Question

A fixed source of sound emitting a certain frequency appears as $f_a$ when the observer is approaching the source with speed v and frequency $f_r$ when the observer recedes from the source with the same speed. The frequency of the source is $\frac{f_r+f_a}{x}$. Find $x$

Answer 1

Using Doppler effect when source is at rest and the observer is moving:

${\nu }^{\prime }=\nu [\frac{v_{sound}\pm v_{observer}}{v_{sound}}]$

When observer is approaching towards source, $f_a=f[\frac{v_{sound}+v}{v_{sound}}]$

$f_a=f[1+\frac{v}{v_{sound}}]$ ..............(1)

When observer is moving away from source, $f_r=f[\frac{v_{sound}-v}{v_{sound}}]$

$f_a=f[1-\frac{v}{v_{sound}}]$ .............(2)

Adding (1) and (2), $f_a+f_r=2f$

$⟹x=2$