Question

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 every 3 seconds, find the velocity of sound in the air.

Answer 1

Hint: To solve this question we need to understand how to use the relationship between velocity and distance. The question demands determining the velocity of sound in the air. To solve this will use the velocity-distance formula by simply substituting the values in the formula to get the desired result.

The formula used:
$Velocity = \dfrac {Distance}{Time}$

Complete answer:
According to the given question distance between the wall and the man=S=50m.
And also, it is given that he has to clap 10 times during every 3 seconds which means the interval between two claps is $\dfrac{3}{{10}}$.

But we need the time taken to reach the sound to the wall which will be only half of the total time. i.e., $\dfrac{3}{{20}}$ seconds.

Now simply substituting the values in the velocity formula,
$\begin{array}{c}
v = \dfrac{S}{t}\\
 \Rightarrow v = \dfrac{{50}}{{\left( {\dfrac{3}{{20}}} \right)}}\\
 \Rightarrow v = 333{\rm{ m}}{{\rm{s}}^{ - 1}}
\end{array}$

Hence the velocity of the sound in air is nothing but $v = 333{\rm{ m}}{{\rm{s}}^{ - 1}}$ .

Note: While solving this question we should always remember that we need to consider half of the total time taken. Because, $\dfrac {3}{10}$ is the total time sound travelled i.e., clap sound to the wall and back to the person, but in this question, it's precisely asked for the velocity of the sound in air, so will consider only half of the total time taken.