Q. 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 ?
NTA AbhyasNTA Abhyas 2020
Solution:
According to Doppler effect we have,
$n^{'}=\frac{v + v_{0}}{v - v_{s}}n$
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 $n^{'}$ is oberved frequency.
as given, $v_{s}=v_{o}=\frac{v}{20}$
$\Rightarrow n^{'}=\frac{v + \frac{v}{20}}{v - \frac{v}{20}}n$
$\Rightarrow n^{'}=\frac{\frac{21}{20} v}{\frac{19}{20} v}=\frac{21}{19}n$
The apparent frequency heard by the listener is $\frac{21}{19}n$ .
$n^{'}=\frac{v + v_{0}}{v - v_{s}}n$
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 $n^{'}$ is oberved frequency.
as given, $v_{s}=v_{o}=\frac{v}{20}$
$\Rightarrow n^{'}=\frac{v + \frac{v}{20}}{v - \frac{v}{20}}n$
$\Rightarrow n^{'}=\frac{\frac{21}{20} v}{\frac{19}{20} v}=\frac{21}{19}n$
The apparent frequency heard by the listener is $\frac{21}{19}n$ .