Question

A source of sound of frequency $n$ and a listener approach each other with a velocity equal to $\frac{1}{20}$ of velocity of sound. What must have been the apparent frequency heard by the listener ?

• A. $(\frac{21}{19})n$
• B. $(\frac{20}{21})n$
• C. $(\frac{21}{20})n$
• D. $(\frac{19}{20})n$

Answer 1

Answer: (A)

According to Doppler effect we have,
$n^{{}^{\prime }}=\frac{v+v_0}{v-v_s}n$
where,
$v$ is speed of sound,
$v_s$ is speed of source,
$v_o$ is speed of observer,
$n$ is natural frequency of source and $n^{{}^{\prime }}$ is oberved frequency.
as given, $v_s=v_o=\frac{v}{20}$
$\Rightarrow n^{{}^{\prime }}=\frac{v+\frac{v}{20}}{v-\frac{v}{20}}n$
$\Rightarrow n^{{}^{\prime }}=\frac{\frac{21}{20}v}{\frac{19}{20}v}=\frac{21}{19}n$
The apparent frequency heard by the listener is $\frac{21}{19}n$ .