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

Velocity of sound in  air is $v=\frac{x}{2}\left(\frac{1}{t_{1}}+\frac{1}{t_{2}}\right)$,

Velocity of wind/air  is $u=\frac{x}{2}\left(\frac{1}{t_{1}}-\frac{1}{t_{2}}\right)$.

Explanation:

Step 1:  

Let velocity of $\text { air=u } \mathrm{m} / \mathrm{s}$

Let the velocity of $\text { sound=v } \mathrm{m} / \mathrm{s}$

Distance between A and B $x m$

Step 2:

Case 1 :

Resultant velocity of $sound =u+v$

$s=\frac{d}{t}$

$\text { Speed }=\mathrm{s}$

$\text { Distance }=\mathrm{d}$

$Time =t$

$(v+u)=\frac{x}{t_{1}}$ $\ldots \ldots \text { eq }^{n}(1)$

Case 2 :

Resultant velocity= $v-u$

$\text {speed}=\frac{\text {distance}}{\text {time}}$

$(v-u)=\frac{x}{t_{2}}$$\ldots \ldots \ldots \text { eq }^{n}(2)$

adding equation 1 and 2  

Step 3:

we get :

$(v+u)+(v-u)=\frac{x}{t_{1}}+\frac{x}{t_{2}}$

$2 v=\frac{x}{t_{1}}+\frac{x}{t_{2}}$

$2 v=x\left(\frac{1}{t_{1}}+\frac{1}{t_{2}}\right)$

$v=\frac{x}{2}\left(\frac{1}{t_{1}}+\frac{1}{t_{2}}\right)$

Step 4:

subtracting equation 1 and 2

we get :

$(v+u)-(v-u)=\frac{x}{t_{1}}-\frac{x}{t_{2}}$

$v+u-v+u=\frac{x}{t_{1}}-\frac{x}{t_{2}}$

$2 u=\frac{x}{t_{1}}-\frac{x}{t_{2}}$

$2 u=x\left(\frac{1}{t_{1}}-\frac{1}{t_{2}}\right)$

$u=\frac{x}{2}\left(\frac{1}{t_{1}}-\frac{1}{t_{2}}\right)$

Velocity of sound in  air =  

$v=\frac{x}{2}\left(\frac{1}{t_{1}}+\frac{1}{t_{2}}\right)$

and

velocity of wind/air =  

$u=\frac{x}{2}\left(\frac{1}{t_{1}}-\frac{1}{t_{2}}\right)$