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I Doppler shift and change in intensity of a sound wave

  1. Jul 4, 2016 #1
    How are the intensity of a sound wave and the Doppler shift of frequency related togheter?
    That is, if the source or the observer are in relative motion, how does the intensity change?

    For a sound wave $$I=\frac{1}{2} \rho \omega^2 A^2 c=2 \pi^2 \rho f^2 A^2c$$

    (##c## is sound speed, ##\rho## is density of air, ##A## is amplitude)

    So, since Doppler effect is only about ##f##, I would say that

    $$I'=I \bigg(\frac{f'}{f}\bigg)^2=I \bigg(\frac{c+v_{oss}}{c+v_{sorg}}\bigg)^2$$

    But I don't think that this is correct, can anyone give suggestion about this?
     
  2. jcsd
  3. Jul 5, 2016 #2

    Simon Bridge

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    What makes you think this is not correct?
     
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