(i) When a tuning fork [vibrating] is held close to ear, one hears a faint hum. The same [vibrating tuning fork] is held such that it's stem is in contact with the table surface, then one hears a loud sound. Explain.
(ii) A man standing in front of a vertical cliff fires a gun. He hears the echo after 3.5 seconds. On moving closer to the cliff by 84 m, he hears the echo after 3 seconds. Calculate the distance of the cliff from the initial position of the man.
Show SolutionHide Solution↓
(i) When the stem of the tuning fork is placed in contact with a table top, it sets the table surface in small vibrations but the large number of air molecules in contact with the table surface also start vibrating, thereby producing a loud sound.
(ii) Let the man be standing at distance d from the cliff and v be the speed of sound in air.
For first echo, t1 = $\frac{2d}{v}$ = 3.5 s [Eqn i ]
and for second echo, t2 = $\frac{2(d - 84)}{v}$ = 3 s [Eqn ii]
Dividing eqn (ii) by eqn (i)
$$\begin{aligned}\frac{\frac{2(d - 84)}{v}}{\frac{2d}{v}} = \frac{3}{3.5} \\[0.5em] \Rightarrow \frac{d - 84}{d} = \frac{3}{3.5} \\[0.5em] \Rightarrow 3.5 (d - 84) = 3d \\[0.5em] \Rightarrow (3.5d) - (3.5 \times 84) = 3d \\[0.5em] \Rightarrow (3.5d) - 3d = 294 \\[0.5em] \Rightarrow 0.5 d = 294 \\[0.5em] \Rightarrow d = \frac{294}{0.5} \\[0.5em] \Rightarrow d = 588 m\end{aligned}$$
Hence, the distance of the cliff from the initial position of the man = 588 m