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marysya [2.9K]
2 years ago
5

A trumpet player on a moving railroad flatcar moves toward a second trumpet player standing alongside the track both play a 490

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? Take the speed of sound to be 343 m/s.
Physics
1 answer:
labwork [276]2 years ago
4 0

Answer:

The value is v_s  =  1.394 \  m/s

Explanation:

From the question we are told that

The frequency of the second player is f_2  =  490 \  Hz

The beat frequency is f_b  =  2.0 \  Hz

The speed of sound is v_s   =  343 \ m/s

Generally the frequency of the note played by the first player is mathematically represented as

f_1  =  f_2 + f_b

=> f_1  =  490 + 2.0

=> f_1  =  492 Hz

From the relation of Doppler Shift we have that

f_1  = \frac{ f_2 (v+ v_o )}{v-v_s }

Here v_o\  is\ the\ velocity\ of\ the\ observer\ with\ value\  0 \ m/s

So

492   = \frac{ 490 (343+0  )}{343 -v_s }

=> v_s  =  1.394 \  m/s

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Answer:

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