- #1
Soren4
- 128
- 2
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?
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?