A source of sound is moving along a circular orbit of radius 3 m with an angular velocity of 10 rad/s. A sound detector located far away from the source is executing linear S.H.M. along the line BD (see Fig. 14.4.13) with an amplitude BC = CD=6m. The frequency of oscillation of the detector is `5//pi` per second. The source is at the point BA when the detector is at the point B. If the source epiits a continuous sound wave of frequency 340 Hz, find the maximum and the minimum frequencies recorded by the detector.

Explanation

Time period of circular motion `T =(2pi//omega) =(2pi //10)`
is same as that of SHM. i.e., `T =(1//f) =(pi//5`, so both will complete one periodic motion in same time.
Futher more, source is moving on a circle, its speed `v_s =romega =3 xx 10 = 30m//s` and as detector is executing `SHM, v_D =omegasqrt(A^2 -y^2) =10sqrt(6^2 -y^2)`
i.e., `(v_D)_(max)= 60 m//s` when y = 0 i.e.,detector is at C.Now in the case of Doppler effect,`f_(Ap) = f[(v pm v_D)/(v pm v_S)]`
So `f_(Ap)` will be maximum when both move towards each other. `f_(max) = f [(v +v_D)/(v -v_S)]` with `v_D =` max i.e., the source is at M and detector at C and moving towards B, so
`f_(max) =340 [ (330 +6)/(330 -30)] =442 Hz`
Similarly `f_(Ap)` will be minimum when both are moving away from each other, `i.e., f_(max) =f[(v-v_D)/(v + v_S)]` with `v_D =` max i.e., the source is at N and detector at C but moving towards D,so
`f_("min") =340 [(330-60)/(330 + 30)]= 255 Hz`