A source of sound is moving along a circular orbit of radius `3 meter` with an angular velocity of `10 rad//s`. A sound detector located far away from the source is executing linear simple harmonic motion along the line `BD` with an amplitude `BC = CD = 6 meters`. The frequency of oscillation of the detector is `(5)/(pi)` per second. The source is at the point `A` when the detector is at the point `B`. If the source emits a continuous sound wave of frequency `340 Hz`, Find the maximum and the minimum frequencies recorded by the detector.

Answer

485.7Hz,
257.3Hz

Explanation

The angular frequency of the detector `2pi f=2pi((5)/(pi))=10 rad//s`
The angular frequency of the detector source is equal,hence , their time period will also be the same.
`implies` When the detector is at C moving towards D,the source is at `A_1` moving leftwards, in this situation ,the frequency heard is minimum ( since `v_s ltlt v_(sound),` so time taken by pulse to reach the detector is neglectined)
`f'=f_0[(V-V_0)/(V-(-V_s))]=f_0[(V-V_0)/(V+V_s))]`
Velocity of source `V_s=omegaA=60 m//s,` `V_s=omegaA=60 m//s,`elocity of detector `V_0=Romega=30 m//s`
`implies f'=340xx((340-60))/((340+30))=257.3 Hz`
Again when the dete tor at C and moving towards B, the sourse is at `A_3` rightwards in the situation that the frequency heard is maximum
`f"= f_0[(V-(-V_0))/(V-V_s)]=f_0[(V+V_0)/(V-V_s)] implies f'=340xx ((340+60))/((340-30))=485.75Hz`