Question

A receiver & a source of sonic oscillations of frequency 200 Hz are located on the x-axis. The receiver is fixed and the source swings harmonically along that axis with a circular frequency $\omega$ and an amplitude 50 cm. At what value of $\omega$ (in rad/sec) will the frequency band width ($f_{max}-f_{min}$) registered by the stationary receiver be equal to 20 Hz. [ The velocity of sound is equal to 340 m/s]

• A. $17$
• B. $34$
• C. $68$
• D. $8.5$

Answer 1

Answer: (B)

$34$

Given, $f=200Hz,angularfrequency=\omega ,Amplitude=50cm=0.5m,f_{max}-f_{min}=20Hz\&v=340m/s$

$f_{max}$ will happen when the source will move towards the listener with maximum velocity and $f_{min}$ will happen when the source

will move away from the listener at maximum speed.

By Doppler's formula:-

$f_{max}=f\left(\frac{v}{v-v_s}\right)$

$f_{min}=f\left(\frac{v}{v+v_s}\right)$

$f_{max}-f_{min}=20Hz$

$f(\frac{v}{v-v_s}-\frac{v}{v+v_s})=20Hz$

$2fv\left(\frac{a\omega }{v^2-a^2{\omega }^2}\right)=20Hz$ where, $v_s=aw$

$10a^2{\omega }^2+avf\omega -10v^2=0$

Upon solving the quadratic equation:

$\omega \approx 34$

So, B is the correct answer.