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A source of sound emitting sound with fr...

A source of sound emitting sound with frequency 540Hz approaches a staionary observer with a speed of `30 ms^-1`. If velocity of sound is taken as `330 ms^-1`, the frequency as heard by the observer will be

A

194 Hz

B

294 Hz

C

394 Hz

D

594 Hz

Text Solution

Verified by Experts

The correct Answer is:
D
`n'=V/(V-V(s))xxn`
n=540Hz,`V=330ms^-1`,`V_s=30ms^-1`
`n'=330/(330-30)xx540`
`330/300xx540`
n'=594Hz
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