 m s^-1 for each °C rise in temperature. A change in pressure alone has no effect, since $\frac{\text P}{\text d}$ then remains constant.

their amplitudes add up

Reason — According to the principle of superposition, when two or more waves overlap at a point, the resultant displacement at that point is the algebraic sum of the displacements produced by the individual waves,

$\text y = \text y_1 + \text y_2$

Each wave, however, retains its own frequency, wavelength and velocity even after the superposition.

one antinode and two nodes

Reason — A string fixed at both ends must have a node at each of the two fixed ends. In the fundamental mode the string vibrates as a whole in one segment, with $\text l = \frac{\lambda}{2}$, so a single antinode is formed at the middle. Hence there are two nodes and one antinode.

the square root of the tension

Reason — The fundamental frequency of a vibrating sonometer wire is

$\text n = \frac{1}{2\text l}\sqrt{\frac{\text T}{\text m}}$

For a given length and a given mass per unit length, $\text n \propto \sqrt{\text T}$. This is the law of tension of the transverse vibrations of a stretched string. The frequency is inversely proportional to the square-root of m, not to m itself.

the tension increases

Reason — The speed of a transverse wave in a stretched string is

$\text v = \sqrt{\frac{\text T}{\text m}}$

so the speed increases with increasing tension and decreases with increasing mass per unit length. The amplitude of the wave has no effect on its speed.

v = nλ

Reason — In one time-period T the wave travels a distance equal to one wavelength λ. Hence the distance travelled in one second is

$\text v = \frac{\lambda}{\text T} = \text n \lambda$

that is, speed = frequency × wavelength.

the tension is increased

Reason — The frequency of a vibrating sonometer wire is

$\text n = \frac{1}{2\text l}\sqrt{\frac{\text T}{\text m}}$

Hence $\text n \propto \sqrt{\text T}$ and $\text n \propto \frac{1}{\text l}$ and $\text n \propto \frac{1}{\sqrt{\text m}}$. The frequency therefore increases only when the tension is increased; increasing the length or the mass per unit length lowers the frequency.

the nature of the boundary conditions (whether the ends are fixed or free)

Reason — The boundary conditions decide which stationary waves can be set up in the medium. A string fixed at both ends must have a node at each end and so gives both even and odd harmonics in the ratio 1 : 2 : 3 : 4 ..., whereas a string fixed at one end and free at the other gives only the odd harmonics. The length and the tension decide the values of the frequencies, not which harmonics are permitted.

two antinodes

Reason — In an open organ pipe the air particles at both the ends have the greatest freedom to vibrate, so an antinode is formed at each end. In the fundamental mode there is one node in the middle of the pipe. Hence the fundamental mode is characterised by two antinodes, one at each open end.

increases

Reason — In constructive interference the two waves meet in the same phase, that is, φ = 0. The resultant amplitude is then

$\text a_{max} = \text a_1 + \text a_2$

which is larger than the amplitude of either wave. Hence the amplitude of the resulting wave increases.

is a stationary wave

Reason — The two waves have the same amplitude and the same frequency and travel in opposite directions. By the principle of superposition,

$$\begin{aligned}\text y = \text y_1 + \text y_2 = \text a \sin (\text k\text x - \omega \text t) + \text a \sin (\text k\text x + \omega \text t) \\[1em] = 2\text a \sin \text k\text x \cos \omega \text t\end{aligned}$$

Here x and t occur in separate harmonic functions, so the resultant is a stationary wave. Its amplitude, 2a sin kx, varies from point to point and is not double the original amplitude everywhere.

decrease by a factor of 3

Reason — The fundamental frequency of a closed organ pipe is

$\text n = \frac{\text v}{4\text l} \qquad \Rightarrow \qquad \text n \propto \frac{1}{\text l}$

If the length of the air column is made 3 l, the fundamental frequency becomes $\frac{\text v}{4(3\text l)} = \frac{\text n}{3}$, that is, it decreases by a factor of 3.

maximum

Reason — A phase difference of 2π corresponds to a path difference of one full wavelength, so the two waves arrive at the point in the same phase. Constructive interference takes place, the resultant amplitude is a₁ + a₂, and since I ∝ a² the resultant intensity is maximum.

Oscillate about their mean positions

Reason — In a progressive wave the energy is transferred from one point to another, but the medium itself does not travel along with the wave. Each particle of the medium merely oscillates about its own mean position, and only the phase of oscillation changes from particle to particle.

four nodes and three antinodes

Reason — In a string of length l fixed at both ends, the pth mode of vibration has (p + 1) nodes and p antinodes. For the third harmonic p = 3, so the string vibrates in three segments with

$\text{nodes at}\ \text x = 0,\ \frac{\text l}{3},\ \frac{2\text l}{3},\ \text l$

that is, four nodes, and three antinodes between them.

$\frac{\pi}{2}$

Reason — The phase difference corresponding to a path difference Δx is

$\Delta \phi = \frac{2\pi}{\lambda} \times \Delta \text x$

Putting $\Delta \text x = \frac{\lambda}{4}$,

$\Delta \phi = \frac{2\pi}{\lambda} \times \frac{\lambda}{4} = \frac{\pi}{2}\ \text{radian}$

1 m

Reason — The distance between a node and the adjacent antinode in a stationary wave is $\frac{\lambda}{4}$. Given that this distance is 0.25 m,

$\frac{\lambda}{4} = 0.25\ \text m \qquad \Rightarrow \qquad \lambda = 4 \times 0.25$

$\lambda = 1\ \text m$

the wave is undergoing destructive interference

Reason — A progressive wave advances in the medium with a definite velocity and all its particles vibrate with the same amplitude, so it has no permanent points of zero amplitude. If the amplitude at a certain point becomes zero, it means that another wave has arrived there in the opposite phase and the two have cancelled each other. Hence the wave is undergoing destructive interference.

tension and mass per unit length

Reason — The speed of a transverse wave in a vibrating string is

$\text v = \sqrt{\frac{\text T}{\text m}}$

so it depends only on the tension T in the string and the mass per unit length m of the string. It does not depend upon the frequency, the amplitude, the length of the string or the number of nodes formed.

varies periodically

Reason — In the phenomenon of beats the resultant displacement is

$\text y = \left[2\text a \cos \pi (\text n_1 - \text n_2)\text t\right]\sin \pi (\text n_1 + \text n_2)\text t$

The quantity within the brackets is the amplitude, and it depends upon t. Hence the amplitude of the resultant wave varies periodically with time between 2a and zero, giving alternate constructive and destructive interference.

$\frac{\text v}{4\text l}$

Reason — A string fixed at one end and free at the other has a node at the fixed end and an antinode at the free end. Hence in the fundamental mode

$\text l = \frac{\lambda}{4} \qquad \Rightarrow \qquad \lambda = 4\text l$

so the fundamental frequency is

$\text n = \frac{\text v}{\lambda} = \frac{\text v}{4\text l}$

Double

Reason — The frequency of a vibrating sonometer wire is

$\text n = \frac{1}{2\text l}\sqrt{\frac{\text T}{\text m}} \qquad \Rightarrow \qquad \text n \propto \sqrt{\text T}$

If the tension is made 4 T, the new frequency is

$\text n' = \frac{1}{2\text l}\sqrt{\frac{4\text T}{\text m}} = 2\left(\frac{1}{2\text l}\sqrt{\frac{\text T}{\text m}}\right) = 2\text n$

that is, the frequency is doubled.

4 Hz

Reason — The number of beats heard per second is equal to the difference in the frequencies of the two tuning forks,

$\text n_1 - \text n_2 = 260\ \text{Hz} - 256\ \text{Hz} = 4\ \text{Hz}$

0.5 m

Reason — Comparing y = 0.05 sin (4πx − 200 πt) with the standard equation y = a sin (kx − ωt),

$\text k = \frac{2\pi}{\lambda} = 4\pi$

$\lambda = \frac{2\pi}{4\pi} = 0.5\ \text m$

double

Reason — The fundamental frequency of a sonometer wire is

$\text n = \frac{1}{2\text l}\sqrt{\frac{\text T}{\text m}} \qquad \Rightarrow \qquad \text n \propto \frac{1}{\text l}$

This is the law of length. If the vibrating length is halved, the new frequency is

$\text n' = \frac{1}{2\left(\frac{\text l}{2}\right)}\sqrt{\frac{\text T}{\text m}} = 2\text n$

that is, the fundamental frequency is doubled.

quadrupled

Reason — The energy carried by a wave is directly proportional to the square of its amplitude,

$\text E \propto \text a^2$

If the amplitude is doubled, the energy becomes (2)² = 4 times its original value, that is, it is quadrupled.

$\text a\sqrt{2}$

Reason — The second wave may be written as

$\text y_2 = \text a \cos (\text k\text x - \omega \text t) = \text a \sin \left(\text k\text x - \omega \text t + \frac{\pi}{2}\right)$

so the phase difference between the two waves is $\phi = \frac{\pi}{2}$, that is, 90°. The resultant amplitude is

$\text A = \sqrt{\text a^2 + \text a^2 + 2\text a \times \text a \cos 90^\circ} = \sqrt{2\text a^2}$

$\text A = \text a\sqrt{2}$

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: The speed of a longitudinal wave in a medium is $\text v = \sqrt{\frac{\text E}{\text d}}$, where E is the modulus of elasticity. Solids possess a very large modulus of elasticity, so the speed of sound is minimum in gases, more in liquids and maximum in solids.

Reason (R) is also correct: In a solid the particles are closely packed and are held in place by strong elastic forces, so a disturbance is passed on from one particle to the next very rapidly.

Because the particles are closely spaced and strongly bound, the energy is transferred more quickly between them, which is exactly why the speed of sound is greater in solids than in gases. The Reason therefore explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: On increasing the tension in a stretched string, the transverse wave travels along it with a greater speed.

Reason (R) is also correct: The speed of a transverse wave in a stretched string is

$\text v = \sqrt{\frac{\text T}{\text m}} \qquad \Rightarrow \qquad \text v \propto \sqrt{\text T}$

for a string of given mass per unit length. Since the speed varies directly as the square-root of the tension, an increase in T necessarily increases v. The Reason therefore explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: From $\text v = \sqrt{\frac{\gamma \text{RT}}{\text M}}$ it follows that $\text v \propto \sqrt{\text T}$, so the speed of sound in air increases with a rise in temperature. Numerically, v_t = (332 + 0.61 t) m s^-1.

Reason (R) is also correct: At higher temperatures the molecules of air move faster, so a disturbance is handed on from one molecule to the next more rapidly and the rate of energy transfer increases.

The faster molecular motion at a higher temperature is the physical cause of the greater speed of sound, so the Reason explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

If both assertion and reason are true but reason is not the correct explanation of assertion.

Explanation

Assertion (A) is correct: A progressive wave advances in the medium with a definite velocity and transmits energy from one point to another, although the particles of the medium merely oscillate about their mean positions.

Reason (R) is also correct, but only for one kind of progressive wave: In a transverse progressive wave the displacement of the particles is perpendicular to the direction of wave propagation.

A progressive wave may be transverse or longitudinal, and a longitudinal progressive wave, in which the particles oscillate along the direction of propagation, also transfers energy. Hence the perpendicular displacement is not the reason for the transfer of energy. The Reason therefore does not explain the Assertion.

Therefore, both assertion and reason are true but reason is not the correct explanation of assertion.

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: Stationary waves do not advance in the medium, and so they do not transmit energy across it. The energy remains confined within the segments between the nodes.

Reason (R) is also correct: Stationary waves are formed by the superposition of two identical waves travelling in opposite directions with the same speed, and this superposition produces permanently fixed nodes and antinodes.

Because the two component waves carry equal energy in opposite directions, and because the nodes remain permanently at rest, there is no net transfer of energy across the medium. The Reason therefore explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

If assertion is true but reason is false.

Explanation

Assertion (A) is correct: From v = nλ, at a constant speed $\lambda \propto \frac{1}{\text n}$. Hence a wave of higher frequency has a shorter wavelength.

Reason (R) is incorrect: The speed of a wave is the product of its frequency and its wavelength, v = nλ, and it is decided by the properties of the medium alone. It is not directly proportional to the frequency; if the frequency of the source is changed, the wavelength changes in such a way that the speed remains unaltered.

Therefore, assertion is true but reason is false.

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: For a given tension, a string of larger linear mass density carries the transverse wave with a smaller speed.

Reason (R) is also correct: The speed of a transverse wave in a stretched string is

$\text v = \sqrt{\frac{\text T}{\text m}} \qquad \Rightarrow \qquad \text v \propto \frac{1}{\sqrt{\text m}}$

Since the speed varies inversely as the square-root of the linear mass density, a larger value of m gives a smaller value of v. The Reason therefore explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: When two waves superpose in the same phase the resultant amplitude is a₁ + a₂, which is constructive interference, and when they superpose in opposite phases the resultant amplitude is a₁ ~ a₂, which is destructive interference.

Reason (R) is also correct: By the principle of superposition, the resultant displacement at any point is the algebraic sum of the displacements produced by the individual waves,

$\text y = \text y_1 + \text y_2$

Whether this algebraic sum is large or small depends upon the phase difference between the waves, which is precisely why both constructive and destructive interference are possible. The Reason therefore explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: Sound waves in air propagate in the form of compressions and rarefactions, and are therefore longitudinal waves.

Reason (R) is also correct: In a longitudinal wave the particles of the medium oscillate parallel to the direction of wave propagation.

In a sound wave the air molecules move back and forth along the direction in which the sound advances, producing the compressions and rarefactions. This is exactly the defining property of a longitudinal wave, so the Reason explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: When two sound waves of nearly equal frequencies are sounded together, the intensity of the resultant sound rises and falls alternately, producing beats.

Reason (R) is also correct: The number of beats heard per second is equal to the difference in the frequencies of the two interfering waves, that is, beat frequency = n₁ ~ n₂.

If the frequencies were exactly equal, the difference would be zero and the phase difference at any point would remain steady, so no beats would be heard. The Reason therefore explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: For an organ pipe the fundamental frequency is $\text n = \frac{\text v}{2\text l}$ for an open pipe and $\text n = \frac{\text v}{4\text l}$ for a closed pipe, so in either case $\text n \propto \frac{1}{\text l}$. Hence a shorter pipe gives a higher fundamental frequency.

Reason (R) is also correct: For a given type of pipe the wavelength of the fundamental mode is a fixed multiple of the length of the pipe, so a shorter pipe supports a shorter wavelength and hence a higher frequency.

The Reason therefore explains the relationship between the length of the pipe and its fundamental frequency.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: For a closed organ pipe, in the fundamental mode

$\text l = \frac{\lambda}{4}$

that is, the length of the pipe is one-fourth of the wavelength of the fundamental note.

Reason (R) is also correct: The closed end of the pipe acts as a node, because the air particles there have no freedom to vibrate, and the open end acts as an antinode, because the air particles there have the greatest freedom to vibrate.

