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Here is a problem I have been asked that I do not know the answer. Consider two ideal wave generators (it can be sound generator or whatever) separated by a distance L and facing each other. At t=0 they start to emit a wave. Basic calculus show a standing wave is created. So the energy flux through any surface on the path of the wave is in average zero. In the same time, the generators continuously feed energy into the system. Question: where does the energy emitted by the generators go ?

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You have a standing wave between the two generators. Are the individual waves supposed to travel beyond the generators once they reach the length L (in which case they move away) or are they supposed to be reflected on the opposite generator (in which case you don't need to feed the idea wave anymore to keep the phenomenon)? – NikolajK Jul 19 '12 at 9:31
It is a thought experiment, you can of course use a reflecting wall at one end instead of two generators. If you imagine a string attached to a wave generator as it is done in any layman classroom, it is clear that when the reflected wave hits the generator, it does not stop shaking up and down... – Shaktyai Jul 19 '12 at 9:54
Second generator aside, what I was implying is that if the waves are ideal, they will just oscillate and once reflected will interfre with whatever waves comes from the other side. Then in this case, what exactly happens will have to do with the wavelength to L ratio. And if everything is reflecive as you say, I see no need to keep on generating waves for ever after the sum of a certain number of wavelengths excedes 2L say. – NikolajK Jul 19 '12 at 10:11
So logically, if the generator keeps pouring energy into the system, we should expect the amplitude of the standing wave to blow up. If the amplitude is bounded then the wave somehow swictches off the generator. We start to be pretty far away from the classical "standing wave" presentation.... – Shaktyai Jul 19 '12 at 10:19
If you increase the power to the speakers/wave generators then the amplitude will start increasing, but as the amplitude increases the losses will increase as well. At some point the losses will increase to the point where they are equal to the input power. – John Rennie Jul 19 '12 at 10:25
up vote 1 down vote accepted

I agree with you that the the issue can be specified theoretically without general losses.

The error in your reasoning is that you assume an actuator has a dynamic nature independent of the wave medium, so that you'd always pour energy into each actuator and thus a paradox ensues since, as you note, a standing wave pattern in a medium has a zero energy transport.

In the limit of a perfect medium where you create the standing wave (or any other wave in general), for example with two actuators and an ideal wire, you get a coupling of the two actuators which disallows you to lose energy in them in total, that is, any voltage drop due to driving current in speaker/actuator A will present as a voltage gain in speaker/actuator B (acting as a microphone) and vice versa. More specifically in the standing-wave situation both ends will lose all driving impedance they had before.

That is, without losses, the setup in total is just an ideal transmission line.

In the setup with one actuator and one perfectly reflecting wall, the sole actuator will lose all its impedance and be infinitely easy to drive.

There are related questions that usually arise about energy conservation and "complete" destructive interference, with the accompanied confusion of thousands of forum posts :)

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Thanks for your answer. That was indeed the most logical answer, though I am curious about the feedback effect of the wave on the actuator in a true experiment. I guess it's just destroying it. – Shaktyai Jul 19 '12 at 15:51
Yeah I'm not sure either, I mean obviously there will be losses at many points.. My answer was built on a purely mathematical analysis of standing waves with an ideal wave-equation. A real loudspeaker has mass that has to be accelerated back and forth and just doing that costs energy so it won't swing "for free", but maybe there is a measurable reduction in driving energy due to the air's feedback effect (in case of a speaker sending the energy to a reflecting wall). Would be an interesting physics experiment! – BjornW Jul 19 '12 at 22:35

The standing wave relies on the travelling waves being reflected at each end of your distance, and this reflection is never 100%. In addition to this you get viscous losses because the propagation of the wave requires shear motion in the fluid i.e. the non-zero velocity gradients.

In the case of your loudspeakers, hi-fi loudspeakers are specifically designed not to reflect sound, so without an energy input your standing wave would be lost very quickly. I would guess this energy loss is far greater than viscous losses in the air.

In the case of water waves, most of us must have discovered that it's easy to set up a zero mode standing wave in a bath (usually to our mother's dismay) and this wave will persist for quite a while. In this case the main energy loss would be viscous losses where the water moves over the walls of the bath and viscous losses in the bulk. I would guess the biggest velocity gradients are at the walls so most of the energy would be lost there.

Response to comment: if you split your standing wave into it's left moving and right moving components there is only no net energy flux if the two components have the same amplitude. However the energy losses due to viscosity mean that the amplitude of the left moving wave decreases with distance to the right, and conversely for the right moving wave. This means there is a net energy flux.

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In my opinion, losses are not the good explanation. By increasing the power output of the generator, one should be able to overcome them. Or more simply, we can use an electric or light generator. A simple laser beam with a mirror would do the trick. So the answer lies probably in the way standing waves are taught or thought about. – Shaktyai Jul 19 '12 at 10:03
Fill your bath and swish your hand to and fro to generate a standing wave. Now stop swishing your hand and you'll see the standing wave decay due to energy losses. Alternatively stand between two walls and clap your hands. You'll hear the echo decay because of energy losses. In both cases, to maintain the wave you need to supply energy, just as you describe in your question. Where else would the energy be going if not into losses? – John Rennie Jul 19 '12 at 10:30
Why do you insist to introduce losses ? The spirit of the question is to determine wether or not the zero energy flux within a standing wave is compatible with a non zero energy flux at the boundaries. It is obviously not and as usual with paradox there is a mistep in the reasoning. I am just asking help to figure it out. – Shaktyai Jul 19 '12 at 10:55
Ah, sorry, I misunderstood you. I think I know what you're asking and I've added a note to my question. Please comment again if I still haven't understood you properly. – John Rennie Jul 19 '12 at 13:55

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