A source `S` of acoustic wave of the frequency `v_0=1700Hz` and a receiver `R` are located at the same point. At the instant `t=0`, the source start from rest to move away from the receiver with a constant acceleration `omega`. The velocity of sound in air is `v=340(m)/(s)`.
If `omega=10(m)/(s^2)` for 10s and then `omega=0` for `tgt10s`, the apparent frequency recorded by the receiver at `t=15s`

Options

A. `1700Hz`
B. `1313Hz`
C. `850Hz`
D. `1.23Hz`

Answer

B

Explanation

Since `omega=0` for `tgt10s` the source will move at constant speed of `u=10xx10=100(m)/(s)` after `t=10s`. At `t=10s`, its distace r from the observer is `r=((1)/(2))omegat^2=((1)/(2))xx10xx100=500m`. Since the wave will take a time `r=10+(500)/(340)` which is less than 15 s, the wave that is received by the observer at `r=15s` would have left the source after 10 s, i.e., when its speed was constant at `100(m)/(s)`. Hence apparent frequency is
`n_a=((340)/(340+100))1700`
`=((17)/(22)xx1700)=1313Hz`