Question

Two friends A and B are standing a distance x apart in an open field and wind is blowing from A to B. A beat a drum and B hears the sound t₁ time after he sees the event. A and B interchange their positions and the experiment is repeated. This time B hears the drum timer after he sees the event. Calculate the velocity of sound in still air v and the velocity of wind u. Neglect the time light takes in travelling between the friends.

Answer 1

Given:
Distance between A and B = x
Velocity of sound in air = v
Velocity of wind = u
First Case:
When A beats the drum from his original position:
Resultant velocity of sound = u + v
⇒ (v + u)t₁ = x
Here, t₁ is the time at which the sound of the drum is heard by B.
$\Rightarrow v+u=\frac{x}{t_1}...\left(\mathrm{i}\right)$
Second Case:
After interchanging the positions:
Resultant velocity of sound = v − u
∴ (v − u)t₂ = x
$\Rightarrow v-u=\frac{x}{t_2}...\left(\mathrm{ii}\right)$
From (i) and (ii), we get:
$$\begin{gathered} 2v=\frac{x}{t_1}+\frac{x}{t_2} \\ \Rightarrow v=\frac{x}{2}\left(\frac{1}{t_1}+\frac{1}{t_2}\right) \end{gathered}$$
From (i), we get:
$u=\frac{x}{t_1}-v$
Putting the value of v in the above equation, we get:
$u\Rightarrow \frac{x}{2t_1}-\frac{x}{2t_2}=\frac{x}{2}\left(\frac{1}{t_1}-\frac{1}{t_2}\right)$
∴ Velocity of sound, $v=\frac{x}{2}\left(\frac{1}{t_1}+\frac{1}{t_2}\right)$
Velocity of wind, $u=\frac{x}{2}\left(\frac{1}{t_1}-\frac{1}{t_2}\right)$