# Superposition of sound waves

1. Sep 8, 2014

### Jackson Lee

The superposition of sound waves puzzled me. Just like the figure below, when two loudspeakers propagate sound waves from two locations to any other locations. We always calculate final wave's amplitude or intensity via considering something just like phase difference or change of amplitude, but we never take into account the angle between the propagation directions. Sound wave is longitudinal wave, thus I suppose the superposition of amplitude should be the superposition of vectors. If it is so, then the final intensity calculation should be done this too. However, it seems that few textbook ever did this. If it is not sound waves but light wave or other electromagnetic waves, it is definitely correct because they are transverse waves. When we calculate final amplitude, we could neglect angles. But we could do this for sound waves????

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2. Sep 8, 2014

### olivermsun

For linear waves, for which the amplitudes are assumed to be "small," then longitudinal vs. transverse doesn't matter. The final amplitude is just whatever you get out the superposed waves at the end point.

3. Sep 8, 2014

### AlephZero

For sound waves in fluids, we define the intensity to correspond to the change in pressure. Pressure is a scalar quantity. It doesn't have any "direction" that needs to be considered, or ignored, so to find the effect of combining waves, you just add up the pressures with the correct phase relationships.

If instead we defined intensity to correspond to the changes in velocity of the fluid, then you would have to take the directions of the velocities into account. That would be more complicated. For a combination of waves the velocity would continuously change direction as well as magnitude, and usually the magnitude never becomes zero. An individual "particle" of fluid would not vibrate back and forth along a straight line, but move around an elliptical path in space.

The relationship between pressure and velocity changes with the sound frequency, and the pressure-based definition of intensity matches the sensitivity of human hearing better than a velocity-based definition would. That is another reason for using it, as well as the fact that it is simpler to work with.

For sound waves in solids, which can be transverse (shear) waves as well as longitudinal waves, in the general case you have to consider the 6 components of stress at a point, not a single scalar pressure value, and you do need to take the directions into account when superimposing different waves.

Last edited: Sep 8, 2014
4. Sep 8, 2014

### olivermsun

Minor note: This doesn't happen because sound "Intensity" is defined (on purpose) as the positive definite quantity of power per unit area, which is only dependent on the pressure (I ~ P2).

The superposed vector velocities be, as you say, more complicated. The velocity-dependent quantity that behaves most like the sound intensity might be kinetic energy density (where KE ~ u2 + u2).

Indeed, you get mixed longitudinal-tranverse motions with gravity waves in fluids (without shear). Strangely enough, wave superposition still holds, but it respects the entire elliptical motion.

5. Sep 8, 2014

### Jackson Lee

Sorry, I was still quite confused. From the view of geometry, it seems the angle between the two waves should be taken into account no matter the amplitude is small or not. Look at the figure below, the sound spreads outward from each of two sources as spherical wave.(I draw only several lines to represent all other waves.)
When two waves interfer at point B, you will find their directions are obviously different. If the angle between them become larger, will final amplitude remain constant? This is longitudinal wave, isn't it? Thus their net displacement will definitely involve vector superposition. Then how could we neglect angles?

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6. Sep 8, 2014

### olivermsun

Because, as AlephZero said above, the intensity at a point is a function of pressure, and pressure is a scalar.

The vector angle will affect the pattern of pressure created by the superposed waves around the point.

Also, the longitudinal displacements are in different directions… but the same would be also be true of transverse waves with different propagation directions. So longitudinal vs. transverse is not really an issue here.

7. Sep 8, 2014

### Jackson Lee

Is pressure a scalar? Not points from source?

8. Sep 8, 2014

### olivermsun

The pressure in a fluid acts in every direction equally, so yes.

9. Sep 8, 2014

### Jackson Lee

So, why couldn't the final pressure at the point B to be a vector superposition?

10. Sep 8, 2014

### olivermsun

I guess I don't really understand what you're asking. The pressure at B is the superposition of the pressures due to the two waves. The pressure itself at B has no directionality.