These very boundary conditions fix the relation between the length of the pipe and the wavelength of the fundamental tone, so the Reason explains the Assertion.

Note: The Assertion as printed reads "a fundamental frequency that is half the wavelength of the pipe's length", which compares a frequency with a length and is therefore not dimensionally meaningful. It has been read here in the sense intended by the book's own explanation, namely the relation between the length of the pipe and the wavelength of the fundamental tone.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: The frequency of a vibrating string is

$\text n = \frac{1}{2\text l}\sqrt{\frac{\text T}{\text m}}$

so the frequency increases when the tension is increased.

Reason (R) is also correct: The speed of a wave on a string is $\text v = \sqrt{\frac{\text T}{\text m}}$, that is, it is directly proportional to the square-root of the tension.

Since $\text n = \frac{\text v}{2\text l}$, and the length of the string is unchanged, an increase in the wave speed produced by the greater tension directly raises the frequency. The Reason therefore explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

If assertion is false but reason is true.

Explanation

Assertion (A) is incorrect: The nodes are the points of a stationary wave at which the displacement of the particles remains permanently zero. The amplitude is maximum at the antinodes, not at the nodes.

Reason (R) is correct: At a node the two component waves always arrive in opposite phases, so destructive interference takes place there and the resultant displacement is always zero.

Therefore, assertion is false but reason is true.

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: A progressive wave carries energy from one point of the medium to another, but the medium itself is not carried forward along with the wave.

Reason (R) is also correct: In a progressive wave each particle of the medium merely oscillates about its own equilibrium position, only the phase of the oscillation changing from particle to particle.

Since every particle returns to its equilibrium position after each oscillation, there is no permanent displacement of matter, and yet the disturbance, and with it the energy, is handed on from particle to particle. The Reason therefore explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: In a closed organ pipe the frequencies of the fundamental tone and the overtones are in the ratio n₁ : n₂ : n₃ ... = 1 : 3 : 5 ..., that is, only the odd harmonics are produced.

Reason (R) is also correct: The closed end of the pipe acts as a node and the open end as an antinode, so that

$\lambda = \frac{4\text l}{2\text m - 1}, \qquad \text m = 1, 2, 3, \dots$

These boundary conditions permit only those wavelengths for which the length of the pipe is an odd multiple of $\frac{\lambda}{4}$, and therefore permit only the odd harmonics. The Reason therefore explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

If assertion is false but reason is true.

Explanation

Assertion (A) is incorrect: The principle of superposition applies to all types of waves, including light waves, water waves and sound waves, provided the medium behaves linearly. It is not restricted to mechanical waves.

Reason (R) is correct: Superposition occurs when two or more waves overlap at a point, and the resultant displacement there is the algebraic sum of the displacements produced by the individual waves.

Therefore, assertion is false but reason is true.

If both assertion and reason are true but reason is not the correct explanation of assertion.

Explanation

Assertion (A) as printed is not correct: In a longitudinal sound wave a compression is a region of maximum pressure and maximum density, and it is a displacement node — the particles there have zero displacement. The maximum displacement of the air molecules occurs midway between a compression and a rarefaction.

Reason (R) is correct: In a longitudinal wave the compressions are indeed the regions of high pressure and high density.

Note: The book's answer key marks option (b), and its own printed explanation states that maximum displacement occurs between compressions and rarefactions — which shows the Assertion to be false. On that reading the correct choice would be If assertion is false but reason is true.

Therefore, both assertion and reason are true but reason is not the correct explanation of assertion.

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: In S.H.M. the velocity of the particle is $\text v = \omega\sqrt{\text a^2 - \text y^2}$, which is maximum, equal to ωa, at the equilibrium position where y = 0.

Reason (R) is also correct: The acceleration in S.H.M. is a_acc. = − ω²y, which is zero at the equilibrium position where y = 0.

Since the restoring force, and hence the retardation, vanishes at the equilibrium position, the particle has been accelerated over the whole of its journey up to that point and there attains its greatest speed. The Reason therefore explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

If assertion is false but reason is true.

Explanation

Assertion (A) is incorrect: Beats can be produced with any type of wave whose frequencies differ slightly and which can superpose. They are not restricted to sound waves alone.

Reason (R) is correct: Beats are indeed the result of the interference of two waves of slightly different frequencies, the intensity of the resultant rising and falling alternately with time.

Therefore, assertion is false but reason is true.

If both assertion and reason are true but reason is not the correct explanation of assertion.

Explanation

Assertion (A) is correct: The frequency of a wave is decided by the source producing it and remains unchanged when the wave passes from one medium into another. Only the speed and the wavelength change.

Reason (R) is also correct: The speed of a wave is determined by the elasticity and the density of the medium through which it travels.

The two statements are about different quantities. The fact that the speed depends upon the medium does not by itself explain why the frequency is fixed by the source; frequency is a characteristic of the wave-emitting source. Hence the Reason does not explain the Assertion.

Therefore, both assertion and reason are true but reason is not the correct explanation of assertion.

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: The intensity of a sound wave goes on decreasing as the wave travels away from the source.

Reason (R) is also correct: The energy sent out by the source spreads over a spherical surface whose area increases as the wave advances, so

$\text I = \frac{\text P}{4\pi \text r^2}$

where P is the power of the source and r the distance from it. Since the same energy is distributed over a larger and larger area, the intensity falls off as $\frac{1}{\text r^2}$. The Reason therefore explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

If assertion is false but reason is true.

Explanation

Assertion (A) is incorrect: A standing wave is formed when two identical waves travel in a bounded medium in opposite directions and superpose. Two waves travelling in the same direction can only produce interference or beats, never a standing wave.

Reason (R) is correct: Standing waves are indeed the result of constructive interference at the antinodes and destructive interference at the nodes.

Therefore, assertion is false but reason is true.

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: The lowest frequency of vibration of a stretched string is called its fundamental frequency, and the corresponding tone is the first harmonic.

Reason (R) is also correct: In the first mode the string vibrates as a whole in a single segment, so that

$\text l = \frac{\lambda_1}{2} \qquad \Rightarrow \qquad \lambda_1 = 2\text l$

which is the longest wavelength that the string can support.

Since $\text n = \frac{\text v}{\lambda}$, the longest wavelength corresponds to the lowest frequency. The Reason therefore explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: At the same temperature the speed of sound in helium is greater than that in air.

Reason (R) is also correct: The speed of sound in a gas is

$\text v = \sqrt{\frac{\gamma \text{RT}}{\text M}} \qquad \Rightarrow \qquad \text v \propto \frac{1}{\sqrt{\text M}}$

that is, it is inversely proportional to the square-root of the molecular mass of the gas.

Helium has a molecular mass of 4, which is much less than the average molecular mass of air, so the speed of sound in helium comes out to be greater. The Reason therefore explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

If assertion is false but reason is true.

Explanation

Assertion (A) is incorrect: For pipes of the same length, the fundamental frequency of an open pipe is $\frac{\text v}{2\text l}$ and that of a closed pipe is $\frac{\text v}{4\text l}$. Hence the fundamental frequency of an open pipe is twice, that is, higher than, that of a closed pipe of the same length, not lower.

Reason (R) is correct: The fundamental frequency of a closed pipe is indeed half that of an open pipe of the same length,

$\frac{\text n_{closed}}{\text n_{open}} = \frac{\text v/4\text l}{\text v/2\text l} = \frac{1}{2}$

Therefore, assertion is false but reason is true.

If assertion is false but reason is true.

Explanation

Assertion (A) is incorrect: The speed of a sound wave is decided by the elasticity and the density of the medium alone, and is completely independent of the amplitude of the wave.

Reason (R) is correct: The amplitude of a wave determines the energy it carries, since E ∝ a², but it has no effect on the speed with which the wave travels.

Therefore, assertion is false but reason is true.

If assertion is false but reason is true.

Explanation

Assertion (A) is incorrect: The beat frequency is n₁ ~ n₂. Hence the frequency of the beats decreases, and not increases, as the frequency difference between the two interfering waves decreases.

Reason (R) is correct: The beat frequency is indeed the absolute difference between the frequencies of the two interfering waves.

Therefore, assertion is false but reason is true.

If assertion is false but reason is true.

Explanation

Assertion (A) is incorrect: In a stationary wave the points of zero displacement are called nodes. The antinodes are the points of maximum displacement.

Reason (R) is correct as regards the confinement of energy: In a stationary wave the energy remains confined within the segments and merely changes back and forth between the kinetic and the potential forms; there is no transfer of energy across the medium from one segment to another.

Therefore, assertion is false but reason is true.

If both assertion and reason are true but reason is not the correct explanation of assertion.

Explanation

Assertion (A) is incorrect as printed: The frequency of a sound wave is fixed by the source and does not change on passing from air into water. Since v = nλ and the speed of sound in water is greater than that in air, the wavelength increases on going from air to water, and does not decrease.

Reason (R) is correct: The speed of sound in water, about 1450 m s^-1, is indeed much greater than that in air, about 332 m s^-1, because water has a far greater bulk modulus.

Therefore, both assertion and reason are true but reason is not the correct explanation of assertion.

Note: The book's answer key marks option (b), and its printed explanation states that the wavelength increases rather than decreases, which makes the Assertion false. On that reading the correct choice would be If assertion is false but reason is true.

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: The speed of a wave on a string does not change when the frequency of the source vibrating it is changed; only the wavelength changes so that the product nλ stays the same.

Reason (R) is also correct: The speed of a wave on a string is

$\text v = \sqrt{\frac{\text T}{\text m}}$

which contains only the tension in the string and its linear mass density.

Since the expression for the speed involves no term of frequency, the speed is necessarily independent of the frequency. The Reason therefore explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

If assertion is false but reason is true.

Explanation

Assertion (A) is incorrect: In S.H.M. the potential energy is $\text U = \frac{1}{2}\text m\omega^2 \text y^2$, which is zero at the equilibrium position where y = 0, and maximum at the extreme positions where y = ± a.

Reason (R) is correct: The potential energy in S.H.M. is indeed proportional to the square of the displacement from the mean position.

Therefore, assertion is false but reason is true.

If assertion is false but reason is true.

Explanation

Assertion (A) is incorrect: The superposition of two waves may result in either constructive or destructive interference, depending upon the phase difference between them. The resultant amplitude lies anywhere between a₁ + a₂ and a₁ ~ a₂.

Reason (R) is correct: Constructive interference does occur when the two waves meet in the same phase, that is, when φ = 0, and then a_max = a₁ + a₂.

Therefore, assertion is false but reason is true.

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: In a closed organ pipe the frequencies present are n₁, 3n₁, 5n₁, ..., so the tone next above the fundamental is the third harmonic. Since the second harmonic 2n₁ is missing altogether, the fundamental is the lowest frequency the pipe can produce and no note of frequency 2n₁ exists below or above it.

Reason (R) is also correct: The second harmonic does not exist in a closed pipe, because the boundary conditions — a node at the closed end and an antinode at the open end — permit only those modes for which the length is an odd multiple of $\frac{\lambda}{4}$.

The absence of the second harmonic is thus the very reason for the statement made in the Assertion, so the Reason explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

If assertion is false but reason is true.

Explanation

Assertion (A) is incorrect: The speed of a mechanical wave in a medium does not depend upon the frequency of the wave. Sound waves of different frequencies travel through air with the same speed, although their wavelengths in air are different.

Reason (R) is correct: The wave speed in a medium is determined by the properties of the medium, that is, by its elasticity and its density, and not by the frequency.

Therefore, assertion is false but reason is true.

If assertion is false but reason is true.

Explanation

Assertion (A) is incorrect: In S.H.M. the velocity of the particle is maximum, equal to ωa, at the equilibrium position, and it is zero at the extreme positions.

Reason (R) is correct in the sense that it identifies where the velocity is not maximum, but as printed it too is wrong: the velocity in S.H.M. is zero, and not maximum, at the extreme positions.

Therefore, assertion is false but reason is true.

Note: The book's answer key marks option (d). Strictly, both the Assertion and the Reason as printed are incorrect statements about S.H.M., since the velocity is maximum at the equilibrium position and zero at the extreme positions.

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: In an open organ pipe the frequencies of the fundamental tone and the overtones are in the ratio n₁ : n₂ : n₃ ... = 1 : 2 : 3 ..., that is, both the even and the odd harmonics are produced.

Reason (R) is also correct: Both the ends of an open organ pipe are open to the atmosphere, so the air particles there have the greatest freedom to vibrate and an antinode is formed at each end.

These boundary conditions give $\lambda = \frac{2\text l}{\text m}$ with m = 1, 2, 3, ..., which permits every integral harmonic. The Reason therefore explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

If assertion is false but reason is true.

Explanation

Assertion (A) is incorrect: A node is a point of a stationary wave at which the displacement is always zero, not maximum. The points of maximum displacement are the antinodes.

Reason (R) is correct: At a node the two component waves always arrive in opposite phases, so destructive interference takes place and the resultant displacement there is zero.

Therefore, assertion is false but reason is true.

If assertion is false but reason is true.

Explanation

Assertion (A) is incorrect: The first harmonic of a string is the fundamental tone itself. The second overtone is the third harmonic, whose frequency is three times the fundamental frequency.

Reason (R) is correct: The overtones are numbered from the second harmonic onwards — the second harmonic is the first overtone, the third harmonic is the second overtone, and so on.

Therefore, assertion is false but reason is true.

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: As the vibrating length of a string is increased, the fundamental frequency of the note emitted by it falls.

Reason (R) is also correct: The fundamental frequency of a stretched string is

$\text n = \frac{1}{2\text l}\sqrt{\frac{\text T}{\text m}} \qquad \Rightarrow \qquad \text n \propto \frac{1}{\text l}$

for a given tension and a given mass per unit length. This is the law of length, and it shows directly that lengthening the string lowers the frequency. The Reason therefore explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

No, the sound of an extremely violent explosion on the moon cannot be heard on the earth.

Sound is a mechanical wave and it requires a material medium for its propagation. There is no atmosphere on the moon, and the space between the moon and the earth is a vacuum. Since a vacuum has no particles to carry the wave's energy, sound cannot travel through it.

Transverse waves cannot be produced in gases because gases do not have rigidity.

The propagation of a transverse wave requires the medium to sustain shearing stress, and only a medium possessing a modulus of rigidity can do so. A gas has no definite shape and yields at once to shearing stress, so a transverse wave cannot be set up in it. Only longitudinal waves, for which the medium is alternately compressed and rarefied, can travel in a gas.

The sound will reach the other end earlier in the steel string.

The speed of a longitudinal wave in a solid rod is

$\text v = \sqrt{\frac{\text Y}{\text d}}$

where Y is the Young's modulus of the material and d its density. The value of $\frac{\text Y}{\text d}$ is much larger for steel than for rubber, so the speed of sound in steel is much greater. Since both the strings are of the same length, the sound takes less time in the steel string.

The speed of sound in a perfectly rigid rod would be infinite.

