# B Moving to a higher harmonic in a standing wave

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1. Nov 8, 2016

### Saw

Imagine that you have plucked a string and it is vibrating as a standing wave at its fundamental tone (frequency f1). You leave it there and later on come back with the intention of bringing it up to the second tone (frequency f2). What should you do? It seems obvious: apply a stimulous oscilating at such second frequency (f2). But what happens then with the energy of the first vibration oscilating at f1? Is it necessarily wasted? Is there no way to benefit from such energy? Can't you build on the existing vibration and apply a stimulous of frequency (f2 - f1)? For example, if I arrive with my arm vibrating at (f2 - f1) and then grasp the string that was already vibrating at f1, will it somehow be raised to vibration level f2?

2. Nov 8, 2016

### Staff: Mentor

If f2 is the frequency of the second harmonic and f1 the fundamental frequency, then what is the value of f2-f1?

3. Nov 8, 2016

### Saw

Well, the difference between f2 and f1... Say f1 = 3 and f2 = 6, then (f2 - f1) would be 6 - 3 = 3.

4. Nov 8, 2016

### Staff: Mentor

Exactly, so any excitation at f2-f1 is simply excitation of the fundamental.

5. Nov 8, 2016

### Saw

Ah..., so frequencies would add up? I had thought that that way I would only increase the amplitude of the original wave. And If instead of moving to an overtone, I just wanted to increase the amplitude of the fundamental one, how should I physically proceed?

6. Nov 8, 2016

### marcusl

When you study vibrations mathematically, you will learn that each mode is independent (their wavefunctions are "orthogonal"). Thus you can put energy into one, the other, both or neither, but what you do in one won't affect the other.

This is true for an ideal string, which is a linear system. Energy converts between modes only in non-linear systems, or in systems specially prepared with some kind of coupling (cross-moding in waveguides, e.g.).

7. Nov 9, 2016

### Saw

Following this guidance, I have read the wiki on normal modes (https://en.wikipedia.org/wiki/Normal_mode), which says: "The most general motion of a system is a superposition of its normal modes. The modes are normal in the sense that they can move independently, that is to say that an excitation of one mode will never cause motion of a different mode. In mathematical terms, normal modes are orthogonal to each other."

So I gather that the answer to my question would be negative at least for such linear systems: there is no way that I could take advantage of the energy of a given mode so as to "excite" it to a higher mode by supplying only the corresponding energy "difference".

If so, when we talk about the excitation of a mode, what do we mean: just setting such mode in motion through a driving force with the right frequency? Or maybe also an increase in the amplitude of the standing wave at such frequency/mode?

Any example of this you could guide me to?

8. Nov 9, 2016

### marcusl

That is correct.

These are different ways of describing the same thing, or rather, the second follows from the first.

There is a vast literature--just Google "waveguide mode conversion" and "waveguide mode coupling". Here's an open-source (i.e., free) article on RF waveguides, as one example:

You'll also see a huge literature on mode conversion in optical fiber guides, relating to the internet and transfer of voice and data across the globe.

9. Nov 9, 2016

### Saw

Thanks a lot, I have looked at that literature, though so far it seems only "inspiring" as it's too advanced for me to follow now. Looking for a simpler model I have thought that I could focus on harmonic oscilators, which are after all a simplified version of standing waves. In this sense, when talking about standing waves, in lecture 49 section 4 (http://www.feynmanlectures.caltech.edu/I_49.html) Feynman discusses the system of two "coupled pendulums" (two pendulums joined by a spring). I think he is saying that the system may have two modes, one where the spring is unstretched and the two pendulums oscilate as if without it, and another where the spring "contributes a restoring force and raises the frequency". I still don't understand how that can physically be done, how this little toy operates but is it relevant for my question? Should I study this system as an example of "coupled" modes or are they also "orthogonal"?

10. Nov 10, 2016

### Saw

I have seen the system mentioned by Feynman in an animation in this site:

I realize now that what is "coupled" is the componentes of the system. But the system has two modes with different frequencies which however would be "normal"...

