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26 May, 22:31

A bird is flying directly toward a stationary bird-watcher and emits a frequency of 1420 Hz. The bird-watcher, however, hears a frequency of 1456 Hz. What is the speed of the bird, expressed as a percentage of the speed of sound

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  1. 26 May, 23:09
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    2.47 %

    Explanation:

    We are given;

    Frequency emitted by source (bird); f_s = 1420 Hz

    Frequency heard by observer; f_o = 1456 Hz

    From doppler shift frequency, we know that;

    f_o = f_s [c / (c - c_s) ]

    Where c_s is speed of source which is the bird and c is speed of sound.

    Thus;

    Rearranging the equation, we have;

    f_s/f_o = (c - c_s) / c

    f_s/f_o = 1 - (c_s/c)

    Plugging in the relevant values to get;

    1420/1456 = 1 - (c_s/c)

    0.9753 = 1 - (c_s/c)

    1 - 0.9753 = (c_s/c)

    (c_s/c) = 0.0247

    Since we want it expressed im percentage,

    Thus, (c_s/c) % = 0.0247 x 100 = 2.47 %
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