Question

A man is standing between two vertical cliffs

685 m

apart. He claps his hands and hears two distinct echoes after

0.9 second

and

1.1 seconds

respectively. What is the speed of sound in air?

Accepted Answer

Let the speed of the sound is $= v m / s$
Let the distance between man and the first cliff $= d m$

The total distance between the cliffs $= 685 m$

Therefore, the distance between the man and the second cliff $= (685 - d) m$
The distance travelled by the wave to reach the man in $0.9 second$ from the first cliff $= 2 d m$

The distance travelled by the wave to reach the man in $1.1 seconds$ from the second cliff $= 2 \times (685 - d) m$

We know that, $S p e e d = \frac{Distance}{Time}$ .

Therefore, for the first cliff, $v = \frac{2 d}{0.9}$ -----------------(1)

and, for the other cliff, $v = \frac{2 (685 - d)}{1.1}$ -------------(2)

The velocity of sound will be equal in both cases, thus, equating equations (1) and (2), we get,

$\frac{2 d}{0.9} = \frac{2 (685 - d)}{1.1} (2 d) \times 1.1 = 2 \times 0.9 \times (685 - d) d \times 1.1 = 0.9 \times (685 - d) 1.1 d = 685 \times 0.9 1.1 d + 0.9 d = 685 \times 0.9 2 d = 616.5 d = \frac{616.5}{2} d = 308.25 m$

Therefore, the distance between the man and the first cliff, $d = 308.25 m$

The time taken by the echo is $= 0.9 s$

Therefore, the speed of sound in air, $v = \frac{d i s \tan c e}{t i m e} = \frac{308.25}{0.9} = 342.5 m s^{- 1}$

Pankaj Bhatt Solutions for Chapter: Sound, Exercise 1: QUESTIONS

Pankaj Bhatt Physics Solutions for Exercise - Pankaj Bhatt Solutions for Chapter: Sound, Exercise 1: QUESTIONS

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