The speed of a longitudinal wave in a rod is $\text v = \sqrt{\frac{\text Y}{\text d}}$. For a perfectly rigid rod the Young's modulus Y is infinite, since such a rod suffers no strain whatever the stress applied. Hence the speed of sound in it comes out to be infinite.

The velocity of sound is maximum in solids and minimum in gases.

The speed of a longitudinal wave in a medium is $\text v = \sqrt{\frac{\text E}{\text d}}$, where E is the modulus of elasticity of the medium. Solids possess a very large modulus of elasticity, liquids a smaller one and gases the smallest. Hence the speed of sound is minimum in gases, more in liquids and maximum in solids.

The sound travelling through the railway track is heard first.

The sound produced by striking the track travels to the second man by two separate paths — through the steel rail and through the air. The speed of sound in steel is far greater than that in air, since the ratio $\frac{\text Y}{\text d}$ is much larger for steel. Hence the sound reaching through the rail arrives earlier, and the sound reaching through the air is heard a little later. This is why two sounds are heard at a small interval of time.

There will be no change in the velocity of sound; it will remain 330 m/s.

The speed of sound in a gas is $\text v = \sqrt{\frac{\gamma \text P}{\text d}}$. From the gas equation, at a constant temperature

$\frac{\text P}{\text d} = \frac{\text{RT}}{\text M} = \text{constant}$

Hence when the pressure is halved, the density is also halved in the same proportion, so the ratio $\frac{\text P}{\text d}$ does not change. Therefore there is no effect of the pressure-change on the speed of sound.

The speed of sound will remain 332 m s^-1.

On doubling the pressure at a constant temperature, the density of the air is also doubled, so that the ratio $\frac{\text P}{\text d}$ remains constant. Since $\text v = \sqrt{\frac{\gamma \text P}{\text d}}$, the speed of sound is unaltered.

As the temperature of the air increases, the speed of sound in air increases.

The speed of sound in a gas is

$\text v = \sqrt{\frac{\gamma \text{RT}}{\text M}} \qquad \Rightarrow \qquad \text v \propto \sqrt{\text T}$

that is, the speed of sound in a gas is directly proportional to the square-root of its absolute temperature.

The velocity of sound in air increases by 0.61 m s^-1 for each 1°C rise in temperature.

Taking the speed of sound in air at 0°C to be 332 m s^-1, the speed at t°C is

$\text v_\text t = (332 + 0.61\text t)\ \text{m s}^{-1}$

so that a rise of 1°C raises the speed by 0.61 m s^-1.

The speed of sound in air increases with increase in humidity.

The density of water-vapour is less than that of air, so the density of moist air, that is, air mixed with water-vapour, is less than that of dry air. Assuming the value of γ for moist air to be the same as that for dry air, it follows from

$\text v = \sqrt{\frac{\gamma \text P}{\text d}} \qquad \Rightarrow \qquad \text v \propto \frac{1}{\sqrt{\text d}}$

that the speed of sound in moist air is slightly greater than in dry air. This is why the sirens of mills and the whistles of trains are heard up to longer distances in the rainy season.

The speed of sound in a gas and the root-mean-square speed of its molecules are related by

$\frac{\text v}{\text v_{rms}} = \sqrt{\frac{\gamma}{3}} \qquad \text{or} \qquad \text v_{rms} = \sqrt{\frac{3}{\gamma}}\ \text v$

where γ (> 1) is the ratio of the two molar specific heats of the gas. Since the value of $\sqrt{\frac{\gamma}{3}}$ is less than 1, the speed of sound in a gas is smaller than the root-mean-square speed of the molecules of that gas.

The equation of a plane progressive wave travelling along the positive direction of the X-axis is

$\text y = \text a \sin 2\pi\left(\frac{\text t}{\text T} - \frac{\text x}{\lambda}\right)$

The symbols used have the following meanings :

• y — the displacement of the particle situated at a distance x from the origin, at any instant t
• a — the amplitude of the wave, that is, the maximum displacement of the particle from its rest position
• T — the time-period of the oscillation of the particle
• λ — the wavelength of the wave
• x — the distance of the particle from the origin, measured along the direction of propagation
• t — the instant of time at which the displacement is measured

If the wave travels along the − X direction, the minus sign inside the bracket is replaced by a plus sign.

In the wave equation y = a sin (ω t − k x),

• ω is the angular frequency of the oscillating particle, related to the frequency n and the time-period T by $\omega = 2\pi \text n = \frac{2\pi}{\text T}$
• k is the propagation constant, or wave number, given by $\text k = \frac{2\pi}{\lambda}$, which defines how rapidly the phase of the wave changes with distance

The ratio $\frac{\omega}{\text k}$ represents the speed of the wave,

$\frac{\omega}{\text k} = \frac{2\pi \text n}{2\pi/\lambda} = \text n \lambda = \text v$

Given, the amplitude is A metre, the velocity is v metre/second and the frequency is n hertz.

The equation of a plane progressive wave travelling along the positive direction of the X-axis is

$\text y = \text A \sin 2\pi \text n\left(\text t - \frac{\text x}{\text v}\right)$

where y is the displacement, at any instant t, of the particle situated at a distance x from the origin.

In the equation y = a sin (ω t − k x), the argument of the sine gives the phase of the particle. Hence

Phase at time t :

$\phi = (\omega \text t - \text k \text x)$

Velocity of the wave : The phase of a given point of the wave remains constant as the wave advances, so

$\text v = \frac{\omega}{\text k}$

since $\omega = 2\pi \text n$ and $\text k = \frac{2\pi}{\lambda}$, which gives $\frac{\omega}{\text k} = \text n \lambda = \text v$.

Two waves meet in opposite phases when the phase difference between them is an odd multiple of π, that is,

$\Delta \phi = \pi,\ 3\pi,\ 5\pi,\ \dots$

or, in general, Δφ = (2m − 1)π, where m = 1, 2, 3, ... At such places destructive interference takes place and the resultant amplitude is a₁ ~ a₂.

The relation between the phase difference (Δφ) and the path difference (Δx) between two points is

$\Delta \phi = \frac{2\pi}{\lambda} \times \Delta \text x$

where λ is the wavelength of the wave. Thus a path difference of one full wavelength corresponds to a phase difference of 2π, which means that the two particles are in the same phase of oscillation.

Beats : When two sound waves of nearly equal frequencies are produced simultaneously, the intensity of the resultant sound produced by their superposition increases and decreases alternately with time. This rise and fall in the intensity of sound is called the phenomenon of beats.

One rise and one fall of the intensity together form one beat, and the number of times the intensity of sound rises and falls in one second is called the beat frequency. It is equal to the difference in the frequencies of the two sound-sources, that is, n₁ ~ n₂.

No, beats cannot be observed with two light-sources.

To observe beats, the phase difference between the two sources must change regularly with time. In light-sources this change takes place at random, because a light-source consists of innumerable atoms and each atom emits wave independently of the others. Hence the phase difference between two light-sources does not vary in a regular manner, and no beats are observed.

The maximum difference in the frequencies of the two sound-sources should be 10.

The sensitivity of our ear is $\frac{1}{10}$ second, that is, the effect of any sound remains on the ear for $\frac{1}{10}$ second. Hence the ear can clearly hear a maximum of 10 beats per second. If the difference in the frequencies exceeds 10, the successive beats overlap on the ear and cannot be distinguished separately.

In a stationary wave,

(i) The distance between a node and the nearest antinode is

$\frac{\lambda}{4}$

(ii) The distance between two successive antinodes is

$\frac{\lambda}{2}$

which is also the distance between two successive nodes.

The phase difference between a node and its nearest antinode in a stationary wave is $\frac{\pi}{2}$ radian.

The distance between a node and its nearest antinode is $\frac{\lambda}{4}$. Hence, using $\Delta \phi = \frac{2\pi}{\lambda} \times \Delta \text x$,

$\Delta \phi = \frac{2\pi}{\lambda} \times \frac{\lambda}{4} = \frac{\pi}{2}\ \text{radian}$

Particles on the two sides of a node : The phase difference is π. At any instant, the phase of vibration of the points on one side of a node is opposite to the phase of vibration of the points on the other side.

Particles lying between two nodes : The phase difference is zero. All the points between two successive nodes vibrate in the same phase — they reach their positions of maximum displacement simultaneously and pass simultaneously through their mean positions.

The nodes and antinodes are not seen because they are far too close together to be resolved.

The wavelength of light is of the order of 10^-7 metre. The distance between two consecutive nodes, or between two consecutive antinodes, is $\frac{\lambda}{2}$, which is therefore also of the order of 10^-7 metre. It is not possible to resolve such a small distance by the eye or by an ordinary optical instrument, and so the nodes and antinodes are not seen.

The pipe is closed at one end.

Given, the fundamental frequency n₁ = 128 hertz and the frequency of the first overtone = 384 hertz. The ratio of the two is

$\frac{384}{128} = 3$

A closed organ pipe produces only odd harmonics, so its first overtone is the third harmonic, that is, 3 n₁. An open pipe would have given the first overtone as the second harmonic, 2 n₁ = 256 hertz. Hence the pipe is closed.

The pipe B, which has the smaller diameter, has the higher frequency.

Applying the end correction, the fundamental frequency of an open organ pipe of radius r is

$\text n = \frac{\text v}{2(\text l + 1.2\text r)}$

As r increases, the denominator increases and so the frequency n decreases. The diameter of A is double that of B, so A has the larger radius and therefore the lower frequency. Hence pipe B has the higher fundamental frequency.

Let n be the fundamental frequency of a stretched string. Since a stretched string gives both the even and the odd harmonics, in the ratio n₁ : n₂ : n₃ ... = 1 : 2 : 3 ...,

Frequency of the second harmonic = 2n

Frequency of the third harmonic = 3n

The second harmonic is also called the first overtone and the third harmonic the second overtone.

The speed of a transverse wave in a stretched string is

$\text v = \sqrt{\frac{\text T}{\text m}}$

where

• T — the tension in the string
• m — the mass per unit length of the string, that is, its linear mass density, and not the mass of the whole string

If r be the radius of the string and d the density of its material, then m = πr²d, so that the speed may also be written as

$\text v = \sqrt{\frac{\text T}{\pi \text r^2 \text d}} = \sqrt{\frac{\text{stress}}{\text{density}}}$

The speed of sound in a medium, in terms of the modulus of elasticity E and the density d of the medium, is

$\text v = \sqrt{\frac{\text E}{\text d}}$

Here E is the appropriate modulus of elasticity of the medium — Young's modulus for a long rod, and the bulk modulus for a liquid or a gas.

The speed of longitudinal waves in a metallic rod is

$\text v = \sqrt{\frac{\text Y}{\text d}}$

where Y is the Young's modulus of the metal and d is its density.

Newton's formula : Newton assumed that when longitudinal waves travel in a gas, the temperature of the gas remains constant, so that the elasticity concerned is the isothermal elasticity, whose value is equal to the pressure P of the gas. Hence

$\text v = \sqrt{\frac{\text P}{\text d}}$

where P is the pressure of the gas and d is its density.

Laplace's formula : Laplace showed that the compressions and rarefactions occur so rapidly that the exchange of heat cannot take place, so the process is adiabatic and the elasticity concerned is the adiabatic elasticity, whose value is γP. Hence

$\text v = \sqrt{\frac{\gamma \text P}{\text d}}$

where $\gamma = \frac{\text C_\text p}{\text C_\text v}$ is the ratio of the specific heat of the gas at constant pressure to that at constant volume, P is the pressure of the gas and d is its density.

The speed of a transverse wave on a stretched string is

$\text v = \sqrt{\frac{\text T}{\text m}}$

where T is the tension in the string and m is the mass per unit length of the string.

The graph will be a straight line parallel to the P-axis.

At a constant temperature the ratio $\frac{\text P}{\text d}$ is a constant, since a change in P is accompanied by a proportional change in d. Hence from $\text v = \sqrt{\frac{\gamma \text P}{\text d}}$ the speed v does not change with P, and the graph between P and v is a straight line parallel to the pressure axis.

Given, γ = 4/3 for the polyatomic gas.

The ratio of the speed of sound in a gas to the root-mean-square speed of its molecules is

$\frac{\text v}{\text v_{rms}} = \sqrt{\frac{\gamma}{3}}$

Substituting the value of γ,

$\frac{\text v}{\text v_{rms}} = \sqrt{\frac{4/3}{3}} = \sqrt{\frac{4}{9}}$

$\frac{\text v}{\text v_{rms}} = \frac{2}{3}$

Hence, the required ratio is 2 : 3.

Hydrogen is a diatomic gas, so $\gamma = \frac{7}{5}$.

The ratio of the speed of sound in a gas to the root-mean-square speed of its molecules is

$\frac{\text v}{\text v_{rms}} = \sqrt{\frac{\gamma}{3}}$

Substituting the value of γ,

$\frac{\text v}{\text v_{rms}} = \sqrt{\frac{7/5}{3}} = \sqrt{\frac{7}{15}}$

Hence, the required ratio is $\sqrt{\frac{7}{15}}$.

Given, the fundamental frequency of the open organ pipe, n_open = 512 Hz.

For an open pipe of length l,

$\text n_{open} = \frac{\text v}{2\text l}$

and for a pipe of the same length closed at one end,

$\text n_{closed} = \frac{\text v}{4\text l} = \frac{1}{2}\text n_{open}$

Substituting the value,

$\text n_{closed} = \frac{512}{2}$

$\text n_{closed} = 256\ \text{hertz}$

Hence, the fundamental frequency becomes 256 hertz when one end is closed.

Given, the frequency of the closed-end organ pipe, n_closed = 50 Hz.

For a closed pipe of length l,

$\text n_{closed} = \frac{\text v}{4\text l} = 50\ \text{Hz}$

When the other end is also opened, the pipe becomes an open pipe of the same length, and its fundamental frequency is

$\text n_{open} = \frac{\text v}{2\text l} = 2 \times 50 = 100\ \text{Hz}$

An open pipe produces both the even and the odd harmonics, in the ratio 1 : 2 : 3 : 4 ... Hence the resonating frequencies are

$100\ \text{Hz},\ 200\ \text{Hz},\ 300\ \text{Hz},\ \dots$

Hence, the resonating frequencies of the opened pipe are 100 Hz, 200 Hz, 300 Hz, and so on.

Given, the frequency of the fundamental note of the closed pipe, n = 150 per second, and the length of the second pipe is half that of the first.

The fundamental frequency of a pipe closed at one end is

$\text n = \frac{\text v}{4\text l} \qquad \Rightarrow \qquad \text n \propto \frac{1}{\text l}$

For the second pipe of length $\frac{\text l}{2}$,

$\text n' = \frac{\text v}{4\left(\frac{\text l}{2}\right)} = 2\left(\frac{\text v}{4\text l}\right) = 2\text n$

$\text n' = 2 \times 150 = 300\ \text{per second}$

Hence, the frequency of the fundamental note of the second pipe is 300 per second.

Let l₁ be the length of the closed organ pipe and l₂ that of the open organ pipe.

