Saved Bookmarks
| 1. |
A source of sound is moving along a circular orbit of radius 3m with an angular velocity of 10 red/s. A sound detector located far away from the source is executing linear simple harmonic motion along the line BD with amplitude BC= CD = 6m. The frequencyof oscillation of the deterctor is (5 //pi) rev/sec. 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 [velocity of sound =330 m/s]. |
|
Answer» Solution :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 =3XX 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` |
|