Question

A man standing in front of a vertical cliff fires a gun. He hears the echo after $3\;s$. On moving closer to the cliff by $82.5\;m$, he fires again and hears the echo after $2.5\;s$. Find:
a.)The distance of the cliff from the initial position of the man, and
b.)The speed of sound

Answer 1

Hint: A good starting point would be to obtain an expression for the velocity of sound in both cases. Since the velocity of sound remains the same for both cases, you can equate the two relations and obtain the distance of the cliff from the initial position of the man. As for the second part, substituting the distance value you get from part one into one of the expressions of sound velocity you will be able to get the desired value. Also, since the sound wave reaches the cliff at a distance from the man and travels the same distance back to the man via an echo, do not forget to account for the total distance that the sound wave has to travel to reach the man.

Formula used:
Velocity of a soundwave producing an echo $v_{sound} = \dfrac{distance\;travelled}{time\;taken} = \dfrac{2d}{t}$, where d is the distance between the source and the reflecting barrier.

Complete answer:
Let us begin by understanding how echoes are produced in the first place.
We know that sound is a mechanical wave that consists of compressions and rarefactions propagating through a medium. This motion through a medium occurs as one particle of the medium interacts with its neighbouring particle transmitting the compressions and rarefactions to it.
Sound waves get reflected off physical barriers. An echo is a reflection of the sound wave that can be heard separately from the original sound, if the sound source is closer than the reflecting surface.
Therefore, the distance travelled by the sound wave to in reaching the receiver back via an echo will be twice the distance between the receiver and the reflecting barrier.
Let us now dive into the context of our question.

Let the cliff be at a distance $d\;m$ from the man. Therefore, the distance travelled by the sound wave to reach the man back will be $2d\;m$ and it takes $3\;s$ to do so.
So, the speed of sound $v_{sound} = \dfrac{distance\;travelled}{time\;taken} = \dfrac{2d}{3}$
Now the man moves $82.5\;m$ close to the cliff, therefore, the distance between the man and the cliff will now be $d – 82.5\;m$, and it takes the sound wave $2.5\;s$ to reach the man back as an echo.
Therefore, the distance travelled by the sound wave will be $2(d-82.5) \;m$

The speed with which the sound travels is now given by
$v_{sound} = \dfrac{2(d-82.5)}{2.5}$

a.)Since the speed of sound remains the same in both the cases,
$\Rightarrow \dfrac{2d}{3} = \dfrac{2(d-82.5)}{2.5} \Rightarrow 5d = 6d – 495 \Rightarrow d =495\;m$
Therefore, the distance of the cliff from the initial position of the man is $495\;m$

b.)Substituting this distance, we can now find the velocity of the sound.
In the initial case, we defined that $ v_{sound} = \dfrac{2d}{3} = \dfrac{2 \times 495}{3} = \dfrac{990}{3} \Rightarrow v_{sound} = 330\;ms^{-1}$
Thus, we have found the velocity of sound to be $330\;ms^{-1}$

Note:
Do not get confused between echo and reverb. An echo is a single reflection of a sound wave off a barrier, whereas reverberation is the sound produced by the superposition of multiple echoes.
An echo can be heard only when the distance between the source and the barrier is 17m, whereas a reverberation can occur by reflection of sound waves off nearby walls also.
An echo can be heard in both open and closed spaces, whereas reverberation is observed only in closed spaces with multiple reflecting surfaces.