Question

A stationary source emits sound of frequency $f_0=492Hz$. The sound is reflected by a large car approaching the source with a speed of $2m{s}^{-1}$. The reflected signal is received by the source and superposed with the original. The beat frequency of the resulting signal is $Hz$. (Given that the speed of sound in air is $330m{s}^{-1}$ and the car reflects the sound at the frequency it has received).

Answer 1

Given, $f_0=492Hz$, $v_{car}=2m/s$ and $u_{sound}=330m/s$
In this condition a source will hear two sounds, one from the original source and other reflected by the car.

Now apparent frequency heard by person in the car is
$f_1=f_0\frac{U_{sound}+U_{car}}{u_{car}}=\frac{330+2}{330}\times 492$
Apparent frequency received by stationery source is
$f_2=f_1\frac{U_{sound}}{U_{sound}-u_{car}}=492\times \frac{332}{330}\left(\frac{330}{330-2}\right)=498Hz$

hence beat frequency= $f_2-f_{0}498-492=6Hz$