Question

The difference between the apparent frequency of a source of sound as perceived by an observer during its approach and recession is 2% of the natural frequency of the source. If the velocity of sound in air is 300 m/sec, the velocity of the source is (It is given that velocity of source << velocity of sound)

• A. 6m/sec
• B. 3m/sec
• C. 1.5m/sec
• D. 12m/sec

Answer 1

The correct option is B

3m/sec

When the source approaches the observer
Apparent frequency n' = $\frac{v}{v-v_s}.n=n\left[\frac{1}{1-\frac{v_s}{v}}\right]$
= $n{[1-\frac{v_s}{v}]}^{-1}=n[1+\frac{v_s}{v}]$
(Neglecting higher powers because $v_s<<v$)

When the source recedes the observed apparent frequency n' = n $[1-\frac{v_s}{v}]$
Given n' - n'' = $\frac{2}{100}n,v=300m/sec$

$\therefore \frac{2}{100}n=n[1+\frac{v_s}{v}]-n[1-\frac{v_s}{v}]=n\left[2\frac{v_s}{v}\right]$

$\Rightarrow \frac{2}{100}=2\frac{v_s}{v}\Rightarrow v_s=\frac{v}{100}=\frac{300}{100}=3m/sec$ .