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. 

\[\Rightarrow \text{ v  + u }  = \frac{x}{t_1} . . . \left( \text{ i }  \right)\]

Second Case:
After interchanging the positions:
Resultant velocity of sound = v − u
∴ (v − u)t₂ = x

\[\Rightarrow \text{ v - u }= \frac{x}{t_2} . . . (\text{ ii } )\] 

From (i) and (ii), we get:

\[2\text{ v}  = \frac{x}{t_1} + \frac{x}{t_2}\]

\[ \Rightarrow \text{ v } = \frac{x}{2}\left( \frac{1}{t_1} + \frac{1}{t_2} \right)\]

From (i), we get:

 

\[\text{ u }= \frac{x}{t_1} - v\]

Putting the value of v in the above equation, we get:

 

\[\text{ u } \Rightarrow \frac{x}{2 t_1} - \frac{x}{2 t_2} = \frac{x}{2}\left( \frac{1}{t_1} - \frac{1}{t_2} \right)\]

∴ Velocity of sound,

\[\text{ v }= \frac{x}{2}\left( \frac{1}{t_1} + \frac{1}{t_2} \right)\]

Velocity of wind,

\[\text{ u }  = \frac{x}{2}\left( \frac{1}{t_1} - \frac{1}{t_2} \right)\]