Question

A man stands in between two cliffs $x$ and $y$, such that he is at a distance of $66m$ from $x$. When he blows a whistle he hears first echo after $0.4s$ and second echo after $1.2s$. Calculate: (i) speed of sound (ii) distance of cliff $y$ from man.

Answer 1

From the given data we have,

$d_x=66m$

$t_1=0.4s$

$t_2=1.2s$

$v=?$

$d_y=?$

When he blows the whistle, the sound generated travels everywhere, including in the direction towards the cliff.

The sound wave travelling towards the cliff, hits it, gets reflected from it, moves backwards and reaches his ear.

This is how he hears the echo.

This shows that the sound wave travels twice the distance between him and the cliff. Hence, while calculating, we'll have to double the distances.

Case 1:- when the 1st echo is due to cliff x and 2nd due to y

$v=\frac{2d_x}{t_1}=\frac{2\times 66}{0.4}=330m/s$

$2d_y=v\times t_2$

$\therefore d_y=\frac{v\times t_2}{2}=\frac{330\times 1.2}{2}=198metres$

Case 2:- when the 1st echo is due to cliff y and 2nd due to x

$v=\frac{2d_x}{t_2}=\frac{2\times 66}{1.2}=110m/s$

$2d_y=v\times t_1$

$\therefore d_y=\frac{v\times t_1}{2}=\frac{110\times 0.4}{2}=22metres$

We know that the speed of sound in air is approx 330m/s

Hence only Case 1 is correct

(i) speed of sound = 330m/s

(ii) distance of cliff y from man = 198m