Question

A siren of frequency $f_0$ approaches a stationary observer and then recedes from the observer. If the velocity of source $v_s$ satisfies the condition $(v_s\ll v)$ where $v$ is the velocity of sound, then the apparent change in frequency is:

[Assume, medium is stationary]

• A. $\frac{2f_0{v}_s}{v}$
• B. $\frac{2f_{0}v}{v_s}$
• C. $\frac{f_0}{v}$
• D. $\frac{2v{v}_s}{f_0}$

Answer 1

The correct option is A $\frac{2f_0{v}_s}{v}$

Given:
Observer is at rest and the source is moving with velocity $v_s$.

If $f$ be the frequency and $v$ be the velocity of sound heard by the observer when both source and observer are at rest, then using Doppler's formula

When source is moving towards observer :

$f_1=f_0\left(\frac{v}{v-v_s}\right)$

When source is moving away from observer:

$f_2=f_0\left(\frac{v}{v+v_s}\right)$

Apparent change in frequency of the sound as heard by the observer during the relative motion is given by

$|f_1-f_2|=f_0[\frac{v}{v-v_s}-\frac{v}{v+v_s}]$

$\Rightarrow |f_1-f_2|=f_0\left(\frac{2v{v}_s}{v^2-v_s^2}\right)$

$\Rightarrow |f_1-f_2|=f_0\left(\frac{2v_s}{v(1-\frac{v_s^2}{v^2})}\right)$

$\because v_s\ll v$, we get

Apparent change in frequency $|f_1-f_2|=\frac{2f_0{v}_s}{v}$

Hence, option $\left(D\right)$ is the correct answer.