Anyhow, this book, Optical Guided Waves and Devices, R.R.A.Syms and J.R.Cozens (http://www.gbv.de/dms/ilmenau/toc/110355911.PDF) has given me an orientation. If I understand it well, I don't need to look for examples of coupled modes from the origin. The example can be initially normal modes, which however become coupled due to some influence.

The relevant text is in chapter 11 and it refers to the cases of stimulated and spontaneous emission in these terms:

"we must remember that the photon is essentially a wave train of electric field with some characteristic frequency, and that the electron states E1 and E2 are simply normal modes of the unperturbed atom. If these modes are perturbed by interacting with an oscillating field, then mode coupling will occur strongly if the frequency of the field matches the difference in frequency between the two modes. This is clearly analogous to the mode coupling situations encountered in Chapter 10, where a spatially periodic perturbation was used to phase-match two propagating modes.

Since the new photon has been emitted in response to the stimulation of the driving field, the phases of the incident and stimulated photons must be linked. In fact, they are identical, so that the two photons are effectively joined together in a single wavetrain. We may describe this as a 'doubly occupied photon mode'. This effect, known as stimulated emission, is the key to the amplification of light in optoelectronic devices. Spontaneous emission can also be interpreted in terms of coupling between higher and lower states, except that in this case the stimulating perturbation is not derived from an external agency (an incident photon) but from fluctuations in the local field within the atom itself."

I think this more or less answers my question: what couples two modes and allows the frequency/energy transition is an influence filling the frequency gap between the two modes. I would just need to visualize how that happens...

11. Nov 15, 2016

### Saw

I am now however coming back to "coupled oscillators" as a good ortientation to the question of the OP. I understand now how the two-pendulum system operates. Its energy is shifting from one pendulum to the other. One separates a first pendulum from equilibrium and releases it, so that it starts oscillating but is progressively damped by the other, which is "stealing" its energy until the first comes to an instantaneous halt and the second reaches maximal amplitude, only to suffer ultimateley the same fate and so on. This can be viewed as the superposition of the two above mentioned normal modes, which by the way have only slightly different frequencies. The same happens in acoustics when sound waves of close frequencies superimpose, first destructively, later on constructively. In the latter case it is said that the ear perceives such combined note at the average frequency of the two waves. In both cases (sound and pendula), the amplitude oscillates at a rate of change equal to the frequency difference between the two modes, which is called the "frequency beat".

The question would then be: what happens if I submit the coupled oscillators system to a driving force oscillating at the beat frequency? Will the amplitude oscillation increase in a typical phenomenon of resonance? And how could that physically happen, how would that driving force operate over the system?

12. Nov 15, 2016

### sophiecentaur

If you 'pluck' a string, it will take up many different modes because the modes relate to the fourier transform of the shape of the string, which would be triangular. To get it to oscillate at any individual frequency, you can either start it off with a sinusoidal shape (difficult) or excite it with a single tone and with a low power signal - allowing it to build up in amplitude.

13. Nov 15, 2016

### Saw

Thanks sophie, I assume that the string is already vibrating, for whatever reason, with one or several (superposed) modes (standing waves since the string is closed) at the fundamental tone (f1) or successive overtones (f2, f3). The question was whether it was possible, through a driving force with frequency (f2 - f1), excite mode 1 into mode 2 and so on. I have learnt that in a linear system, where modes are normal, that is not possible: the new wave thus created will be again superposed (combined) with the others in terms of amplitude but it cannot alter the frequency of any existing waves.

So I have re-formulated the question in another way. We imagine that the superposed waves are just two, with different frequencies, f1 and f2. In the beats phenomenon, those frequencies would be very close and their difference (the "beats difference") very small. But I suppose that the question can still be formulated equally and the answer is the same, no matter the size of such "beats difference". The question is then: if you apply a driving force oscillating with the "beats frequency" (f2 - f1), what happens? I tend to believe that the answer is simply that you increase the amplitude of the envelope. This would be equivalent to (in the similar coupled oscillators example = two pendula joined by a spring) to giving a little push to pendulum 1 every time that a cycle is completed. Could anyone confirm that this is right?

