Doppler shift and change in intensity of a sound wave

  • #1
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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?
 

Answers and Replies

  • #2
Simon Bridge
Science Advisor
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What makes you think this is not correct?
 

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