At an air show a jet flies directly toward the stands at a speed of 1230 km/h, emitting a frequency of 3140 Hz, on a day when the speed of s

Question

At an air show a jet flies directly toward the stands at a speed of 1230 km/h, emitting a frequency of 3140 Hz, on a day when the speed of sound is 342 m/s.

a. What frequency (in Hz) is received by the observers?
b. What frequency do they receive as the plane flies directly away from them?

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Thông Đạt 6 months 2021-07-29T23:46:16+00:00 2 Answers 17 views 0

Answers ( )

  1. giabao
    0
    2021-07-29T23:47:33+00:00
    Reply

    Explanation:

    Given that,

    Speed of the jet, v_s=1230\ km/h=341.67\ m/s

    Frequency of the sound produced by jet, f = 3140 Hz

    Speed of sound in air, v = 342 m/s

    (a) The formula of the observed frequency is given in terms of Doppler’s effect as :

    f'=f(\dfrac{v+v_o}{v-v_s})

    v_o is the speed of observer

    f'=3140\times (\dfrac{342}{342-341.67})\\\\f'=3254181.8\\\\f'=3254\ kHz

    (b) Let f’ is the frequency received as the plane flies directly away from them. So,

    f'=f(\dfrac{v+0}{v-(-v_s)})\\\\f'=f(\dfrac{v+0}{v+v_s})

    f'=3140\times (\dfrac{342}{342+341.67})\\\\f'=1570.75

    Hence, this is the required solution.

  2. minhkhoi
    0
    2021-07-29T23:47:49+00:00
    Reply

    Answer:

    a. f_o=1570.75\ Hz

    b. f_o=3254181.82\ Hz which won’t be audible to the humans as the audible capacity of the human ear is from 20 Hz to 20,000 Hz.

    Explanation:

    Given:

    • speed of jet, v=1230\ km.hr^{-1}=341.67\ m.s^{-1}
    • speed of sound in air, s=342\ m.s^{-1}
    • frequency of sound, f=3140\ Hz

    a.

    Since the sound source is moving therefore we can determine the observed frequency by the Doppler’s effect given as:

    \frac{f_o}{f} =\frac{s-v_o}{s+v}

    where:

    f_o= observed frequency when the jet plane approaches the observers

    v_o= observer’s velocity = 0

    \frac{f_o}{3140} =\frac{342-0}{342+341.67}

    f_o=1570.75\ Hz

    b.

    When the plane flies away from the observers then we take its velocity as negative.

    \frac{f_o}{f} =\frac{s-v_o}{s+v}

    where:

    f_o= observed frequency when the jet plane flies away from the observers

    v_o= observer’s velocity = 0

    \frac{f_o}{3140} =\frac{342-0}{342+(-341.67)}

    f_o=3254181.82\ Hz which won’t be audible to the humans as the audible capacity of the human ear is from 20 Hz to 20,000 Hz.

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