The fundamental frequencies are

$\text n_1 = \frac{\text v}{4\text l_1} \qquad \text{and} \qquad \text n_2 = \frac{\text v}{2\text l_2}$

It is given that the two fundamental frequencies are the same, so n₁ = n₂,

$\frac{\text v}{4\text l_1} = \frac{\text v}{2\text l_2}$

$\frac{\text l_1}{\text l_2} = \frac{2}{4} = \frac{1}{2}$

Hence, the ratio of the length of the closed pipe to that of the open pipe is 1 : 2.

In the two diagrams the stretched string vibrates between the fixed points P and Q, so a node is formed at each of these ends.

First case : The string vibrates in one segment, that is, in the fundamental mode. If l be the length PQ, then

$\text l = \frac{\lambda_1}{2} \qquad \Rightarrow \qquad \text n_1 = \frac{\text v}{2\text l}$

Second case : The string vibrates in three segments, with two additional nodes between P and Q. Hence

$\text l = \frac{3\lambda_2}{2} \qquad \Rightarrow \qquad \text n_2 = \frac{3\text v}{2\text l} = 3\text n_1$

The tension in the string remains unchanged, so the speed v is the same in both the cases.

$\frac{\text n_1}{\text n_2} = \frac{1}{3}$

Hence, the ratio of the frequencies of the string in the two cases is 1 : 3.

Given, the maximum particle velocity is twice the wave velocity.

For a plane progressive wave the maximum particle velocity is

$\text u_{max} = \text a \omega = \text a (2\pi \text n)$

and the wave velocity is

$\text v = \text n \lambda$

It is given that u_max = 2v, so

$\text a (2\pi \text n) = 2\text n \lambda$

$\frac{\lambda}{\text a} = \pi$

Hence, the ratio of the wavelength to the amplitude of the wave is π : 1.

Given,

• Frequency of the sound waves, n = 660 Hz
• Speed of sound in air, v = 330 m/s

The wave falls normally on a perfectly reflecting wall, so the incident and the reflected waves superpose and form a stationary wave. The wall is a rigid boundary, so a node is formed at the wall.

The wavelength of the wave is

$\lambda = \frac{\text v}{\text n} = \frac{330\ \text{m/s}}{660\ \text{Hz}} = 0.5\ \text m$

The air particles have a maximum amplitude of vibration at an antinode, and the distance of the nearest antinode from the node at the wall is $\frac{\lambda}{4}$. Hence the shortest distance is

$\frac{\lambda}{4} = \frac{0.5}{4}$

$= 0.125\ \text m$

Hence, the shortest distance from the wall at which the air particles have a maximum amplitude of vibration is 0.125 m.

The equation of a stationary wave is

$\text y = 2\text a \cos \frac{2\pi \text x}{\lambda} \sin \frac{2\pi \text t}{\text T}$

The symbols used have the following meanings :

• y — the resultant displacement, at the instant t, of the particle situated at a distance x from the origin
• a — the amplitude of each of the two component progressive waves, so that 2a is the maximum amplitude at an antinode
• λ — the wavelength of the component waves
• T — the time-period of the component waves
• x — the distance of the particle from the origin
• t — the instant of time

In this wave the positions x = 0, $\frac{\lambda}{2}$, $\frac{2\lambda}{2}$, ... are the antinodes, and the positions x = $\frac{\lambda}{4}$, $\frac{3\lambda}{4}$, $\frac{5\lambda}{4}$, ... are the nodes.

Given, the fundamental frequency of the pipe closed at one end, n_closed = 200 hertz.

For a pipe of length l closed at one end,

$\text n_{closed} = \frac{\text v}{4\text l} = 200\ \text{hertz}$

For a similar pipe of the same length open at both ends,

$\text n_{open} = \frac{\text v}{2\text l} = 2\left(\frac{\text v}{4\text l}\right) = 2 \times 200$

$\text n_{open} = 400\ \text{hertz}$

Hence, the fundamental frequency of the pipe open at both ends is 400 hertz.

In the figure the two vibrating air columns are those of pipes closed at one end, since each has a node at the closed end and an antinode at the open end.

First case : The figure shows the first overtone, that is, the third harmonic, of the closed pipe. Its frequency is

$\text n_1 = \frac{3\text v}{4\text l}$

Second case : The figure shows the second overtone, that is, the fifth harmonic, of the closed pipe. Its frequency is

$\text n_2 = \frac{5\text v}{4\text l}$

Therefore,

$\frac{\text n_1}{\text n_2} = \frac{3\text v/4\text l}{5\text v/4\text l} = \frac{3}{5}$

Hence, the ratio of the frequencies in the two cases is 3 : 5.

Given, the two strings are of equal length, are stretched by equal forces, are made of the same material, and their diameters are in the ratio 3 : 2, that is, r₁ : r₂ = 3 : 2.

The fundamental frequency of a stretched string is

$\text n = \frac{1}{2\text l}\sqrt{\frac{\text T}{\text m}} = \frac{1}{2\text l}\sqrt{\frac{\text T}{\pi \text r^2 \text d}} \qquad \Rightarrow \qquad \text n \propto \frac{1}{\text r}$

since l, T and d are the same for both the strings. Therefore,

$\frac{\text n_1}{\text n_2} = \frac{\text r_2}{\text r_1} = \frac{2}{3}$

Hence, the ratio of their fundamental frequencies is 2 : 3.

Given, the frequency of the third harmonic of the stretched string, n₃ = 255 hertz.

A stretched string gives both the even and the odd harmonics, whose frequencies are in the ratio n₁ : n₂ : n₃ ... = 1 : 2 : 3 ... Hence the third harmonic is three times the fundamental frequency,

$\text n_3 = 3\text n_1$

$\text n_1 = \frac{\text n_3}{3} = \frac{255}{3}$

$\text n_1 = 85\ \text{hertz}$

Hence, the fundamental frequency of the string is 85 hertz.

Yes. If sounds are produced simultaneously by different musical instruments in an orchestra, they are all heard by the ear at the same time.

The notes produced by the various instruments have widely different frequencies. If the speed of sound depended upon the frequency, then the notes of different frequencies would reach the listener at different instants of time and the music would be heard jumbled. Since we hear them together and in the proper sequence, the speed of sound in air must be the same for all frequencies. This is why we can enjoy an orchestra even when seated far away.

The frequency of the note emitted by an organ pipe increases with a rise in temperature.

The frequency of the note emitted by an organ pipe is

$\text n \propto \frac{\text v}{\text l}$

With a rise in temperature the speed of sound in air increases, since $\text v \propto \sqrt{\text T}$. The length of the pipe is practically unchanged, so the frequency of the note also increases.

The fundamental frequency of the closed organ pipe will decrease.

The fundamental frequency of a closed organ pipe is

$\text n = \frac{\text v}{4\text l}$

where v is the speed of sound in the gas filling the pipe. Since $\text v \propto \frac{1}{\sqrt{\text d}}$, a gas heavier than air has a greater density and so the speed of sound in it is smaller. The length of the pipe being unchanged, the fundamental frequency therefore decreases.

This happens because of the end correction.

Applying the end correction e, the fundamental frequencies of a closed pipe and of an open pipe of the same length l are

$\text n_{closed} = \frac{\text v}{4(\text l + \text e)} \qquad \text{and} \qquad \text n_{open} = \frac{\text v}{2(\text l + 2\text e)}$

Hence the fundamental frequency of the open pipe is not exactly twice that of the closed pipe of the same length, but slightly less than twice. Since the harmonics of the open pipe therefore do not coincide exactly with the frequency of the fork that resonates with the closed pipe, the same fork does not produce resonance in the open pipe.

The final velocity of the wave will be twice its initial velocity.

The speed of a transverse wave in a stretched string is

$\text v = \sqrt{\frac{\text T}{\text m}} \qquad \Rightarrow \qquad \text v \propto \sqrt{\text T}$

If the tension is made 4 T, then

$\text v' = \sqrt{\frac{4\text T}{\text m}} = 2\sqrt{\frac{\text T}{\text m}} = 2\text v$

The new wave speed will be double its initial value.

The speed of a transverse wave in a stretched wire of radius r is

$\text v = \sqrt{\frac{\text T}{\pi \text r^2 \text d}} \qquad \Rightarrow \qquad \text v \propto \frac{1}{\text r}$

for a given tension and a given material. If the radius is reduced to $\frac{\text r}{2}$, then

$\text v' = 2\text v$

The wire will be in unison with a fork of frequency 2 n.

The fundamental frequency of a sonometer wire is

$\text n = \frac{1}{2\text l}\sqrt{\frac{\text T}{\text m}} \qquad \Rightarrow \qquad \text n \propto \frac{1}{\text l}$

for a given tension and a given wire. If the length is reduced to $\frac{\text l}{2}$, the frequency becomes

$\text n' = 2\text n$

(i) A path difference of $\frac{\lambda}{4}$ corresponds to

$\text{phase difference} = \frac{2\pi}{\lambda} \times \frac{\lambda}{4} = \frac{\pi}{2}$

$\text{time difference} = \frac{\text T}{\lambda} \times \frac{\lambda}{4} = \frac{\text T}{4}$

(ii) A phase difference of $\frac{\pi}{3}$ corresponds to

$\text{time difference} = \frac{\text T}{2\pi} \times \frac{\pi}{3} = \frac{\text T}{6}$

$\text{path difference} = \frac{\lambda}{2\pi} \times \frac{\pi}{3} = \frac{\lambda}{6}$

Given, the equation of the progressive wave is

$\text y = 0.09 \sin 8\pi\left[\text t - \frac{\text x}{20}\right]$

and the amplitude of the reflected wave is $\frac{2}{3}$ of that of the incident wave.

Amplitude of the reflected wave :

$\text a' = \frac{2}{3} \times 0.09 = 0.06$

Direction of travel : The incident wave travels along the positive direction of the X-axis, so the reflected wave travels along the negative direction. Hence x is replaced by − x, that is, the minus sign inside the bracket becomes a plus sign.

Phase change : The wave is reflected back from a rigid wall, so it suffers a phase change of π. Therefore

$\text y = 0.06 \sin \bigg\lbrace 8\pi\left[\text t + \frac{\text x}{20}\right] + \pi \bigg\rbrace$

Using sin (θ + π) = − sin θ,

$\text y = - 0.06 \sin 8\pi\left(\text t + \frac{\text x}{20}\right)$

Hence, the equation of the reflected wave is $\text y = - 0.06 \sin 8\pi\left(\text t + \frac{\text x}{20}\right)$.

Newton assumed that when longitudinal waves travel in a gas, the temperature of the gas remains constant. On this assumption the elasticity concerned is the isothermal elasticity, whose value is equal to the pressure P of the gas, and the speed of sound is

$\text v = \sqrt{\frac{\text P}{\text d}}$

Taking P = 1.01 × 10⁵ N m^-2 and d = 1.29 kg m^-3 for air at 0°C, this gives v = 280 m s^-1, whereas the experimental value is nearly 331 m s^-1. The discrepancy is far too large.

Laplace pointed out that Newton's assumption is wrong. When sound waves travel in a gaseous medium, the states of compression and rarefaction occur alternately at every point. At a compression some heat is developed and at a rarefaction some heat is lost. The compressions and rarefactions occur so rapidly that the heat produced during compression cannot go out into the surroundings and the heat lost during rarefaction cannot come in from them. Besides this, gases are bad conductors of heat, so no exchange of heat takes place at all. Hence the temperature at a point rises during compression and falls during rarefaction, and the process is adiabatic, not isothermal.

Therefore the elasticity concerned is the adiabatic elasticity, whose value is γP, and the corrected formula is

$\text v = \sqrt{\frac{\gamma \text P}{\text d}}$

With γ = 1.41 for air, this gives $\text v = \sqrt{1.41} \times 280 = 332$ m s^-1, which agrees closely with the experimental value.

The speed of sound in moist air is greater than in dry air.

The density of water-vapour is less than that of air, so the density of air mixed with water-vapour, that is, moist air, is less than that of dry air. Since

$\text v = \sqrt{\frac{\gamma \text P}{\text d}} \qquad \Rightarrow \qquad \text v \propto \frac{1}{\sqrt{\text d}}$

at a given pressure, the smaller density of moist air gives a greater speed of sound.

No, the speed of sound in moist hydrogen will be less than in dry hydrogen.

The density of water-vapour is more than that of hydrogen, which is the lightest of all gases. Hence the density of moist hydrogen is greater than that of dry hydrogen, and by the same relation $\text v \propto \frac{1}{\sqrt{\text d}}$ the speed of sound in moist hydrogen is less than in dry hydrogen.

In hydrogen the speed will be more, and in carbon-dioxide it will be less than 300 m s^-1.

At a given temperature the speed of sound in a gas is

$\text v = \sqrt{\frac{\gamma \text{RT}}{\text M}} \qquad \Rightarrow \qquad \text v \propto \frac{1}{\sqrt{\text M}}$

that is, the speed of sound in a gas is inversely proportional to the square-root of its molecular-mass.

The molecular mass of hydrogen, 2, is much less than the average molecular mass of air, about 29, so the speed of sound in hydrogen is much more than in air.

The molecular mass of carbon-dioxide, 44, is more than that of air, so the speed of sound in carbon-dioxide is less than in air.

Stationary waves : When two identical transverse, or longitudinal, progressive waves travel in a bounded medium with the same speed but in opposite directions, then by their superposition a new type of wave is produced which appears stationary in the medium. This wave is called a stationary (or standing) wave.

The characteristic of the stationary wave is that some particles of the medium remain permanently at rest, while some other particles undergo maximum displacement compared to others. The former are called the nodes and the latter the antinodes.

Equation : If two waves of amplitude a travelling in opposite directions superpose, the equation of the resulting stationary wave is

$\text y = 2\text a \cos \left(\frac{2\pi \text x}{\lambda}\right)\sin \left(\frac{2\pi \text t}{\text T}\right)$

where λ is the wavelength and T the time-period of the component waves.

The conditions for the formation of a stationary wave are :

(i) The medium should not be unlimited; it should have a boundary. That is, the medium must be a bounded medium, so that the wave propagating in it is reflected at the boundary.

(ii) The two superposing waves must travel in opposite directions along the same line.

(iii) The two waves must be of the same kind, that is, both transverse or both longitudinal.

(iv) The two waves must have the same frequency (or wavelength) and the same speed.

(v) The amplitudes of the two waves should be equal, or nearly equal, so that the nodes are well defined.

The air column above the water level in the vessel behaves as an air column closed at one end, the water surface being the closed end. The frequency of the note emitted by such an air column is

$\text n = \frac{\text v}{4\text l} \qquad \Rightarrow \qquad \text n \propto \frac{1}{\text l}$

that is, the frequency is inversely proportional to the length of the air column.

As the water level in the vessel rises, the length l of the air column above it goes on decreasing, so the frequency of the emitted note goes on increasing, that is, the note becomes more and more shrill.

