Question

An observer move towards a stationary source of sound with a velocity equal to one fifth of the velocity of sound. The apparent change in frequency is

• A. $35\%$
• B. $30\%$
• C. $25\%$
• D. $20\%$

Answer 1

The correct option is D $20\%$

From Doppler's effect, the apparent frequency heard by the observer is,
$f^{{}^{\prime }}=f_0\left(\frac{v\pm v_0}{v\pm v_s}\right)$, here,
$f_0$ is the original frequency produced by the source, $v$ is the speed of sound, $v_0$ is the speed of observer and $v_s$ is the speed of source.
Given, $v_0=\frac{v}{5}$
$\Rightarrow f^{{}^{\prime }}=f_0\left(\frac{v+\frac{v}{5}}{v-0}\right)$
$\Rightarrow f^{{}^{\prime }}=\frac{6}{5}{f}_0=1.2f_0$
Now,
Apparent change in frequency $=\frac{\Delta f}{f_0}\times 100$
$=\frac{1.2f_0-f_0}{f_0}\times 100=20\%$