A boy standing in front of a wall at a distance of 85 m produces 2 claps per second -He noticed that the sound of his clapping coincides with the echo- The echo it-s hard only once when clapping it-s stopped- Calculate speed of sound-

Since 2 claps are produced in one second, Time interval between two claps,t=1/2 sec=0.5 sec the distance of the wall from the boy=d=85m Then speed of sound( v)= ...

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Standard IX

Physics

Echo

A boy standin...

Question

A boy standing in front of a wall at a distance of 85 m produces 2 claps per second. He noticed that the sound of his clapping coincided with the echo. The echo is heard only once when clapping is stopped. Calculate the speed of sound.

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Solution

Since 2 claps are produced in one second,
The time interval between two claps,
$t = \frac{1}{2} s = 0.5 s$
The distance of the wall from the boy(d) = 85 m
Then speed of sound( v)= $\frac{2 d}{t}$
Putting all the values
Speed of the sound = $\frac{2 \times 85}{0.5} = 340$ m/s

Hence the speed of the sound is 340 m/s

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A boy standing in front of a wall at a distance of 85 m produces 2 claps per second. He noticed that the sound of his clapping coincided with the echo. The echo is heard only once when clapping is stopped. Calculate the speed of sound.

Since 2 claps are produced in one second, The time interval between two claps,t=12s=0.5s The distance of the wall from the boy(d) = 85 m Then speed of sound( v)= 2dt Putting all the valuesSpeed of the sound = 2×850.5=340 m/s Hence the speed of the sound is 340 m/s