B Resonance of air between two loudspeakers

Tags:
1. Nov 13, 2016

resurgance2001

A student I am working with showed me a problem they have been working on.

In the first part of the question there are two loudspeaker facing each other. The student has been told that they can treat this situation as an closed tube. The frequency of the note is such that its wavelength is equal to the distance between the loudspeakers.

In this situation we determined that there are nodes at the speakers and another node midway between the speakers. The amplitude of the anti-nodes found 1/4 and 3/4 of the distance from one speaker would be twice the amplitude from one speacker. Would one accept that this is correct? At least if one is told to assume that the situation works as a closed tube. (I am not 100% convinced of this)

If one assumes that we asnwered correctly in the situation above, what would be the effect of reversing the polarity of the wires in one loudspeaker? We answered that this would make the 2 speakers exactly 180 degrees out of phase which would result in a diminishing of the volume at all places. Is that correct?

thanks

Peter

2. Nov 13, 2016

CWatters

I would expect an anti-node at the speakers and in the middle, with nodes at 1/4 and 3/4. Are you sure they were told to treat it as a closed tube?

I think that's correct for an anti-node at the speakers.

Last edited: Nov 13, 2016
3. Nov 13, 2016

resurgance2001

Thanks - I want to try to set this up and test it experimentally.

4. Nov 13, 2016

Cutter Ketch

I don't think you should think of this as the same as a closed tube. These are two independent sources, and really this is just a superposition, a simple addition, of two independent counter propagating waves. OK, That IS what happens in a closed tube, but I think the idea of the speakers is to get you thinking in terms of just adding counter propagating waves so you can understand that that is the right way to think of the tube. Also, in the tube they aren't independent. One is a reflection of the other with a fixed and definite phase relation. Here you get to see what other phase relations will do. So think in terms of making counter propagating waves independently and just adding.

Another interesting thing about this superposition, when the speakers are "in phase" they are actually 180 degrees out of phase. That is because the positive direction for one of them is to the right and the positive direction for the other is to the left. (facing each other you see). That is why you get nodes at the speakers when they are "in phase" and one wavelength apart. If they were actually in phase (both traveling left at the same time and right at the same time) you would get antinodes at the speakers not nodes.

However, just adding the two counter propagating waves together you will indeed get all of the features you describe in a good analogy to a closed tube. The lesson being that the standing wave in a closed tube is just the sum of the two counter propagating waves.

The fact that it isn't a closed tube is highlighted by what you do next. In a closed tube you certainly can't alter the phase relationship of the end reflections!

Changing the phase of one speaker by 180 degrees certainly does NOT give diminished volume at all places. The two waves are counter propagating and could never cancel everywhere. That can only happen if they propagate together in the same direction. In changing the phase all you do is move the nodes. As I mentioned, you had nodes at the speakers in the first situation not because the are in phase, but rather because they are perfectly out of phase. How else could the two waves add to zero? By reversing one speaker now they really ARE in phase (both moving left and right at the same time. You will now find double amplitude antinodes at each speaker, and you will now have nodes at 1/4 and 3/4 with a third antinode right in the middle. Effectively what you've done is shifted the phase of the standing wave.

In fact, if you could vary the phase of the speakers arbitrarily you could walk the nodes and antinodes, the whole standing waveform left and right any arbitrary distance. Note that that means the speakers don't have to be nodes or antinodes. Try THAT with a tube!

On the back side of either speaker where the sound from both are traveling toward you together you will find the global enhancement or diminishment you expected. However they won't completely cancel because the volume falls with distance ( here's where a tube would have been handy). Or you could put them next to each other facing the same way. That would make the cancelation dramatic. With them out of phase covering and uncovering one speaker should make a neat demo.

I strongly suggest you try adding counter propagating waves in a computer program. Make two counter propagating sine waves and add them and see what the standing wave looks like. Adjust the phase and see how the standing wave pattern shifts. Alternatively glue together 3 or 4 slinkies and try it with a student at each end waving in phase or out of phase or in between.

5. Nov 14, 2016

resurgance2001

Thanks - that's a bit clearer. As I said earlier I would like to try this out experimentally so I can see what is actually happening physically.

I don't suppose you know of any text resources I could look at? I have dug around a bit on the net but haven't found anything yet. Cheers

6. Nov 14, 2016

Cutter Ketch

I'm afraid not. I know practically every general physics text covers standing waves in a tube and mentions that it's the sum of counter propagating waves. However I'm not aware of anything beyond that.

I think once you see how two counter propagating waves make a standing wave it will all be clear. If you don't feel you can do it on a computer, try drawing sine waves on clear plastic sheets, moving them across each other and doing the addition by eye.

If you would like to plot it on a computer here are equations that will give you a wave propagating left and right.

