An ambulance is traveling down a highway at `34.5 ms^(-1)`. The ambulance having siren emits sound of frequency 380 Hz. If a passenger in a car travelling with the speed of `26.4 ms^(-1)` in the opposite direction as the car approaches the ambulance and after the two vehicles passed each other, then find the frequency heared by that passenger.
Take speed of sound in air is `343 ms^(-1)`

Options

A. 100 Hz
B. 120 Hz
C. 136 Hz
D. None of these

Answer

C

Explanation

Given, speed of sound in air, `V=343 ms^(-1)`
speed of the observer (i.e. pasenger in this case), `v_(o)=26.4 ms^(-1)`
speed of the source (i.e. ambulance in this case), `v_(s)=34.5 ms^(-1)`
emitting frequency from the ambulance, `n_(o)=380 Hz`
The apparent frequency when ambulance and car approach each other is given by,
`n'=n_(o)((v+v_(o))/(v-v_(s)))=380 ((343+26.4)/(343-34.5))`
`=380xx1.197`
`~~454.86 Hz`
If both the source and observer recede from each other, then the passenger in the car hears the frequency, given as
`n'=n_(o)((v-v_(o))/(v+v_(s)))`
`n'=380 ((343-26.4)/(343+34.5))`
`=380xx0.838`
`n'=318.44 Hz`
The change in frequency which is detected by the passenger sitting in the car is `=454.86-318.44~~136 Hz`