A man stands before a large wall at a distance of 50-0 m and claps his hands at regular intervals. Initially, the interval is large. He gradually reduces the interval and fixes it at a value when the echo of a clap merges with the next clap. If he has to clap 10 times during every 3 seconds, find the velocity of sound in air.

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Solution

He has to clap 10 times in 3 seconds. So, time interval between two clap

$= (\frac{3}{10 second})$

So, the time taken to go the wall

$= v = (\frac{3}{12 second})$

We know, s =vt

$\Rightarrow V = \frac{s}{t} = \frac{50}{(\frac{3}{20})} = 333 m / s$

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He has to clap 10 times in 3 seconds. So, time interval between two clap =(310 second) So, the time taken to go the wall =v=(312 second) We know, s =vt ⇒V=st=50(320)=333m/s

He has to clap 10 times in 3 seconds. So, time interval between two clap

$= (\frac{3}{10 second})$

So, the time taken to go the wall

$= v = (\frac{3}{12 second})$

We know, s =vt

$\Rightarrow V = \frac{s}{t} = \frac{50}{(\frac{3}{20})} = 333 m / s$