An observer and a source emitting sound of frequency 120 Hz are the X-axis . The observer is stationary while the source of sound is in motion given by the equation x=3 sin `omega` (x is in metres and t is in seconds). If the diffrence between the maximum and minimum frequencies of the sound observed by the observers is 22Hz , then the value of `omega` is (speed of sound in air = 330 `ms^(-1)`

Options

A. 33 rad `s^(-1)`
B. 36 rad `s^(-1)`
C. 20 rad `s^(-1)`
D. 10 rad `s^(-1)`

Answer

D

Explanation

Instantaneous speed of source is `v=(dx)/(dt)= 2 omega sin omega t`
Difference between maximum and minimum frequencies is 22 Hz.
So `f_(max)-f_(min)= f((v)/(v-v_s))-f((v)/(v+v_s))=22`
`rArr f{ (v)/(v-v_s)-(v)/(v+v_s)}=22`
Now here , f= 120 Hz, v = 330 `ms^(-1)v_s = 3 omega`
Substituting these values in Eq (i) , we get `120 ( 330/(330-3 omega)-(330)/(330+3 omega))=22 rArr omega = 10s^(-1)`