# Doppler shift and change in intensity of a sound wave

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

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

Edit I report an example exercise (I'm not looking for the solution, my doubt is conceptual and it is explained above)

A source emits a spherical sound wave at frequency $f=400Hz$ with power $P=1 W$ in a solid angle of $\frac{\pi}{4} sr$. An observer $A$ is at distance $R=228m$ and does not move, a second observer $B$ is at the same distance and moves with velocity $v_{B}=200 km/h$ towards the source. Determine the sound intensity level received by the two observers. Use speed of sound at $20 ° C$, $v_{sound}=343 m/s$.

Answer : $\bigg[L_{A}=73.9 dB \, \, , \, \, L_{B}=L_{A}+0.65 dB=74.5 dB \bigg]$

I have no problem for $A$ $$I_{A}=\frac{P}{\frac{\pi}{4} R^2}=2.45 \cdot 10^{-5} W/m^2 \implies L_{A}=10 Log \frac{I_A}{10^{-12}}=73.9 dB$$

But I do have problems for $B$. Using the formula proposed in my question I get the wrong result

$$I_{B} =I_{A}(\frac{343+55.55}{343})^2=3.31 \cdot 10^{-5} W/m^2 \implies L_{B}=10 Log \frac{I_B}{10^{-12}}=75.1 dB$$

I do not know why, but without squaring the ratio of frequecy I do get the correct result.

$$I_{B} =I_{A}(\frac{343+55.55}{343})=2.85 \cdot 10^{-5} W/m^2 \implies L_{B}=10 Log \frac{I_B}{10^{-12}}=74.5 dB$$

So I found a way to get the result but I do not understand why should not be correct to square the ratio of frequencies. Furthermore, in the answer the result is given as an adding sound level. I would really like to know how can one get that $+0.65 dB$ directly, so that one knows what to add to the result, without doing a lot of calculations.

• Why does that not seem correct? Jul 5 '16 at 16:55
• @KyleArean-Raines I found an exercise with a similar question, I used the formula above and found a wrong result. Furthermore the intensity is not only dependent on the (change of) frequency but it is also linked to the magnitude of the surface over which the power is spread and I think that the radius of the sphere may change because of the velocity of observer or source. Nevertheless I cannot really undestand this. Jul 6 '16 at 13:16
• Could you provide a reference to this exercise? Jul 6 '16 at 13:18
• @KyleArean-Raines I included it in the question Jul 6 '16 at 13:51
• 1st: Shouldn't the intensity of observer B be a function of distance from the object? 2nd: The Doppler-shift says nothing about changing intensity, just the frequency of the observed signal. One may hear a louder signal but that would be due to the non-constant frequency response of the human ear, not a consequence of a changing intensity. Jul 8 '16 at 12:27

In this type of problem one has to take great care in defining intensities. In this case there are 4 different intensities: 1. $I_{ss}$, the intensity received by the static observer as perceived by himself, 2. $I_{ms}$, the intensity received by the moving observer as perceived by a static observer. 3 $I_sm$, the intensity received by the static observer as perceived by the moving observer and 4 $I_mm$, the intensity received by the moving observer as perceived by himself.

You are trying to compare $I_{mm}$ to $I_{ss}$. These are intensities from two different reference frames and thus incomparable. You should be comparing $I_{ms}$ to $I_{ss}$ or $I_{mm}$ to $I_{sm}$.

The easiest way to see what happens to the intensity when one approaches a source is to compare it with someone shooting paint balls at two observers. One standing still and the other approaching the shooter. At $t=0$ the two observers are at the same distance from the shooter. After a time $\Delta t$, the static observer has received $N=\Delta t F$ paint balls., where the flux $F$ is the number of paint balls per second shot at the observer. The moving observer will have been hit by more paint balls, because during the time $\Delta t$ he has moved $\Delta x$ closer to the shooter. There are thus some paint balls which have already reached the position of the moving observer but not yet that of the static observer. The number of paintballs in mid air between the two observers at $t=\Delta t$ is: $N_{diff}=F\frac{\Delta x}{c}=F\frac{v_{obs} \Delta t}{c}$. The number of paintballs received by the moving observer is thus $N'=F \Delta t + F\frac{v_{obs}}{c} \Delta t$. The relative flux is thus $$\frac{F'}{F}=1+\frac{v_{obs}}{c}=\frac{c+v_{obs}}{c}.$$

The intensity $I$, the amount of energy per second is then given by multiplying the flux with the amount of energy per paint ball. The balls arrive at both observers with the same velocity. However, since both observers will assign a different value for this velocity, they will also perceive different intensities. Still, the ratio between the intensities as perceived by one observer, will be the same for both observers and is identical to the ratio of the Fluxes.

• Thanks for the clear answer! I think that in the case I reported in the question we are comparing $I_{ss}$ and $I_{ms}$, but I'm not sure. Furthermore I don't really get the difference between the moving and the static reference frame. If I got that right, in the end, intensity changes with velocity of the observer just as the frequency do, so $$\frac{I_{ms}}{I_{ss}}=\frac{c+v_{obs}}{c}$$ Is this true also for $I_{mm}$ and $I_{sm}$? That is $$\frac{I_{mm}}{I_{sm}}=\frac{c+v_{obs}}{c}$$ Jul 12 '16 at 12:58

Intensity is energy per unit area; over short distances, the intensity can be considered constant. In a time period dt, the sound wave travels a distance c.dt, so the total energy passing through an area A will be equal to the sound energy present in a volume cAdt. (think of water flowing through a pipe, if it travels at 10 m/s through a pipe with a 0.1 m cross section, the energy per second passing through is the energy of the water in 10 m pipe.)

The energy density $w(R)$ is the same for both observers since they are at the same distance R from the source. The total energy passing through area $A$ is for observer A: $I_AAdt=w(R)Acdt$, and for observer B: $I_BAdt=w(R)A(c+v_B)dt$ since observer B moves towards the source with velocity $v_B$

Which gives you $\frac{I_A}{c}=\frac{I_B}{(c+v_B)}$