14. Nov 16, 2016

### sophiecentaur

You mean that they are both very high overtones? 12th and 13th, perhaps? In this situation, you would be dealing with a pretty non-ideal situation. Any argument should also apply to a situation with the fundamental and the first overtone. (The idea of just "sliding" form one overtone to the next would still apply.)

15. Nov 16, 2016

### sophiecentaur

Do you mean here that the energy at one frequency would turn up at another frequency by supplying energy at a third frequency? If the system is linear then how could this happen? If the exciter is at any frequency f' then it can't do anything but to cause the oscillator to move at f'. If there is a natural mode in the oscillator at f' then you can get a resonance at f'.
There is an important factor that is not often considered in this sort of discussion and that is the Q of the resonant response. There is always some resistance / friction in any real oscillator. If there were not, there would be a constant build up of energy, if the exciter is exactly at f, with no limit. In fact we always find that the loss will limit the total energy held in the oscillator when the loss equals the added power from the exciter. A real resonator will have a response curve and it will be possible to excite it at a frequency other than the centre response frequency. With a Q of 100, you get a significant response at +/- 1% of the centre frequency. See this link which relates to a simple RLC resonator. When you turn off an off-centre excitation, the 'off centre' oscillations will just die down but cannot change frequency.
All this relates to classical, linear oscillators. What goes on in a quantum system doesn't need to be the same at all - in QM, you have non linearity and an energy source 'within' the system. Are you trying to bring the two together? I don't think that's valid.

16. Nov 16, 2016

### Saw

Yes, I was told that above by marcusl and it is assumed in the way that I have re-formulated the question in post #13.

Ok, although in principle this is not related to the question. We can assume that the oscillator is ideally undamped.

The question is only if impinging upon an oscillator (the two-pendulum system described above or a complex sound wave made up of two waves with different frequencies) with a force oscillating at the "beat difference" (difference between the frequencies of the two modes of the coupled oscillator or the two waves of the complex sound wave) is of any use. As commented, the answer seems obvious: you can increase the amplitude of such thing that is vibrating precisly at that "beat frequency", which is the amplitude of the envelope embracing the two coupled oscillators or the complex wave. You can modulate it, so to speak. I just wanted now confirmation that this is ok.

But even in that case, the nonlinear oscillation can be resolved into components, into harmonics by Fourier theorem, right?

No, I am just trying to conclude if you can do anyhthing with the above mentioned "frequency difference" in a classical system. It is true however that, in a QM system, a photon for example (external energy source) excites the atom also at a "frequency difference". But we had better not get distracted with the QM here...

17. Nov 17, 2016

### Saw

I think I mixed things there. Maybe I should have said the following: whilst a linear oscillation can be decomposed into harmonics, a nonlinear oscillation can be decomposed into an infinite series of sine and cosine functions? Or how else?

18. Nov 17, 2016

### Staff: Mentor

I think this is the crux of the "problem." If you take a quantum harmonic oscillator, it has a single frequency $\omega$. When you excited it, you are always exciting it at $\omega$, not at a "frequency difference." What is quantized is the energy, not the frequency.

19. Nov 17, 2016

### Saw

Well, I didn't want to talk about the QM analog, but what you say...

My understanding is that both at classical and QM level you have simple oscillators with 1 degree of freedom (and one natural frequency), coupled oscillators with 2 or more degrees of freedom (and correlative frequencies) and waves with plenty of degrees of freedom (and frequencies).

Also that at quantum level, when an electron transition happens, because a photon is absorbed or released, the frequency of such photon (which is tracked by way of absorption or emission lines) multiplied by Planck's constant must correspond to the energy difference between energy levels. Mathematically, that looks identical to saying that such frequency must be equal to the frequency difference between frequency levels.

20. Nov 17, 2016

### Staff: Mentor

That was my point: you are confusing the frequency of the quantum oscillator and that of the photon related to the transition. The transition appears at the frequency of the oscillator, not at a difference of frequencies in the oscillator. It is best to think of it in terms of energy even when considering the photon, and only convert (if needed) to the frequency of the photon afterwards.