Hence, simply by listening to the sound and noting how shrill it has become, we can estimate from a distance the height up to which the vessel has been filled with water.

The organ pipe is in resonance with the tuning fork, so its frequency $\text n \propto \frac{\text v}{\text l}$ must be kept unchanged.

(i) Temperature rises : With a rise in temperature the speed of sound in air increases, since $\text v \propto \sqrt{\text T}$. To keep the frequency of the pipe unchanged, the length l of the pipe must therefore be increased.

(ii) Hydrogen is filled in place of air : Hydrogen is far lighter than air, and since $\text v \propto \frac{1}{\sqrt{\text d}}$, the speed of sound is more in hydrogen than in air. Hence again, to keep the frequency unchanged, the length l must be increased.

(iii) Pressure becomes higher : There is no effect of a pressure rise on the speed of sound, because at a constant temperature the ratio $\frac{\text P}{\text d}$ remains constant. Hence no change is required in l.

Harmonics : If the frequencies of the fundamental tone and the overtones produced by a source of sound are in a harmonic series, then these tones are called harmonics. The fundamental tone is the first harmonic, and the tone whose frequency is m times the fundamental frequency is the mth harmonic.

Examples :

(i) Closed organ pipe : The frequencies of the fundamental tone and the overtones are in the ratio

$\text n_1 : \text n_2 : \text n_3 \dots = 1 : 3 : 5 : 7 \dots$

that is, a closed pipe produces only the odd harmonics.

(ii) Open organ pipe : The frequencies of the fundamental tone and the overtones are in the ratio

$\text n_1 : \text n_2 : \text n_3 : \text n_4 \dots = 1 : 2 : 3 : 4 \dots$

that is, an open pipe produces both the even and the odd harmonics. A stretched string also gives both types of harmonics in this same ratio.

Resonance : When an external periodic force is applied over a body and the frequency of the force is different from the natural frequency of the body, then the body executes forced vibrations with the frequency of the applied force but with a small amplitude. If, however, the frequency of the external force is equal to the natural frequency of the body, or to its integral multiple, then the amplitude of the forced vibrations of the body becomes quite large. This phenomenon is called resonance. Thus, resonance is a particular case of forced vibrations.

Explanation : When the frequency of the external force is equal to the natural frequency of the body, then at each step the force is in phase with the oscillating body. Hence the successive impulses given by the periodic force to the body are added up and increase the amplitude of oscillation continuously. But with increasing amplitude the air resistance and the internal friction also increase, so that the loss of energy from the body also increases. Finally a stage is reached when the energy supplied by the external force becomes equal to the energy lost by the body, and the amplitude becomes steady.

Example : If a tuning fork of frequency 256 Hz is vibrated over the mouth of a resonance tube and the length of the air column is gradually adjusted, then at one particular length the air column emits a loud sound. This happens when the natural frequency of the air column becomes equal to the frequency of the fork.

The speed of sound in a gas is directly proportional to the square-root of its absolute temperature,

$\text v \propto \sqrt{\text T}$

If v₀ be the speed of sound at 0°C and v_t the speed at t°C, then

$\text v_0 = \sqrt{\frac{\gamma \text R}{\text M}} \times \sqrt{0 + 273} \qquad \dots(\text i)$

$\text v_\text t = \sqrt{\frac{\gamma \text R}{\text M}} \times \sqrt{\text t + 273} \qquad \dots(\text{ii})$

Dividing equation (ii) by equation (i),

$\frac{\text v_\text t}{\text v_0} = \sqrt{\frac{\text t + 273}{273}} = \sqrt{1 + \frac{\text t}{273}}$

$\text v_\text t = \text v_0\left(1 + \frac{\text t}{273}\right)^{1/2}$

Since t is small compared with 273, the binomial theorem may be applied and the terms beyond the first power of $\frac{\text t}{273}$ neglected,

$\text v_\text t = \text v_0\left(1 + \frac{1}{2} \times \frac{\text t}{273}\right) = \text v_0\left(1 + \frac{\text t}{546}\right)$

Taking the speed of sound in air at 0°C as v₀ = 332 m s^-1,

$\text v_\text t = 332\left(1 + \frac{\text t}{546}\right) = 332 + \frac{332\text t}{546}$

$\text v_\text t = (332 + 0.61\text t)\ \text{m s}^{-1}$

Hence, the velocity of sound in air increases by 0.61 m s^-1 for each °C rise in the temperature of the air.

The graph will be a straight line passing through the origin.

The speed of sound in a gas is

$\text v = \sqrt{\frac{\gamma \text{RT}}{\text M}} \qquad \Rightarrow \qquad \text v^2 = \left(\frac{\gamma \text R}{\text M}\right)\text T$

For a given gas the quantity $\frac{\gamma \text R}{\text M}$ is a constant, so

$\text v^2 \propto \text T$

Since v² is directly proportional to T, and v² = 0 when T = 0, the graph between the absolute temperature T and the square of the speed of sound v² is a straight line passing through the origin.

Given, the equation of the progressive wave is

$\text y = \text a \sin 2\pi\left(\frac{\text t}{\text T} - \frac{\text x}{\lambda}\right)$

First other form : In wave motion $\frac{1}{\text T} = \frac{\text v}{\lambda}$. Substituting this,

$\text y = \text a \sin 2\pi\left(\frac{\text v}{\lambda}\text t - \frac{\text x}{\lambda}\right)$

$\text y = \text a \sin \frac{2\pi}{\lambda}(\text{vt} - \text x)$

Second other form : In wave motion $\text n = \frac{\text v}{\lambda}$, so that $\frac{2\pi}{\lambda} = \frac{2\pi \text n}{\text v}$. Substituting this in the first form,

$\text y = \text a \sin \frac{2\pi \text n}{\text v}(\text{vt} - \text x)$

$\text y = \text a \sin 2\pi \text n\left(\text t - \frac{\text x}{\text v}\right)$

(i) Phases of the points P and Q :

In one complete wave cycle the phase changes by 2π radian. From the curve, the point P is at the position of maximum positive displacement, that is, at a crest, which corresponds to one-quarter of a cycle. Hence

$\text{Phase of P} = \frac{\pi}{2}$

The point Q is at the position of maximum negative displacement, that is, at a trough, which corresponds to three-quarters of a cycle. Hence

$\text{Phase of Q} = \frac{3\pi}{2}$

(ii) Time-period of the wave :

The time-period is the time taken for one complete oscillation. From the graph the displacement curve repeats itself after every 4 seconds.

$\text T = 4\ \text{seconds}$

Hence, the phases of P and Q are $\frac{\pi}{2}$ and $\frac{3\pi}{2}$ respectively, and the time-period of the wave is 4 seconds.

Given, the two progressive waves are

$\text y_1 = \text a \sin \frac{2\pi}{\lambda}(\text{v t} - \text x) \qquad \text{and} \qquad \text y_2 = \text a \sin \frac{2\pi}{\lambda}(\text{v t} + \text x)$

The two waves are identical and travel in opposite directions, so their superposition gives a stationary wave.

By the principle of superposition, the resultant displacement is

$\text y = \text y_1 + \text y_2 = \text a\left[\sin \frac{2\pi}{\lambda}(\text{vt} - \text x) + \sin \frac{2\pi}{\lambda}(\text{vt} + \text x)\right]$

Using $\sin \text C + \sin \text D = 2 \sin \frac{\text C + \text D}{2}\cos \frac{\text C - \text D}{2}$,

$$\begin{aligned}\text y = 2\text a \sin\left[\frac{1}{2}\bigg\lbrace\frac{2\pi}{\lambda}(\text{vt} - \text x) + \frac{2\pi}{\lambda}(\text{vt} + \text x)\bigg\rbrace\right] \\[1em] \times \cos\left[\frac{1}{2}\bigg\lbrace\frac{2\pi}{\lambda}(\text{vt} - \text x) - \frac{2\pi}{\lambda}(\text{vt} + \text x)\bigg\rbrace\right]\end{aligned}$$

$\text y = 2\text a \sin \frac{2\pi \text{vt}}{\lambda}\cos \frac{2\pi \text x}{\lambda}$

Since $\frac{\text v}{\lambda} = \frac{1}{\text T}$,

$\text y = 2\text a \cos \frac{2\pi \text x}{\lambda}\sin \frac{2\pi \text t}{\text T}$

Hence, the equation of the resultant wave is $\text y = 2\text a \cos \frac{2\pi \text x}{\lambda}\sin \frac{2\pi \text{vt}}{\text T}$, which is the equation of a stationary wave.

In an open organ pipe an antinode (A) is formed at each of the two open ends.

The second overtone of an open pipe is its third harmonic, corresponding to m = 3. In this mode

$\text l = \frac{3\lambda_3}{2} \qquad \Rightarrow \qquad \lambda_3 = \frac{2\text l}{3}$

The air column therefore vibrates with four antinodes — one at each of the two ends and two in between — and three nodes (N) between them, as shown in the diagram.

The frequency of this mode is

$\text n_3 = \frac{\text v}{\lambda_3} = \frac{3\text v}{2\text l} = 3\text n_1$

In an organ pipe closed at one end, a node (N) is always formed at the closed end and an antinode (A) at the open end.

The third harmonic of a closed pipe is its first overtone, corresponding to m = 2. In this mode

$\text l = \frac{3\lambda_2}{4} \qquad \Rightarrow \qquad \lambda_2 = \frac{4\text l}{3}$

In this mode one additional antinode and one additional node are formed between the closed and the open ends of the pipe, so the air column has two nodes and two antinodes, as shown in the diagram.

The frequency of this mode is

$\text n_2 = \frac{\text v}{\lambda_2} = \frac{3\text v}{4\text l} = 3\text n_1$

A stretched string is clamped at both its ends, so a node (N) is always formed at each end.

Fundamental tone (first harmonic), p = 1 : The string vibrates as a whole in one segment. There are two nodes, one at each end, and one antinode (A) in the middle.

$\text l = \frac{\lambda_1}{2} \qquad \Rightarrow \qquad \text n_1 = \frac{\text v}{2\text l}$

First overtone (second harmonic), p = 2 : The string vibrates in two segments. There are three nodes — one at each end and one in the middle — and two antinodes.

$\text l = \frac{2\lambda_2}{2} \qquad \Rightarrow \qquad \text n_2 = \frac{2\text v}{2\text l} = 2\text n_1$

Second overtone (third harmonic), p = 3 : The string vibrates in three segments. There are four nodes and three antinodes.

$\text l = \frac{3\lambda_3}{2} \qquad \Rightarrow \qquad \text n_3 = \frac{3\text v}{2\text l} = 3\text n_1$

Given,

• Length of the iron rod = 100 cm
• The rod is clamped at 25 cm from one end

On rubbing the rod with a sand-paper, longitudinal stationary waves are established in it. At the clamped point the rod cannot vibrate, so a node (N) is formed there. At each free end-point the rod has the greatest freedom to vibrate, so an antinode (A) is formed there.

The distance between an antinode and its nearest node is $\frac{\lambda}{4}$. The clamp is at 25 cm from one end, so the distance between the antinode at that end and the node at the clamp is 25 cm. Hence

$\frac{\lambda}{4} = 25\ \text{cm}$

$\lambda = 4 \times 25$

$\lambda = 100\ \text{cm}$

Hence, the wavelength of the wave produced in the rod is 100 cm.

Given, two diatomic gases at the same temperature and pressure, of densities d₁ and d₂.

The speed of sound in a gas is

$\text v = \sqrt{\frac{\gamma \text P}{\text d}}$

Both the gases are diatomic, so the value of γ is the same for both, namely $\frac{7}{5}$. The pressure P is also the same for both. Hence

$\text v_1 = \sqrt{\frac{\gamma \text P}{\text d_1}} \qquad \text{and} \qquad \text v_2 = \sqrt{\frac{\gamma \text P}{\text d_2}}$

Dividing,

$\frac{\text v_1}{\text v_2} = \sqrt{\frac{\text d_2}{\text d_1}}$

Hence, the ratio of the speeds of sound in the two gases is $\sqrt{\text d_2} : \sqrt{\text d_1}$.

Given, one gas is monoatomic of density d₁ and the other is diatomic of density d₂, both at the same temperature and pressure.

The speed of sound in a gas is

$\text v = \sqrt{\frac{\gamma \text P}{\text d}}$

Since the pressure is the same for both,

$\frac{\text v_1}{\text v_2} = \sqrt{\frac{\gamma_1 \text d_2}{\gamma_2 \text d_1}}$

For a monoatomic gas $\gamma_1 = \frac{5}{3}$ and for a diatomic gas $\gamma_2 = \frac{7}{5}$. Substituting these values,

$\frac{\text v_1}{\text v_2} = \sqrt{\frac{\left(\frac{5}{3}\right)\text d_2}{\left(\frac{7}{5}\right)\text d_1}} = \sqrt{\frac{5}{3} \times \frac{5}{7} \times \frac{\text d_2}{\text d_1}}$

$\frac{\text v_1}{\text v_2} = \sqrt{\frac{25\text d_2}{21\text d_1}}$

Hence, the ratio of the speeds of sound in the two gases is $\sqrt{\frac{25\text d_2}{21\text d_1}}$.

Given, hydrogen and oxygen at a constant temperature.

Both hydrogen and oxygen are diatomic gases, so the value of γ is the same for both. At the same temperature the speed of sound in a gas is

$\text v = \sqrt{\frac{\gamma \text{RT}}{\text M}} \qquad \Rightarrow \qquad \text v \propto \frac{1}{\sqrt{\text M}}$

The molecular masses of hydrogen and oxygen are M_H = 2 and M_O = 32. Hence

$\frac{\text v_\text H}{\text v_\text O} = \sqrt{\frac{\text M_\text O}{\text M_\text H}} = \sqrt{\frac{32}{2}} = \sqrt{16}$

$\frac{\text v_\text H}{\text v_\text O} = 4$

Hence, the speed of sound in hydrogen is four times that in oxygen, that is, the ratio is 4 : 1.

Given, the equation of the transverse wave is

$\text y = \text y_0 \sin 2\pi\left(\text n \text t - \frac{\text x}{\lambda}\right)$

and the maximum particle velocity is four times the wave velocity.

Particle velocity : Differentiating y with respect to t, keeping x constant,

$$\begin{aligned}\text u = \frac{\text{dy}}{\text{dt}} = \frac{\text d}{\text{dt}}\left[\text y_0 \sin 2\pi\left(\text{nt} - \frac{\text x}{\lambda}\right)\right] \\[1em] = \bigg\lbrace\text y_0 \cos 2\pi\left(\text{nt} - \frac{\text x}{\lambda}\right)\bigg\rbrace(2\pi \text n)\end{aligned}$$

The maximum value of the cosine term is 1, so the maximum particle velocity is

$\text u_{max} = \text y_0 (2\pi \text n)$

Wave velocity :

$\text v = \text n \lambda$

It is given that u_max = 4 v, so

$\text y_0 (2\pi \text n) = 4\text n \lambda$

$\lambda = \frac{2\pi \text y_0}{4} = \frac{\pi \text y_0}{2}$

Hence, the wavelength of the wave is $\frac{\pi \text y_0}{2}$.

