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A trumpet player on a moving railroad flatcar moves toward a second trumpet player standing alongside the track both play a 660 Hz note. The sound waves heard by a stationary observer between the two players have a beat frequency of 2.0 beats/s. What is the flatcar's speed?

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Answer:

Since the beat frequency is 2/sec and one of the players is moving towards the stationary observer (he will hear 660 / sec from the trumpet player that is also stationary), the player moving towards the observer will produce a frequency of 662/sec as heard by the observer.

f' = f * v / (v - vs) where vs is the speed of the source

We can use 331 m/s as the speed of sound in dry air

f' v - f' vs = f v rearranging

vs = v (f' - f) / f'

vs = 331 m/s * (662 - 660) / 662 = 1 m/s

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