A car emitting sound of frequency 500 Hz speeds towards a fixed wall at `4(m)/(s)`. An observer in the car hears both the source frequency as well as the frequency of sound reflected from the wall. If he hears 10 beats per second between the two sounds, the velocity of sound in air will be

Options

A. `330(m)/(s)`
B. `387(m)/(s)`
C. `404(m)/(s)`
D. `340(m)/(s)`

Answer

D

Explanation

The frequency that the observer receives directly from the source has frequency `n_1=500Hz`. As the observer and source both move towards the fixed wall with velocity `u`, the apparent frequency of the reflected wave coming from the wall to the observer wii have frequency
`n_2=((V)/(V-u))500Hz`
Where `V` is the velocity of sound wave in air. The apparent frequency of this reflected wave as heard by the oberver will then be
`n_3=((V+u)/(V))n_2=((V+u)/(V))((V)/(V-u))500=((V+u)/(V-u))500`
It is given that the nubmer of beat per second is `n_3-n_1=10`
`(n_3-n_1)=10=((V+u)/(V-u))500-500`
`=500[(V+u)/(V-i)-1]`
`10=(2xxuxx500)/(V-u)`
Hence,
`10V=1000u+10u=1010u`
putting `u=4(m)/(s)`
we have `V=(1)/(10)[4040]=404(m)/(s)`