A source of sound and a detector are placed at the same place on ground At `t=0`, the source `S` is projected towards reflector with velocity `v_0` in vertical upward directon and reflector starts moving down with constant velocity `v_0` At `t-0`, the vertical separation between the reflector and source is `H((gtv_0^2)/(2g))`. The speed of sound in air is `v(gtgtv_0)`, Take `f_0` as the frequency of sound waves emitted by source. Based on above information answer the following questions.
Q. Frequency of sound received by detector after being reflected by reflector at `t=(v_0)/(2g)` is

Options

A. `(2f_0(v+v_0))/(2v-v_0)`
B. `(2f_0v)/(v-v_0)`
C. `2f_0[(v+v_0)/(2v-v_0)]xx[(v)/(v-v_0)]`
D. `2f_0xx(v+v_0)/(v-v_0)`

Answer

C

Explanation

As we known velocity at `t=(v_0)/(2g)` is `(v_0)/(2)`.
Number of waves reaching the reflector per second.
`f_1=(v+v_0)/(v-(v_0)/(2))f_0`
Now the wall acts as the source with above frequency and the detector is at rest. So, frequency heard by the detetor is
`f=(v)/(v-v_0)f_1=(v)/(v-v_0)xx(v+v_0)/(v-(v_0)/(2))f_0`
`=(v)/(v-v_0)xx(v+v_0)/(2v-v_0)2f_0`