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)`.
Q. If `omega=10(m)/(s^2)`, the apparent frequency that will be recorded by the stationary receiver at `t=10s` will be

Options

A. `1700Hz`
B. `1.35Hz`
C. `850Hz`
D. `1.27Hz`

Answer

B

Explanation

Source frequency `n_0=1700Hz`. Source (coinciding with observer at `t=0`) moves away with uniform acceleration `omega`. Consider the wave which is received by the observer at instant `t=r` It will have left the source at an earlier instant of time, say `t(ltr)`, when the distance of source was r(say). If u be velocity of source at instant t, then `r=((1)/(2))omegat^2` and `u=omegat`. We then have the relation between `r` and `t`
`r=t+(r )/(V)=t+(omegat^2)/(2V)`
This is quadratic equation in t, giving the solution
`omegat=(-2+sqrt(4V+8Vomegat))/(2)`
`u=omegat=V[sqrt(1+(2omegar)/(V)-1)]=340`
`xx[(sqrt1+(2xx10xx10)/(340)-1)]=340[sqrt((27)/(17))-1]`
Then apparent frequency is given by
`n_a=((V)/(V+u))n_0`
Putting values `V=340(m)/(s)`,`r=10s`,`omega=10(m)/(s^2)`,
We have
`n_a=((340)/(340+u))1700`
`=1700xxsqrt((17)/(27))=1.35kHz`