The equation of a stationary wave is

$\text y = 2\text a \cos \frac{2\pi \text x}{\lambda}\sin \frac{2\pi \text t}{\text T}$

where a is the amplitude of each of the two component progressive waves, λ their wavelength and T their time-period.

Positions of the antinodes : Substituting x = 0, $\frac{\lambda}{2}$, $\frac{2\lambda}{2}$, $\frac{3\lambda}{2}$, ... in the equation, the value of $\cos \frac{2\pi \text x}{\lambda}$ becomes alternately + 1 and − 1. At these points the displacement y is always maximum in comparison with the other points. These points are the antinodes, and they are separated from each other by a distance $\frac{\lambda}{2}$.

Positions of the nodes : Substituting x = $\frac{\lambda}{4}$, $\frac{3\lambda}{4}$, $\frac{5\lambda}{4}$, ... the value of $\cos \frac{2\pi \text x}{\lambda}$ becomes zero. At these points the displacement y is always zero. These points are the nodes, and they too are separated from each other by a distance $\frac{\lambda}{2}$.

Given, the amplitude of the incident wave = 0.5 cm, and the wave is incident on a hard (rigid) surface.

(i) Reflected wave : In the absence of the hard surface the incident wave would have advanced further. On reflection at a rigid surface, however, the wave suffers a phase change of 180°, that is, the directions of displacement of the particles in the reflected wave become inverted with respect to the incident wave.

Hence the reflected wave is a similar curve of amplitude 0.5 cm, but turned upside down with respect to the incident wave.

(ii) Resultant wave : For the figure drawn, the displacements of the incident wave and of the reflected wave are similar at all points. Hence, by the principle of superposition, the resultant wave obtained by adding the two displacements algebraically is also a similar curve, but with an amplitude of

$0.5\ \text{cm} + 0.5\ \text{cm} = 1.0\ \text{cm}$

Hence, the reflected wave has an amplitude of 0.5 cm and is inverted with respect to the incident wave, while the resultant wave is a similar curve of amplitude 1.0 cm.

A.

(i) the tension in the string is increased.

The speed of a wave on a string is $\text v = \sqrt{\frac{\text T}{\text m}}$. Hence the speed increases when the tension T is increased, and it decreases when the mass per unit length m is increased. The frequency and the wavelength of the wave have no effect on the speed, since the speed is decided by the properties of the medium alone.

(ii) The particles in solids are closer together.

In a solid the particles are closely packed and are bound by strong elastic forces, so a disturbance is handed on from one particle to the next very rapidly. Hence the energy is transferred more quickly and the speed of sound in a solid is greater than in a gas.

(iii) Speed increases as temperature increases.

The speed of sound in a gas is $\text v = \sqrt{\frac{\gamma \text{RT}}{\text M}}$, so that $\text v \propto \sqrt{\text T}$. At a higher temperature the molecules move faster and the rate of energy transfer between them increases, so the speed of sound increases.

(iv) properties of the medium.

The speed of sound in a medium is $\text v = \sqrt{\frac{\text E}{\text d}}$, where E is the modulus of elasticity of the medium and d its density. The frequency, the amplitude and the wavelength of the wave do not enter into this expression.

(v) increases.

Since $\text v \propto \sqrt{\text T}$, a rise in the temperature of the air raises the speed of sound. Numerically, v_t = (332 + 0.61 t) m s^-1, so the speed increases by about 0.61 m s^-1 for each °C rise.

B.

(i) Effect of pressure of gas on the speed of longitudinal waves in a gas :

The speed of longitudinal waves in a gas is

$\text v = \sqrt{\frac{\gamma \text P}{\text d}} \qquad \dots(\text i)$

From the ideal gas equation for one mole,

$\text{PV} = \text{RT}$

If M be the molecular weight and d the density of the gas, then $\text V = \frac{\text M}{\text d}$, so

$\text P \times \frac{\text M}{\text d} = \text{RT} \qquad \Rightarrow \qquad \frac{\text P}{\text d} = \frac{\text{RT}}{\text M}$

At a constant temperature, $\frac{\text P}{\text d} = \text{constant}$. Hence, if P changes, d also changes in the same proportion and the ratio $\frac{\text P}{\text d}$ does not change.

Hence, at a constant temperature there is no effect of pressure on the speed of longitudinal waves in a gas.

(ii) Effect of temperature of gas on the speed of longitudinal waves in a gas :

Substituting $\frac{\text P}{\text d} = \frac{\text{RT}}{\text M}$ in equation (i),

$\text v = \sqrt{\frac{\gamma \text{RT}}{\text M}} \qquad \Rightarrow \qquad \text v \propto \sqrt{\text T}$

Hence, the speed of longitudinal waves in a gas is directly proportional to the square-root of its absolute temperature, so it increases with a rise in temperature.

(iii) Given,

• Initial temperature, t₁ = 97°C, so T₁ = 97 + 273 = 370 K
• Speed at this temperature, v₁ = v (say)
• Required speed, v₂ = 2 v

Since $\text v \propto \sqrt{\text T}$,

$\frac{\text v_1}{\text v_2} = \sqrt{\frac{\text T_1}{\text T_2}}$

Substituting the values,

$\frac{\text v}{2\text v} = \sqrt{\frac{370}{\text T_2}} \qquad \Rightarrow \qquad \frac{1}{2} = \sqrt{\frac{370}{\text T_2}}$

Squaring both sides,

$\frac{1}{4} = \frac{370}{\text T_2}$

$\text T_2 = 4 \times 370 = 1480\ \text K$

Converting into the celsius scale,

$\text t_2 = 1480 - 273$

$\text t_2 = 1207^\circ \text C$

Hence, the speed of sound in air will be double of its speed at 97°C at a temperature of 1207°C.

(i) Transverse wave.

In a transverse wave the particles of the medium oscillate along a direction perpendicular to the direction of propagation of the wave, and the wave travels in the form of crests and troughs. A wave on a stretched string is a common example.

(ii) 300 m/s.

Given, λ = 1.5 m and n = 200 Hz. From the relation between speed, frequency and wavelength,

$\text v = \text n \lambda = 200 \times 1.5$

$\text v = 300\ \text{m/s}$

(iii) Sound wave in air.

Sound waves in air are longitudinal waves, in which the air particles oscillate back and forth along the same direction in which the wave advances, producing compressions and rarefactions. Light waves are transverse electromagnetic waves, and a wave on a string is a transverse mechanical wave.

(iv) the speed and wavelength.

The three quantities are related by v = nλ, so that

$\text n = \frac{\text v}{\lambda}$

Thus the frequency is fixed by the speed of the wave and its wavelength. The amplitude has no bearing on the frequency.

(v) Particle motion is parallel to wave direction.

This is the defining property of a longitudinal wave. Such waves do not require a vacuum — on the contrary they need a material medium — and nodes and antinodes belong to stationary waves, not to progressive longitudinal waves.

(i) They create a wave of larger amplitude.

When two waves of the same frequency and amplitude meet in phase, the phase difference between them is φ = 0. The resultant amplitude is then

$\text a_{max} = \text a_1 + \text a_2 = 2\text a$

which is larger than the amplitude of either wave. This is constructive interference.

(ii) Beats.

When two waves of slightly different frequencies interfere, the phase difference between them at any point varies with time, so the intensity of the resultant sound rises and falls alternately. This periodic variation in amplitude is called the phenomenon of beats, and the beat frequency is n₁ ~ n₂.

(iii) a wave with decreased amplitude.

In destructive interference the two waves meet in opposite phases, that is, φ = π, and the resultant amplitude is

$\text a_{min} = \text a_1 \sim \text a_2$

which is smaller than the amplitude of either wave. It becomes zero only when the two amplitudes are exactly equal.

(iv) The principle that wave displacement is the sum of individual displacements.

According to the principle of superposition, when two or more waves overlap at a point, the resultant displacement at that point is the algebraic sum of the displacements produced by the individual waves,

$\text y = \text y_1 + \text y_2$

(v) greater than the original amplitudes.

In constructive interference the waves add together, so the resultant amplitude a₁ + a₂ is greater than either of the original amplitudes. Since I ∝ a², the intensity at such a place is maximum.

A.

(i) the displacement is minimum.

The nodes are the points of a stationary wave at which the particles of the medium remain permanently at rest, so their displacement is always zero, that is, minimum. Destructive interference occurs at these points.

(ii) Two waves of the same frequency travelling in opposite directions must interfere.

For the formation of a stationary wave, two identical waves of the same frequency, wavelength and speed must travel in a bounded medium in opposite directions and superpose. A bounded medium is essential, since the wave must be reflected at the boundary to produce the oppositely travelling wave.

(iii) A point of maximum displacement.

The antinodes are the points at which the particles of the medium undergo maximum displacement in comparison with the others. Constructive interference occurs at these points, and the amplitude there is 2a.

(iv) The length, tension, and mass per unit length of the string.

The fundamental frequency of a string fixed at both ends is

$\text n = \frac{1}{2\text l}\sqrt{\frac{\text T}{\text m}}$

so it is decided by the length l, the tension T and the mass per unit length m of the string.

(v) A stationary wave does not transfer energy, while a progressive wave does.

A stationary wave remains steady between the two boundaries of the medium and does not advance, so it does not transmit energy across the medium. A progressive wave, on the other hand, advances with a definite velocity and carries energy from one point to another.

B.

(i) In a stationary wave the nodes are situated at x = 0, $\frac{\lambda}{2}$, $\frac{2\lambda}{2}$, $\frac{3\lambda}{2}$, ...

Hence, the distance between two successive nodes is $\frac{\lambda}{2}$.

(ii) The antinodes lie exactly midway between two successive nodes.

Hence, the distance between consecutive nodes and antinodes is $\frac{\lambda}{4}$.

(iii) Given,

• Length of the closed organ pipe, l = 0.80 m
• Speed of sound in air, v = 330 m/s

In a closed organ pipe a node is formed at the closed end and an antinode at the open end, so in the first mode of vibration

$\text l = \frac{\lambda}{4} \qquad \Rightarrow \qquad \lambda = 4\text l$

The frequency of the fundamental tone is

$\text n = \frac{\text v}{\lambda} = \frac{\text v}{4\text l}$

Substituting the values,

$\text n = \frac{330\ \text{m/s}}{4 \times 0.80\ \text m} = \frac{330}{3.2}$

$\text n = 103.12\ \text{Hz}$

Hence, the frequency of the fundamental tone of the closed organ pipe is 103.12 Hz.

(i) have slightly different frequencies superpose.

Beats are produced by the superposition of two sound waves of nearly equal frequencies. If the frequencies were exactly equal, the phase difference at any point would remain steady and the intensity of sound would remain constant, so no beats would be heard.

(ii) the difference between the two frequencies.

The number of times the intensity of sound rises and falls in one second is equal to the difference in the frequencies of the two sound-sources, that is, beat frequency = n₁ ~ n₂.

(iii) 5 Hz.

Given, n₁ = 355 Hz and n₂ = 350 Hz. Hence

$\text{Beat frequency} = \text n_1 - \text n_2 = 355 - 350$

$= 5\ \text{Hz}$

(iv) 196 Hz.

Given, fifty tuning forks are arranged in the order of increasing frequency, any two successive forks giving 4 beats per second, and the frequency of the 50^th fork is one octave higher than that of the 1^st fork.

Let the frequency of the first fork be n. Since each successive fork is 4 Hz higher,

• Frequency of the second fork = n + 4 = n + 1 × 4
• Frequency of the third fork = n + 8 = n + 2 × 4
• Frequency of the fourth fork = n + 12 = n + 3 × 4

Continuing in this way, the frequency of the 50^th fork is

$\text n + 49 \times 4$

One octave higher means double the frequency, so the frequency of the 50^th fork is also 2n. Therefore,

$\text n + 49 \times 4 = 2\text n$

$\text n = 196\ \text{Hz}$

(v) tune musical instruments.

Musicians sound two instruments together and adjust the frequency of one until the number of beats per second goes on decreasing and finally the beats disappear. When no beats are heard, the frequencies of the two instruments are exactly equal.

(i) A wire clamped at both its ends and vibrating in two loops is vibrating in its second mode, that is, p = 2. In this mode

$\text l = \frac{2\lambda}{2} = \lambda$

Hence the frequency of vibration of the string is

$\text n = \frac{\text v}{\lambda} = \frac{\text v}{\text l} = \frac{1}{\text l}\sqrt{\frac{\text T}{\text m}}$

Hence, the formula for the frequency of vibration in two loops is $\text n = \frac{1}{\text l}\sqrt{\frac{\text T}{\text m}}$, which is twice the fundamental frequency.

(ii) Given,

• Density of the material of the wire, d = 4.8 × 10³ kg/m³
• Stress in the wire = 3 × 10⁸ N/m²

The speed of a transverse wave in a stretched wire is

$\text v = \sqrt{\frac{\text T}{\text m}} = \sqrt{\frac{\text T}{\pi \text r^2 \text d}} = \sqrt{\frac{\text{stress}}{\text{density}}}$

since $\frac{\text T}{\pi \text r^2}$ is the tension per unit area of cross-section, that is, the stress.

Substituting the values,

$\text v = \sqrt{\frac{3 \times 10^8\ \text{N m}^{-2}}{4.8 \times 10^3\ \text{kg m}^{-3}}} = \sqrt{6.25 \times 10^4}$

$\text v = 250\ \text{m/s}$

Hence, the speed of transverse waves in the wire is 250 m/s.

(iii) Given,

• Length of the stretched wire, l = 2 m
• Fundamental frequency, n = 400 Hz

The fundamental frequency of a stretched wire is

$\text n = \frac{\text v}{2\text l} \qquad \Rightarrow \qquad \text v = \text n \times 2\text l$

Substituting the values,

$\text v = 400 \times 2 \times 2$

$\text v = 1600\ \text{Hz}\ \text{m} = 1600\ \text{m/s}$

Hence, the speed of the transverse waves in the wire is 1600 m/s.

Characteristics of a harmonic wave :

(i) The wave is produced by a source executing simple harmonic motion, so every particle of the medium also executes simple harmonic motion about its own mean position.

(ii) All the particles of the medium oscillate with the same amplitude and the same frequency.

(iii) The phase of oscillation changes continuously from one particle to the next, so that at any instant all the particles are in different phases of their oscillation.

(iv) The wave advances in the medium with a constant velocity, which is decided by the properties of the medium.

(v) The wave carries energy from one point of the medium to another without any permanent movement of matter.

Derivation of the relation :

Suppose a simple harmonic progressive wave is advancing in a medium along the positive direction of the X-axis. Let the time be counted from the instant when the particle at the origin O passes through the mean position in the positive direction of the Y-axis. Then the displacement y of the particle at O at any time t is

$\text y = \text a \sin \omega \text t \qquad \dots(\text i)$

where a is the amplitude and ω is the angular frequency of the S.H.M. executed by the particle.

