The correct options are
B Minimum frequency recorded by detector will be
257 Hz.
C Maximum frequency recorded by detector will be
485.7 Hz.
The speed of sound in air
=340 m/s. r=3 m, ω=10 rad/s. The linear speed of the source along the circular orbit is
vs=rω=3×10 m/s ![](https://search-static.byjusweb.com/question-images/byjus/infinitestudent-images/ckeditor_assets/pictures/665737/original_original_11.png)
The detector is executing
SHM of amplitude
a=6 m and frequency
n=5π Hz. The angular velocity of the detector is
ω′=2πn=2π×5π=10 rad/s and the linear velocity for the detector is
u0=aω′=6×10=60 m/s Since the source and the detector have the same angular velocity (i.e.,
ω=ω′), they have the same time period
T. At time
t=T4, the source will be at point
P and the detector will be at point
C.
![](https://search-static.byjusweb.com/question-images/byjus/infinitestudent-images/ckeditor_assets/pictures/663890/original_12.png)
Since they are receding from each other at the maximum relative velocity, the recorded frequency will be minimum which is given by
νmin=ν(v−u0v+us)=340×(340−60340+30) νmin=257 Hz At a time
t=3T4, the source will be at point
Q and the detector will be at point
C approaching the source with the maximum relative velocity. Hence the recorded frequency will be maximum which is given by
νmax=ν(v+u0v−u0)=340×(340+60340−30) νmax=485.7 Hz.