Sum
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 every 3 seconds, find the velocity of sound in air.
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Solution
Given:
Distance of the large wall from the man S = 50 m
He has to clap 10 times in 3 seconds.
So, time interval between two claps will be \[= \frac{3}{10}\text { second }\]
Therefore, the time taken \[\left( t \right)\] by sound to go the wall is \[t = \frac{3}{20}\text { second }\]
\[\text { We know that: } \]
\[\text { Velocity } v = \frac{S}{t}\]
\[\Rightarrow v = \frac{50}{\left( \frac{3}{20} \right)} = 333 m/s\]
Hence, the velocity of sound in air is 333 m/s.
Concept: Wave Motion
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