physics - doppler effect

Two identical tuning forks can oscillate at 329.6 Hz. A person is located somewhere on the line between them. The speed of sound in the air is 343 m/s. Calculate the beat frequency as measured by this individual under the following conditions.
(a) the listener is standing still and the tuning forks both move to the right at 30 m/s
(b) the tuning forks are stationary and the listener moves to the right at 30 m/s

I tried using:
f' = f [(v +/- vD)]/[(v +/- vS)]
f' = (329.6)[(343 - 0)/(343-30)]
f' = 31.6 Hz

Is this correct?

The results will be slghtly different for the two cases. You have the correct formula but did not use it properly.
For case (b), the two frequencies received are
329.6(343/313) and 329.6(343/373) = 361.2 and 303.1 and the beat frequency is 58.1 Hz.
For case (a), the frequencies are
329.6(313/343) and 329.6(373/343) = 300.8 and 358.4m and the beat frequency is 57.6 Hz.

ok thanks a lot!

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asked by COFFEE

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