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 beats a drum and B hears the sound ${{t}_{1}}$time after he sees the event. A and B interchange their positions and the experiment is repeated. This time B hears the drum ${{t}_{2}}$time 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 friends.
$A.\,\dfrac{1}{2}\left( \dfrac{1}{{{t}_{1}}}+\dfrac{1}{{{t}_{2}}} \right),\dfrac{x}{2}\left( \dfrac{1}{{{t}_{1}}}-\dfrac{1}{{{t}_{2}}} \right)$
$B.\,\dfrac{x}{2}\left( \dfrac{1}{{{t}_{1}}}+\dfrac{1}{{{t}_{2}}} \right),\dfrac{x}{2}\left( \dfrac{1}{{{t}_{1}}}-\dfrac{1}{{{t}_{2}}} \right)$
$C.\,\dfrac{x}{2}\left( \dfrac{1}{{{t}_{1}}}+\dfrac{1}{{{t}_{2}}} \right),\dfrac{3x}{2}\left( \dfrac{1}{{{t}_{1}}}-\dfrac{1}{{{t}_{2}}} \right)$
$D.\,\dfrac{3x}{2}\left( \dfrac{1}{{{t}_{1}}}+\dfrac{1}{{{t}_{2}}} \right),\dfrac{x}{2}\left( \dfrac{1}{{{t}_{1}}}-\dfrac{1}{{{t}_{2}}} \right)$

Answer 1

Using the formula that relates the speed, distance and time, this problem can be solved. This problem solution is mainly based on the formation of the equations of distances and then finding the difference between the same.

Formula used: $\text{Speed}=\dfrac{\text{Distance}}{\text{Time}}$

Complete step by step answer:
From given, we have the data,
The distance between the friends’ A and B = x
The velocity of sound in still air = v m/s
The velocity of wind/air = u m/s

As per the given statement, we have to solve two cases.
In the first case, A beats a drum and B hears the sound ${{t}_{1}}$time after he sees the event.
In the second case, A and B interchange their position, then A beats a drum and B hears the sound ${{t}_{2}}$ time after he sees the event.
Let us consider each case one by one.
Case I:
A beats a drum and B hears the sound ${{t}_{1}}$time after he sees the event.
Using the data given, the resultant of the sound is given as follows.
$R=u+v$
Use the formula that relates speed, distance and time for the further calculation.
So, we get,
$v+u=\dfrac{x}{{{t}_{1}}}$…… (1)
Case II:
A beats a drum and B hears the sound ${{t}_{2}}$time after he sees the event but after interchanging the positions.
Using the data given, the resultant of the sound is given as follows.
$R=v-u$
Use the formula that relates speed, distance and time for the further calculation.
So, we get,
$v-u=\dfrac{x}{{{t}_{2}}}$…… (2)
Perform the addition operation on the equations (1) and (2).
So, we get,

& 2v=\dfrac{x}{{{t}_{1}}}+\dfrac{x}{{{t}_{2}}} \\\ & \Rightarrow v=\dfrac{1}{2}\left( \dfrac{x}{{{t}_{1}}}+\dfrac{x}{{{t}_{2}}} \right) \\\ & \Rightarrow v=\dfrac{x}{2}\left( \dfrac{1}{{{t}_{1}}}+\dfrac{1}{{{t}_{2}}} \right) \\\ \end{aligned}$$ Hence this is an expression for velocity of the sound in the air. Perform the subtraction operation on the equations (1) and (2). So, we get, $$\begin{aligned} & 2u=\dfrac{x}{{{t}_{1}}}-\dfrac{x}{{{t}_{2}}} \\\ & \Rightarrow u=\dfrac{1}{2}\left( \dfrac{x}{{{t}_{1}}}-\dfrac{x}{{{t}_{2}}} \right) \\\ & \Rightarrow u=\dfrac{x}{2}\left( \dfrac{1}{{{t}_{1}}}-\dfrac{1}{{{t}_{2}}} \right) \\\ \end{aligned}$$ Hence this is an expression for velocity of the wind. As the expressions for the velocity of the sound in the air and the velocity of the wind are $$\dfrac{x}{2}\left( \dfrac{1}{{{t}_{1}}}+\dfrac{1}{{{t}_{2}}} \right)$$and $$\dfrac{x}{2}\left( \dfrac{1}{{{t}_{1}}}-\dfrac{1}{{{t}_{2}}} \right)$$ **So, the correct answer is “Option B”.** **Note:** The things to be on your finger-tips for further information on solving these types of problems are: As in this problem, the positions of the bodies are changed, so the difference and the sum of the velocity sound and the wind are considered. These velocities should be added or subtracted based on the direction of these parameters.