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.

Explanation

The speed of sound in air `=340 m//s`
`r=3m`, `omega=10rad//s`. The linear speed of the source along the circular orbit is
`v_(s)-romega=3xx10m//s`
The detector is executing `SHM` of amplitude `a=6m` and frequency `n=5//pi Hz`. The angular velocity of the detector is
`omega'=2pin=2pixx5//pi=10rad//s`
and the linear velocity of the detector is
`u_(0)=aomega'=6xx10=60m//s`
Since the source and the detector have the same angular velocity (i.e. `omega=omega'`), they have the same time period `T`. At time `t=T//4`, the source will be at point `P` and the detector will be at point `C`.
Since they are receding from each other at the maximum relative velocity, the recorded frequency will be minimum which is given by
`n_(min)=n((v-u_(0))/(v+u_(s)))=340xx((340-60)/(340+30))=257Hz`
At a time `t=3T//4`, the source will be at point `Q` and the detector will be at point `C` approaching the source with the maximum relative. Hence the recorded frequency will be maximum which is given by
`n_(max)=n((v+u_(0))/(v-u_(s)))=340xx((340+60)/(340-30))=485.7Hz`