A sin(2*pi*f*t + 2*pi*x/L)
And
A sin(-2*pi*f*t + 2*pi*x/L + ph)

You can even do this in excel if that is more comfortable.
Make a column of increasing numbers for x
Make a column of eqn 1 with t pointing to a single cell somewhere (using $R$C syntax in the formula) and a column of eqn 2 with the same cell referenced for t and another single cell referenced for ph (relative phase). Start with the ph cell set to 0
Make a column of the sum of the 2 columns
Plot all three columns vs x
Now change t to watch the two waves propagate and see the resulting standing wave
Put ph anywhere between 0 and 2 pi to see how relative phase affects the standing wave

7. Nov 14, 2016

CWatters

Google Rubens Tube if not already done. Most have only one speaker but someone must have built it with two.

8. Nov 19, 2016

resurgance2001

Thanks for both the last two replies. I am still working on this but slowly due to other commitments. I will let you know how I get on

9. Nov 20, 2016

sophiecentaur

There is really no significant "resonance" involved when two loudspeakers are feeding energy into a free space situation. This is because there is no build up of energy anywhere. That, to my mind, is the essential factor that the word Resonance implies. Before being challenged on this, I have to acknowledge that the physical space taken up by each speaker will cause a very small amount of reflected sound power to get back to its source and that could (when the spacing is right) produce an extremely low Q resonance. The standing wave / interference pattern, produced by two directional sources, pointing away from each other, would not involve any stored energy. All that happens is that the energy is channeled into certain directions and canceled in others.
The closed tube situation is very different if you assume minimal loss and, hence, well defined resonance peaks, with antinodes where energy builds up. I would suggest that the best way to study standing waves on transmission lines with multiple sources would probably be best done in the context of electrical transmission lines. Same sums and thousands of texts (I would expect) about theory, measurements and even practical implementations with transmitters.

10. Nov 20, 2016

resurgance2001

Yes that seems like a reasonable assumption. I am still not sure about what happens in any case when a wave reflects off the surface of a moving speaker. I would have thought that it should be treated as a soft reflection not a hard reflection. I am going to try to do an experiment now using ultrasonic transducers and an electret microphone. I think I will be able to set the whole thing up nicely on a table. I will post the results I get. I really like/insist on seeing things with my own eyes as far as possible. Cheers

11. Nov 20, 2016

sophiecentaur

By that, I imagine you mean a term like scattering coefficient? That would depend on the material and size of the speaker. You could get an estimate of that coefficient by moving a microphone around between the source and the test speaker (unpowered). If there is actually a standing wave (i.e. a measurable variation of sound level as peaks and troughs. The RF equivalent would be the SWR (Standing Wave Ratio) which would tell you the percentage of the incident wave that returns on the reciprocal path on a transmission line. That could be very small indeed unless the speaker being tested produces specular reflections from a large flat surface. A unity SWR would actually tell you there is no resonance - something worth checking on.

12. Nov 20, 2016

Cutter Ketch

I agree with Sophie: this is not a case of resonance. I would not waste your time thinking too much about the reflections at the speakers. There may be some, but that isn't really the point.

The speakers are meant to be an analogy to the tube. In the tube reflections at the ends allow a standing wave to grow in resonance so long as there are nodes at the end. We are told to think of this resonance as the sum of counter propagating waves: waves reflect off the ends and the standing wave is the interference of these counter propagating waves.

The speakers make this idea of counter propagating waves explicit, and, since you aren't relying on reflection and resonance, you are freed from the constraint of having nodes at the ends. You then can explore. You can see that independent counter propagating waves do give you standing waves. You can see that only particular distances and phase relations put the nodes at the speakers. And all of this helps you think about standing waves and resonance in tubes and on strings etc etc.

13. Nov 20, 2016

sophiecentaur

Actually, this could be a fairly difficult experiment to achieve with accuracy. The advantage of channeling the sound back and forth in a parallel sided tube make it very attractive but you would need to eliminate the effect of reflections at the remote ends of the tube, where the loudspeakers are each sited. You want to make sure that the source at each end is at the amplitude and phase of your choice and what actually comes out will be a combination of the speaker cone motion and the reflection of any incident power. You could probably achieve what I say by having a monitoring microphone at each end and an amplitude and phase control on each speaker to try to get what you want. But you could find yourself chasing your tail to set up any situation other than a symmetrical one. A way to deal with that could be to have an open ended tube with some damping in it, so eliminate any reflections and to feed the two sounds via small holes in the tube sides or small probes. That way, there would be no build of of sound energy in the tube and the Q of the system would be very low so it would not get unmanageable. Conical / horn shaped ends could also reduce reflections. Your advantage would be that you can choose just one frequency and tune the ends to eliminate reflections.

I suggested an RF approach, higher up the thread. It may be worth considering because there is so much more available information and suitable equipment. The two situations are very strongly analogous. But then, I would say that, wouldn't I, having spent my professional life at RF.