Chapter - 5 Sound

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Q
Sound

Question:

A stone is dropped from the top of a tower 500 m high into a pond of water at the base of the tower. When is the splash heard at the top? Given, g = 10 ms^{-2} and speed of sound = 340 ms^{-1}.

Answer:

Firstly, we calculate the time taken by stone to reach the pool and then the time is taken by the sound of a splash to reach the top.

Case 1: For the time taken by the stone to reach the surface of the pool.

Given: s = 500 m

g = 10 ms^{-2}

Initially, the stone is at rest, so u = 0

Now, using the second equation of motion

\ S = ut + frac { 1 }{ 2 } g{ t }^{ 2 } \ 500 = 0 	imes t + frac { 1 }{ 2 } 	imes 10 	imes { t }^{ 2 }\ 500 =quad 5{ t }^{ 2 }\ { t }^{ 2 } =100\ t = sqrt { 100 } = 10sec

Therefore, t = 10 seconds

Hence, the time taken by the stone to reach the surface of the pond will be 10 seconds.

 

Case 2: For the time taken by the sound to reach the top.

Given: Speed of sound = 340 m/s

s = 500 m

Now, using the relation

Speed = frac{Distance}{Time}

Time = frac{Distance}{Speed}

The time taken by the sound to reach the top = frac{500}{340} = 1.47 sec 

So, the total time taken for the splash to be heard = Time taken for case 1 + Time taken for case 2

The total time taken for the splash to be heard = 10 + 1.47 = 11.47 seconds.


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