In a wave motion the successive particles start oscillating a definite time later than the preceding particle. Hence, as we move away from O, the phase lag of the oscillation of a particle with respect to the particle at O goes on increasing. If φ be the phase lag of a particle P distant x from the origin O, then the displacement of P at any instant t is

$\text y = \text a \sin (\omega \text t - \phi) \qquad \dots(\text{ii})$

For a distance λ, the phase-change is 2π radian. Hence the phase-change for a distance x will be

$\phi = \frac{2\pi}{\lambda}\text x = \text k\text x \qquad \dots(\text{iii})$

where $\text k = \frac{2\pi}{\lambda}$ is called the propagation constant.

Substituting equation (iii) in equation (ii),

$\text y = \text a \sin (\omega \text t - \text k\text x) \qquad \dots(\text{iv})$

Now $\omega = \frac{2\pi}{\text T}$, where T is the period of an oscillation of the particles, and $\text k = \frac{2\pi}{\lambda}$. Substituting these values in equation (iv),

$\text y = \text a \sin\left(\frac{2\pi}{\text T}\text t - \frac{2\pi}{\lambda}\text x\right)$

$\text y = \text a \sin 2\pi\left(\frac{\text t}{\text T} - \frac{\text x}{\lambda}\right)$

Meaning of the symbols :

• y — the displacement, at the instant t, of the particle situated at a distance x from the origin
• a — the amplitude of the wave
• T — the time-period of the oscillation
• λ — the wavelength of the wave
• x — the distance of the particle from the origin
• t — the instant of time

If the wave is travelling along the − X direction, the minus sign inside the bracket is replaced by a plus sign,

$\text y = \text a \sin 2\pi\left(\frac{\text t}{\text T} + \frac{\text x}{\lambda}\right)$

Wave velocity : Wave velocity, or propagation velocity, is the speed with which the wavefronts, or the energy of the wave, travels through the medium. It depends upon the properties of the medium and the type of the wave, and is given by

$\text v = \text n \lambda$

Particle velocity : Particle velocity refers to the velocity of the individual particles of the medium as they oscillate due to a wave passing through them. It is the rate of change of the displacement of a particle with time.

Variation with time :

The equation of a travelling wave along the positive X-axis is

$\text y = \text a \sin 2\pi\left(\frac{\text t}{\text T} - \frac{\text x}{\lambda}\right) \qquad \dots(\text i)$

Differentiating equation (i) with respect to t, keeping x constant, the instantaneous particle velocity is

$$\begin{aligned}\text u = \frac{\text{dy}}{\text{dt}} = \frac{\text d}{\text{dt}}\left[\text a \sin 2\pi\left(\frac{\text t}{\text T} - \frac{\text x}{\lambda}\right)\right] \\[1em] = \frac{2\pi \text a}{\text T}\cos 2\pi\left(\frac{\text t}{\text T} - \frac{\text x}{\lambda}\right) \qquad \dots(\text{ii})\end{aligned}$$

The particle velocity therefore changes harmonically with time and with distance, whereas the wave velocity v remains constant. From equation (ii) the particle velocity is maximum when the cosine term is ± 1, so

$\text u_{max} = \frac{2\pi \text a}{\text T} = 2\pi \text n\text a = \omega \text a$

There is a phase difference of $\frac{\pi}{2}$ between the displacement and the particle velocity.

Relation between particle and wave velocity :

Differentiating equation (i) with respect to x, keeping t constant,

$\frac{\text{dy}}{\text{dx}} = -\frac{2\pi}{\lambda}\text a \cos 2\pi\left(\frac{\text t}{\text T} - \frac{\text x}{\lambda}\right) \qquad \dots(\text{iii})$

Dividing equation (ii) by equation (iii),

$\frac{\frac{\text{dy}}{\text{dt}}}{\frac{\text{dy}}{\text{dx}}} = \frac{\frac{2\pi \text a}{\text T}\cos 2\pi\left(\frac{\text t}{\text T} - \frac{\text x}{\lambda}\right)}{-\frac{2\pi}{\lambda}\text a \cos 2\pi\left(\frac{\text t}{\text T} - \frac{\text x}{\lambda}\right)} = -\frac{\lambda}{\text T} = -\text v$

$\text u = \frac{\text{dy}}{\text{dt}} = -\text v\frac{\text{dy}}{\text{dx}}$

Hence, particle velocity = − wave velocity × slope of the displacement curve.

Adiabatic process in the propagation of a wave :

When longitudinal waves travel in a gaseous medium, the states of compression and rarefaction occur alternately at every point. At the time of compression some heat is developed at the point, and at the time of rarefaction some heat is lost from the medium at that point.

The compressions and rarefactions occur so rapidly that the heat produced during compression cannot go out into the surroundings, and the heat which disappeared during rarefaction cannot come in from the surroundings. Besides this, gases are bad conductors of heat, so the exchange of heat does not take place at all.

Hence, during the propagation of sound waves, the temperature at a point of the medium rises during compression and falls during rarefaction, that is, temperature-changes take place in the gaseous medium. Therefore the process is adiabatic and not isothermal, as Newton had assumed.

Adiabatic elasticity of a gas :

For an adiabatic change in a gas,

$\text{PV}^\gamma = \text{constant}$

where $\gamma = \frac{\text C_\text p}{\text C_\text v}$ is the ratio of the specific heat of the gas at constant pressure to that at constant volume.

Differentiating both sides,

$\text V^\gamma \text{dP} + \text P \times \gamma \text V^{\gamma - 1}\text{dV} = 0$

Dividing throughout by V^{γ - 1},

$\text V\text{dP} + \gamma \text P\text{dV} = 0$

$\text{dP} = -\gamma \text P\frac{\text{dV}}{\text V}$

By definition, the elasticity of a gas is

$\text E = \frac{\text{normal stress}}{\text{volume strain}} = \frac{-\text{dP}}{\left(\frac{\text{dV}}{\text V}\right)}$

Substituting the value of dP,

$\text E_{adiabatic} = \gamma \text P$

Speed of sound in a gas :

The speed of a longitudinal wave in a medium of elasticity E and density ρ is

$\text v = \sqrt{\frac{\text E}{\rho}}$

Substituting the adiabatic elasticity E = γP,

$\text v = \sqrt{\frac{\gamma \text P}{\rho}}$

where P is the pressure of the gas and ρ is its density. This is Laplace's formula.

For air at 0°C, taking γ = 1.41, P = 1.01 × 10⁵ N m^-2 and ρ = 1.29 kg m^-3,

$\text v = \sqrt{1.41} \times 280 = 332\ \text{m s}^{-1}$

which agrees closely with the experimental value.

Superposition of sound waves : When two or more waves travel through the same medium, each individual wave behaves as though it were travelling alone. The resultant wave that we observe in the medium is the sum of the displacements of all the individual waves at any point. Even after the superposition, each wave retains its own nature — its amplitude, phase, frequency and direction — and continues its journey as though nothing had happened.

Mathematically, if y₁ and y₂ are the displacements produced by two waves at a point, then the resultant displacement is

$\text y = \text y_1 + \text y_2 \qquad \dots(\text i)$

Application to interference of waves :

Let the displacements at a point due to the arrival of two waves of the same frequency $\frac{\omega}{2\pi}$ and amplitudes a₁ and a₂ be

$\text y_1 = \text a_1 \sin (\omega \text t - \text k\text x) \qquad \text{and} \qquad \text y_2 = \text a_2 \sin (\omega \text t - \text k\text x + \phi)$

where the second wave is ahead of the first by a phase angle φ.

According to the superposition principle,

$\text y = \text y_1 + \text y_2 = \text a_1 \sin (\omega \text t - \text k\text x) + \text a_2 \sin (\omega \text t - \text k\text x + \phi)$

Expanding the second term,

$\text y = \text a_1 \sin (\omega \text t - \text k\text x) + \text a_2 \sin (\omega \text t - \text k\text x)\cos \phi + \text a_2 \cos (\omega \text t - \text k\text x)\sin \phi$

$\text y = \left[\text a_1 + \text a_2 \cos \phi\right]\sin (\omega \text t - \text k\text x) + \left[\text a_2 \sin \phi\right]\cos (\omega \text t - \text k\text x) \qquad \dots(\text{ii})$

Let us put

$\text a_1 + \text a_2 \cos \phi = \text a \cos \theta \qquad \dots(\text{iii})$

$\text a_2 \sin \phi = \text a \sin \theta \qquad \dots(\text{iv})$

From equations (ii), (iii) and (iv),

$\text y = \text a \sin (\omega \text t + \theta) \qquad \dots(\text v)$

This shows that the resultant wave has the same frequency as the component waves.

Amplitude of the resultant wave : Squaring and adding equations (iii) and (iv),

$\text a^2 \cos^2 \theta + \text a^2 \sin^2 \theta = (\text a_1 + \text a_2 \cos \phi)^2 + \text a_2^2 \sin^2 \phi$

$\text a^2 = \text a_1^2 + \text a_2^2 \cos^2 \phi + 2\text a_1 \text a_2 \cos \phi + \text a_2^2 \sin^2 \phi$

$\text a = \sqrt{\text a_1^2 + \text a_2^2 + 2\text a_1 \text a_2 \cos \phi} \qquad \dots(\text{vi})$

Maximum intensity : The intensity of a wave is directly proportional to the square of its amplitude, that is, I ∝ a².

At places where φ = 0, that is, where the two waves arrive in the same phase, cos φ = + 1 and equation (vi) gives the maximum amplitude,

$\text a_{max} = \sqrt{\text a_1^2 + \text a_2^2 + 2\text a_1 \text a_2} = \sqrt{(\text a_1 + \text a_2)^2} = \text a_1 + \text a_2$

Hence the maximum intensity is

$\text I_{max} = (\text a_1 + \text a_2)^2$

Similarly, at places where φ = π, cos φ = − 1 and the amplitude is minimum,

$\text a_{min} = \text a_1 \sim \text a_2 \qquad \Rightarrow \qquad \text I_{min} = (\text a_1 \sim \text a_2)^2$

so that

$\frac{\text I_{max}}{\text I_{min}} = \frac{(\text a_1 + \text a_2)^2}{(\text a_1 \sim \text a_2)^2}$

Beats : When two sound waves of nearly equal frequencies are produced simultaneously, the intensity of the resultant sound produced by their superposition increases and decreases alternately with time. This rise and fall in the intensity of sound is called the phenomenon of beats. One rise and one fall together form one beat, and the number of times the intensity of sound rises and falls in one second is called the beat frequency.

Analytical proof :

Let us consider two harmonic waves of the same amplitude a but of slightly different frequencies n₁ and n₂, where ω₁ = 2πn₁ and ω₂ = 2πn₂, moving with the same velocity in the same direction. Their wave functions are

$\text y_1 = \text a \sin (\omega_1 \text t - \text k\text x) \qquad \text{and} \qquad \text y_2 = \text a \sin (\omega_2 \text t - \text k\text x)$

Let the observation point be located at x = 0. Then

$\text y_1 = \text a \sin \omega_1 \text t = \text a \sin 2\pi \text n_1 \text t$

$\text y_2 = \text a \sin \omega_2 \text t = \text a \sin 2\pi \text n_2 \text t$

By the principle of superposition, the resultant displacement of the particle is

$\text y = \text y_1 + \text y_2 = \text a \sin 2\pi \text n_1 \text t + \text a \sin 2\pi \text n_2 \text t$

Using $\sin \text C + \sin \text D = 2 \sin \frac{\text C + \text D}{2}\cos \frac{\text C - \text D}{2}$,

$$\begin{aligned}\text y = 2\text a \sin \left[\pi(\text n_1 + \text n_2)\text t\right]\cos \left[\pi(\text n_1 - \text n_2)\text t\right] \\[1em] = 2\text a \cos \left[\pi(\text n_1 - \text n_2)\text t\right]\sin \left[\pi(\text n_1 + \text n_2)\text t\right]\end{aligned}$$

Putting $2\text a \cos \lbrace\pi(\text n_1 - \text n_2)\text t\rbrace = \text A$,

$\text y = \text A \sin \left[\pi(\text n_1 + \text n_2)\text t\right]$

Thus the particle vibrates in simple harmonic motion with amplitude A. But the value of A depends upon t, so the amplitude is not constant but varies periodically with time.

For maximum amplitude :

$\cos \pi(\text n_1 - \text n_2)\text t = \pm 1 \qquad \Rightarrow \qquad \pi(\text n_1 - \text n_2)\text t = \text k\pi$

$\text t = \frac{\text k}{\text n_1 - \text n_2}, \qquad \text k = 0, 1, 2, 3, \dots$

Substituting k = 0, 1, 2, 3, ...,

$\text t = 0,\ \frac{1}{\text n_1 - \text n_2},\ \frac{2}{\text n_1 - \text n_2},\ \frac{3}{\text n_1 - \text n_2},\ \dots$

Hence the time-interval between two consecutive maximum intensities is $\frac{1}{\text n_1 - \text n_2}$ second, so in one second the intensity will be maximum (n₁ − n₂) times.

For minimum amplitude :

$\cos \pi(\text n_1 - \text n_2)\text t = 0 \qquad \Rightarrow \qquad \pi(\text n_1 - \text n_2)\text t = \frac{\text k\pi}{2}$

$\text t = \frac{\text k}{2(\text n_1 - \text n_2)}, \qquad \text k = 1, 3, 5, \dots$

Substituting k = 1, 3, 5, ...,

$\text t = \frac{1}{2(\text n_1 - \text n_2)},\ \frac{3}{2(\text n_1 - \text n_2)},\ \frac{5}{2(\text n_1 - \text n_2)},\ \dots$

Hence the time-interval between two consecutive minimum intensities is also $\frac{1}{\text n_1 - \text n_2}$ second, so in one second the intensity will be minimum (n₁ − n₂) times, the minima occurring in between the maxima.

Thus, in one second the intensity of sound will be (n₁ − n₂) times maximum and (n₁ − n₂) times minimum, that is, (n₁ − n₂) beats will be heard in 1 second.

Hence, the number of beats per second, that is, the beat frequency, is equal to the difference in the frequencies of the two sound-sources, n₁ ~ n₂.

Waves produced in a bounded medium :

A bounded medium is one which has definite boundaries and whose boundaries are separated from other media by distinct surfaces. Such a medium can vibrate with only certain definite frequencies, and these frequencies are the characteristic frequencies of that medium. The string of a sitar, the air column of a flute and the membrane of a tabla are all bounded media, and when any of these is sounded, notes of only certain definite frequencies are produced from them.

The boundary of a bounded medium may be of two types — rigid (or closed) and free (or open). When a wave travels in a medium having no boundary, it continues to travel as such. If the medium has a boundary, then the wave is reflected from the boundary, and the superposition of the incident and the reflected waves gives rise to a stationary wave.

Conditions for the formation of stationary waves :

(i) The medium must be a bounded medium, so that the wave is reflected at the boundary and produces a wave of the same kind travelling in the opposite direction.

