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

Question

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.

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Thiên Ân 2 months 2021-07-22T12:29:30+00:00 1 Answers 6 views 0

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    2021-07-22T12:31:11+00:00

    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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