Question

The difference between the apparent frequencies of a source of sound perceived by a stationary observer during its approach and recession is $2\%$ of the actual frequency of the source. If the speed of sound is $300\text{m/s}$, the speed of the source is

• A. $12\text{m/s}$
• B. $6\text{m/s}$
• C. $1.5\text{m/s}$
• D. $3\text{m/s}$

Answer 1

The correct option is D $3\text{m/s}$

Let,
$f_o=$ actual frequency of source;
$f_A=$ frequency heard during approach;
$f_R=$ frequency heard during recession:
$V_s=$ speed of source:

Given:
Speed of observer, $V_o=0;$
Speed of sound, $V=300\text{m/s}$;

$f_A-f_R=\frac{2}{100}{f}_o.......\left(1\right)$

Now using formula of apparent frequency (Doppler effect)

$f_{app}=f_o(\frac{V\pm V_o}{V\pm V_s})$

During approach;

$f_A=f_o(\frac{V+V_o}{V-V_s})$

$\Rightarrow f_A=f_o(\frac{300+0}{300-V_s})$

$\Rightarrow f_A=f_o(\frac{300}{300-V_s}).....\left(2\right)$

During recession;

$f_R=f_o(\frac{V+V_o}{V+V_s})=f_o(\frac{300+0}{300+V_s})$

$\Rightarrow f_R=f_o(\frac{300}{300+V_s}).....\left(3\right)$

from eqs.(1), (2) and (3)

$f_o(\frac{300}{300-V_s})-f_o(\frac{300}{300+V_s})=\frac{2}{100}{f}_o$

$\Rightarrow (\frac{300}{300-V_s})-(\frac{300}{300+V_s})=\frac{2}{100}$

$\Rightarrow \frac{300\times 2V_s}{{300}^2-V_s^2}=\frac{2}{100}$

Since, ${300}^2>>>V_s^2,$ so we can neglect this term

$\Rightarrow V_s=3\text{m/s}$

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