What is the frequency a stationary observer hears when a train approaches her with a speed of 30 m/s. The frequency of the train horn is 0.6

Question

What is the frequency a stationary observer hears when a train approaches her with a speed of 30 m/s. The frequency of the train horn is 0.600 kHz and the speed of sound is 340 m/s. A) 551 Hz B) 600 Hz C) 653 Hz D) 658 Hz

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Euphemia 5 months 2021-08-24T01:17:16+00:00 1 Answers 7 views 0

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    2021-08-24T01:18:43+00:00

    Answer:

    The frequency a stationary observer hears when a train approaches her with a speed of 30 m/s is 658 Hz (option D)

    Explanation:

    The Doppler effect is defined as the change in the apparent frequency of a wave produced by the relative movement of the source with respect to its observer. In other words, this effect is the change in the perceived frequency of any wave motion as the emitter and receiver, or observer, move relative to each other.

    The following expression is considered the general case of the Doppler effect:

    f'=f*\frac{v+-vR}{v-+vE}

    Where:

    f ’, f: Frequency perceived by the receiver and frequency emitted by the emitter respectively. Its unit of measurement in the International System (S.I.) is hertz (Hz), which is the inverse unit of the second (1 Hz = 1 s-1)

    v: propagation speed of the wave in the medium. It is constant and depends on the characteristics of the medium.

    vR, vE: Receiver and emitter speed respectively. Your unit of measure in the S.I. is the m / s

    ±, ∓:

    The + sign is used:

    • In the numerator if the receiver approaches the sender
    • In the denominator if the sender moves away from the receiver

    The sign – is used:

    • In the numerator if the receiver moves away from the sender
    • In the denominator if the sender approaches the receiver

    In this case:

    • f= 0.600 kHz= 600 Hz (being 1 kHz= 1000 Hz)
    • v= 340 m/s
    • vR= 0 m/s  because the observer is in a stationary state.
    • vE= 30 m/s
    • The emitter (the train) approaches the receiver, so the sign in the denominator is negative.

    Then:

    f'=600 Hz*\frac{340 \frac{m}{s} +-0 \frac{m}{s} }{340 \frac{m}{s} -30 \frac{m}{s} }

    Solving:

    f'=600 Hz*\frac{340 \frac{m}{s} }{340 \frac{m}{s} -30 \frac{m}{s} }

    f'=600 Hz*\frac{340 \frac{m}{s} }{310 \frac{m}{s} }

    f’= 658 Hz

    The frequency a stationary observer hears when a train approaches her with a speed of 30 m/s is 658 Hz (option D)

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