(ii) The two superposing waves must be of the same kind, that is, both transverse or both longitudinal.

(iii) The two waves must have the same frequency and the same speed.

(iv) The amplitudes of the two waves should be equal or nearly equal.

Reflection from a free end :

Suppose a transverse progressive wave is sent along a string whose one end is free to move perpendicular to the length of the string. The wave is reflected from this end, but in the reflected wave the directions of displacements of the particles of the string remain the same as in the incident wave.

When the pulse of the incident wave reaches the free end, there is no change in its phase and the wave returns in the same way as it would have advanced in the absence of the reflecting surface. In the figure, AB is the incident wave and BC is the reflected wave.

Hence, the wave does not suffer any phase change on reflection from a 'free' end.

Reflection from a rigid end :

When a transverse progressive wave is sent along a string whose one end is tied to a rigid surface, then in the reflected wave the directions of displacements of the particles of the string become inverted with respect to the incident wave.

In the figure, AB is the incident wave and BC is the reflected wave, and the reflected curve is seen to be turned upside down with respect to the incident one.

Hence, the wave undergoes a phase change of 180° or π on reflection from a 'rigid' end.

Stationary wave : When two identical transverse, or longitudinal, progressive waves travel in a bounded medium with the same speed but in opposite directions, then by their superposition a new type of wave is produced which appears stationary in the medium. This wave is called a stationary (or standing) wave.

The characteristic of a stationary wave is that some particles of the medium remain permanently at rest, while some other particles undergo maximum displacement compared to others. The former are called the nodes and the latter the antinodes.

Derivation of the equation :

Suppose a plane progressive wave of amplitude a is travelling along the positive direction of the X-axis. The equation of this wave is

$\text y_1 = \text a \sin (\omega \text t - \text k\text x)$

where ω is the angular frequency of the wave and k is the propagation constant.

Suppose this wave is reflected from a free boundary and the reflected wave advances in the negative direction of the X-axis. Since there is no phase change on reflection from a free end, the equation of the reflected wave is

$\text y_2 = \text a \sin (\omega \text t + \text k\text x)$

Let y₁ and y₂ be the displacements at a point x at any instant t due to the incident and the reflected waves respectively. Then, by the principle of superposition, the resultant displacement at that point is

$\text y = \text y_1 + \text y_2 = \text a\left[\sin (\omega \text t - \text k\text x) + \sin (\omega \text t + \text k\text x)\right]$

Using $\sin \text C + \sin \text D = 2 \sin \frac{\text C + \text D}{2}\cos \frac{\text C - \text D}{2}$,

$\text y = \text a\left[2 \sin \omega \text t \cos \text k\text x\right] = 2\text a \cos \text k\text x \sin \omega \text t$

Substituting $\text k = \frac{2\pi}{\lambda}$ and $\omega = \frac{2\pi}{\text T}$,

$\text y = 2\text a \cos \left(\frac{2\pi \text x}{\lambda}\right)\sin \left(\frac{2\pi \text t}{\text T}\right)$

which may be written as

$\text y = \text a \sin \frac{2\pi \text t}{\text T}\cos \frac{2\pi \text x}{\lambda}$

with 2a replaced by a for brevity. This is the equation of a stationary wave.

Nodes and antinodes : Substituting x = 0, $\frac{\lambda}{2}$, $\frac{2\lambda}{2}$, $\frac{3\lambda}{2}$, ... in this equation, the value of $\cos \left(\frac{2\pi \text x}{\lambda}\right)$ becomes alternately + 1 and − 1. At these points the displacement y is always maximum in comparison with the other points. These points are the antinodes, and they are separated from each other by a distance $\frac{\lambda}{2}$.

Similarly, substituting x = $\frac{\lambda}{4}$, $\frac{3\lambda}{4}$, $\frac{5\lambda}{4}$, ... the value of $\cos \left(\frac{2\pi \text x}{\lambda}\right)$ becomes zero, so the displacement y becomes zero. These points are the nodes, and these are also separated from each other by a distance $\frac{\lambda}{2}$.

Formation of standing waves in a closed organ pipe :

Let us consider a closed organ pipe of length l lying along the X-axis, the closed end being at x = 0 and the open end at x = l. When we blow air at its open end, a longitudinal (sound) wave travels in the air of the pipe towards the closed end, which reflects it back towards the open end. The incident and the reflected waves travelling in the air-column in opposite directions superpose and form stationary longitudinal waves.

At the closed end of the pipe the particles of air have no freedom to vibrate. Hence there is always a node at the closed end. At the open end the air particles have the greatest freedom to vibrate. Hence there is always an antinode at the open end.

The incident wave moving along the negative direction of the X-axis is represented by

$\text y_1 = \text a \sin (\omega \text t + \text k\text x)$

The wave reflected at the closed end suffers a phase change of π and moves along the positive direction of the X-axis, so it is represented by

$\text y_2 = \text a \sin (\omega \text t - \text k\text x + \pi) = -\text a \sin (\omega \text t - \text k\text x)$

On superposition the resultant displacement is

$$\begin{aligned}\text y = \text y_1 + \text y_2 = \text a\left[\sin (\omega \text t + \text k\text x) - \sin (\omega \text t - \text k\text x)\right] \\[1em] = 2\text a \cos \omega \text t \sin \text k\text x \qquad \dots(\text i)\end{aligned}$$

At the closed end (x = 0), equation (i) gives y = 0 for all values of t, that is, a node is formed there.

At the open end (x = l),

$\text y = 2\text a \cos \omega \text t \sin \text k\text l = 2\text a \cos \omega \text t \sin \frac{2\pi \text l}{\lambda}$

The displacement y will always be maximum if

$\sin \left(\frac{2\pi \text l}{\lambda}\right) = \pm 1 \qquad \Rightarrow \qquad \frac{2\pi \text l}{\lambda} = (2\text m - 1)\frac{\pi}{2}$

$\lambda = \frac{4\text l}{2\text m - 1}, \qquad \text m = 1, 2, 3, \dots \qquad \dots(\text{ii})$

First mode of vibration (m = 1) :

$\lambda_1 = 4\text l \qquad \Rightarrow \qquad \text n_1 = \frac{\text v}{\lambda_1} = \frac{\text v}{4\text l}$

This is the lowest frequency of vibration and is called the fundamental frequency. The note produced is the fundamental tone or the first harmonic.

Second mode of vibration (m = 2) :

$\lambda_2 = \frac{4\text l}{3} \qquad \Rightarrow \qquad \text n_2 = \frac{\text v}{\lambda_2} = \frac{3\text v}{4\text l} = 3\text n_1$

The note produced is called the first overtone or the third harmonic.

Third mode of vibration (m = 3) :

$\lambda_3 = \frac{4\text l}{5} \qquad \Rightarrow \qquad \text n_3 = \frac{\text v}{\lambda_3} = \frac{5\text v}{4\text l} = 5\text n_1$

The note produced is called the second overtone or the fifth harmonic.

In general, the frequency of the mth mode of vibration is

$\text n_\text m = \frac{\text v}{\lambda_\text m} = \frac{(2\text m - 1)\text v}{4\text l}$

Hence the frequencies of the fundamental tone and the overtones have the relationship

$\text n_1 : \text n_2 : \text n_3 \dots = 1 : 3 : 5 : 7 \dots$

Hence, a closed organ pipe produces only odd harmonics, in the ratio 1 : 3 : 5 : 7.

Different modes of vibration in air columns : The air column in an organ pipe can vibrate in a number of different ways, each way corresponding to a different stationary wave pattern with its own arrangement of nodes and antinodes. Each such pattern is called a mode of vibration, and each mode has a definite frequency of its own. The mode of lowest frequency is the fundamental mode, and the higher modes give the overtones.

Open organ pipe :

Let us consider a pipe of length l open at both ends, lying along the X-axis with ends at x = 0 and x = l. Since the pipe is open at both ends, the air particles at both the ends have the greatest freedom to vibrate, so there is an antinode at each end.

The incident and the reflected waves are

$\text y_1 = \text a \sin (\omega \text t + \text k\text x) \qquad \text{and} \qquad \text y_2 = \text a \sin (\omega \text t - \text k\text x)$

since reflection at an open end introduces no change in phase. On superposition,

$\text y = \text y_1 + \text y_2 = 2\text a \sin \omega \text t \cos \text k\text x$

At both the ends the resultant displacement should always be maximum. At x = l,

$\cos \left(\frac{2\pi \text l}{\lambda}\right) = \pm 1 \qquad \Rightarrow \qquad \frac{2\pi \text l}{\lambda} = \text m\pi$

$\lambda = \frac{2\text l}{\text m}, \qquad \text m = 1, 2, 3, \dots$

Hence the frequency of the mth mode is

$\text n_\text m = \frac{\text v}{\lambda_\text m} = \frac{\text m\text v}{2\text l}$

For m = 1, 2, 3, ... the frequencies are

$\text n_1 = \frac{\text v}{2\text l}, \qquad \text n_2 = \frac{2\text v}{2\text l} = 2\text n_1, \qquad \text n_3 = \frac{3\text v}{2\text l} = 3\text n_1$

so that

$\text n_1 : \text n_2 : \text n_3 \dots = 1 : 2 : 3 \dots$

Hence, an open pipe produces both the even and the odd harmonics.

Closed organ pipe :

In a pipe closed at one end there is always a node at the closed end and an antinode at the open end. On superposing the incident wave and the wave reflected at the closed end, which suffers a phase change of π,

$\text y = 2\text a \cos \omega \text t \sin \text k\text x$

At the open end x = l the displacement must always be maximum, so

$\sin \left(\frac{2\pi \text l}{\lambda}\right) = \pm 1 \qquad \Rightarrow \qquad \lambda = \frac{4\text l}{2\text m - 1}, \qquad \text m = 1, 2, 3, \dots$

Hence the frequency of the mth mode is

$\text n_\text m = \frac{(2\text m - 1)\text v}{4\text l}$

For m = 1, 2, 3, ... the frequencies are

$\text n_1 = \frac{\text v}{4\text l}, \qquad \text n_2 = \frac{3\text v}{4\text l} = 3\text n_1, \qquad \text n_3 = \frac{5\text v}{4\text l} = 5\text n_1$

so that

$\text n_1 : \text n_2 : \text n_3 \dots = 1 : 3 : 5 \dots$

Hence, in a closed organ pipe only the odd harmonics can be produced, the even harmonics being missing. This is why the musical sound produced by an open organ pipe is richer than that produced by a closed organ pipe.

Vibrations of a stretched string :

When a wire clamped to rigid supports at its ends is plucked in the middle, transverse progressive waves travel towards each end of the wire. These waves are reflected at the ends of the wire, and by the superposition of the incident and the reflected waves, transverse stationary waves are set up in the wire.

Since the ends of the wire are clamped, there is a node (N) at each end. In the simplest mode of vibration the wire vibrates as a whole in a single segment, with an antinode (A) in the middle.

Expression for the velocity of a transverse wave :

The speed of a transverse wave in a flexible stretched string depends upon the tension in the string and the mass per unit length of the string. It is given by

$\text v = \sqrt{\frac{\text T}{\text m}} \qquad \dots(\text i)$

where T is the tension in the wire and m is the mass per unit length of the wire, and not the mass of the whole wire.

If r be the radius of the wire and d the density of the material of the wire, then

$\text m = \text{volume per unit length} \times \text{density} = (\pi \text r^2 \times 1) \times \text d = \pi \text r^2 \text d$

so that the speed may also be written as

$\text v = \sqrt{\frac{\text T}{\pi \text r^2 \text d}} = \sqrt{\frac{\text{stress}}{\text{density}}}$

Expression for the fundamental frequency :

We know that the distance between two consecutive nodes is $\frac{\lambda}{2}$, where λ is the wavelength. Hence if l be the length of the wire between the clamped ends, then in the simplest mode

$\text l = \frac{\lambda}{2} \qquad \text{or} \qquad \lambda = 2\text l$

If n be the frequency of vibration of the wire, then

$\text n = \frac{\text v}{\lambda} = \frac{\text v}{2\text l}$

Substituting the value of v from equation (i),

$\text n = \frac{1}{2\text l}\sqrt{\frac{\text T}{\text m}} \qquad \dots(\text{ii})$

This is the fundamental frequency, or the first harmonic, of the stretched string, and the sound note produced by the string in this mode is called the fundamental tone.

Equation (ii) shows that the frequency of the sound emitted from a stretched string can be changed in two ways — by changing the length of the string, or by changing the tension in the string. In a sitar and a violin the frequencies of the notes are adjusted by tightening or loosening the pegs of the wires.

Fundamental frequency of an open organ pipe :

Let us consider a pipe of length l, open at both ends, lying along the X-axis with its ends at x = 0 and x = l. When we blow air at one end, a longitudinal (sound) wave travels in the air of the pipe towards the other end, which reflects it back. The incident and the reflected waves travelling in opposite directions superimpose and form stationary waves.

Since the pipe is open at both ends, the air particles at each end have the greatest freedom to vibrate, so there is an antinode at each end.

The incident and the reflected waves are represented by

$\text y_1 = \text a \sin (\omega \text t + \text k\text x) \qquad \text{and} \qquad \text y_2 = \text a \sin (\omega \text t - \text k\text x)$

since the reflection at an open end introduces no change in phase.

On superposition the resultant displacement is

$$\begin{aligned}\text y = \text y_1 + \text y_2 = \text a\left[\sin (\omega \text t + \text k\text x) + \sin (\omega \text t - \text k\text x)\right] \\[1em] = 2\text a \sin \omega \text t \cos \text k\text x \qquad \dots(\text i)\end{aligned}$$

At both the ends of the pipe the resultant displacement should always be maximum. At the end x = l,

$\text y = 2\text a \sin \omega \text t \cos \frac{2\pi \text l}{\lambda}$

The displacement y will always be maximum if

$\cos \left(\frac{2\pi \text l}{\lambda}\right) = \pm 1 \qquad \Rightarrow \qquad \frac{2\pi \text l}{\lambda} = \text m\pi$

$\lambda = \frac{2\text l}{\text m}, \qquad \text m = 1, 2, 3, \dots \qquad \dots(\text{ii})$

First mode of vibration : For m = 1, equation (ii) gives

$\lambda_1 = 2\text l \qquad \text{or} \qquad \text l = \frac{\lambda_1}{2}$

There are antinodes at the two ends and a node in the middle of the pipe. Hence the fundamental frequency is

$\text n_1 = \frac{\text v}{\lambda_1} = \frac{\text v}{2\text l}$

Which pipe gives the more musical sound :

In a closed organ pipe only the odd harmonics are present and the even harmonics are missing, whereas in an open organ pipe all the harmonics are present, the frequencies being in the ratio 1 : 2 : 3 : 4 ...

Hence, the musical sound produced by an open organ pipe is richer than that produced by a closed organ pipe, because a larger number of overtones blend with the fundamental tone in it.