The angle between the two rays is taken into account when you determine the difference in path lengths.

11. Sep 8, 2014

### Jackson Lee

No, you seems got what I mean, maybe I misunderstand other part. Does the loudspeaker（below）'s pressure on B points from loudspeaker to point B? I suppose this because pressure is force divides area. Besides, pressure here seems a little different from pressure in static liquid, because the fluid is moving. The static pressure is same in all directions, but it is not for kinetic pressure(the second term of Bernoulli's equation).

12. Sep 8, 2014

### Jackson Lee

Sorry, I was confused. Why does pressure has no directionality? Does the pressure applied by loudspeaker below on B point from loudspeaker to point B?

13. Sep 8, 2014

### olivermsun

There is not a steady flow (as you might think of in a Bernoulli equation sense) where there is a pressure gradient with the speaker at the center and flow emanating outward (including pointing toward B).

Instead what you have is an oscillatory displacement that is driven by oscillatory pressures -- basically, concentric spherical "shells" of alternating high and low pressure are radiating outward from the speaker -- with zero net displacement of the air, averaged over cycles.

The shells have a definite outward propagation, hence at B there is a definite "direction" to the propagation of pressures and displacements. However, if you measure the sound at B with a microphone, which is just the opposite of a speaker, then you will pick up the (oscillatory) pattern of acoustic pressure exerted on the microphone diaphragm. There is no "directionality" to the pressure, i.e., the pressure does not point from the speaker to B. However, the pattern of pressure waves that reach B are certainly propagating in a direction which points from the speaker toward B.

14. Sep 9, 2014

### Jackson Lee

Thanks a lot to both of you. I got it.

Last edited: Sep 9, 2014
15. Sep 10, 2014

### Jackson Lee

Hey, guys.There is something else maybe we need to delve again. If we go to extreme, for example, the two loudspeaks above are put face to face, while the microphone is put in the middle point between them. (Actually, this is the condition of angle between them 180 degrees while path difference is zero.) Then the wave will be the standing wave, for the energy from both sides are same. Even though the amplitude could be twice of the original one, the net averge energy intensity is zero now, does it appropriate to still define it as constructive interference or destructive interference just by phase difference or wave amplitude? And is the result of standing wave shown by microphone same as constructive interference?

16. Sep 10, 2014

### olivermsun

Along the axis between the speakers, the superposed waves will form a standing wave.

At the center point, the net energy FLUX, which is a vector quantity, is zero. However, there is perfect constructive interference, and thus the amplitude of the pressure will be doubled at that point, with quadrupling of the energy intensity.

17. Sep 10, 2014

### Jackson Lee

You mean the result of standing wave shown by microphone is same as two traveling waves. But the power intensity is zero, eh? Because it is divided by average net power to area.

18. Sep 10, 2014

### olivermsun

A standing wave IS the superposition of two, opposing traveling waves, so the result recorded by a microphone has to be the same.

The net energy flux is zero because one vector flux is +1 and the other is -1 in order to form the standing wave.

The (sound) pressure intensity at the center would not be zero: if the pressure at the center due to one wave is exactly p(x0, t), and the pressure due to the other wave is also p(x0, t) because the path lengths are the same. The total pressure would therefore be 2*p(x0, t), and since I $\propto$ p2, the total intensity would be four times as large.

19. Sep 10, 2014

### Jackson Lee

Isn't the intensity power to area ratio? Besides, if the wave is not sinusoidal, does the I~P^2 still exist?

20. Sep 10, 2014

### olivermsun

Well, properly speaking the intensity of a plane wave is, as you say, the energy flux per unit area normal to the propagation, so basically I~pv. This works even if the wave is not sinusoidal.

There is a strange consequence for the standing wave, since if you place a test (normal) surface at the center position, then both superposed waves contribute positive energy fluxes in their respective directions of propagation.

The conventional usage I have seen in cases where there is superposition is just to treat call the intensity at the location some function of the pressure squared (usually relative to a reference intensity).

Maybe someone here knows of a different convention for the intensity of a wave field