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When you pluck a guitar string, initially the vibration is chaotic and complex, but the components of the vibration that aren't eigenmodes die out over time due destructive interference. This ostensibly explains why the harmonics quickly become dominant.

The problem with this model is that there infinitely more non-eigenmodes than eigenmodes. (If you formalize the space of all possible modes, the subspace of eigenmodes will have measure 0.) So the naive mathematical extrapolation suggests that if the non-eigenmodes cancel themselves out, only an infinitesimal part of the original energy remains. Obviously this is not the case, given that the initial energy from a pluck is finite (actually pretty small), and we can only hear vibrations above some threshold of energy.

So why do the harmonics constitute a significant fraction of the total energy? Does it have to do with the discreteness of the universe (particles having nonzero size and mass)? Or is it that vibrations that are close-to-but-not-exactly eigenmodes also die out slowly?

Qmechanic
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silver
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3 Answers3

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First of all, one should notice that the dichotomy "eigenmode $-$ non-eigenmode" is a false one. The correct statement is that any generic motion of the string can be decomposed into superposition of natural vibration modes of the string (eigenmodes).

Second, in the course of dissipation, the form (i.e. the set of participating modes) of vibration is changing not because of destructive interference. Interference by definition is a linear phenomenon and therefore cannot affect energy distribution between eigenmodes. In order to decay, one needs real dissipation.

In my understanding your observation has a different interpretation. First, when you pluck your string, you excite many different modes in it (the main tone plus overtones). Now, as long as your system can be treated as linear, after this point on, each modes evolves independently. Obviously, each mode has its own frequency but in addition, generally speaking, each mode has its own dissipation/decay rate. In other words, different modes dissipate different amounts of energy to the environment per unit time. Naturally, the faster modes produce heat faster, therefore, dying off quicker. Therefore, after a while, only the mode with the lowest dissipation rate survives. In a well-designed instrument, this lone survivor mode is the main tone of the string.

John
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    Seeme like GPT-4 has given me false info RE intereference. boo! But I still don't think your reply resolves the problem fully. Even if we ignore non-eigenmodes, you can rephrase the question as why there is a significant part of the energy in the first harmonic (or the first few), given that there are infinitely many. In fact, the decomposition thing alone doesn't seem that meaningful since there are countless other ways (other than harmonics) to form a basis such that everything can be decomposed into elements of that basis, only no one uses them. – silver Oct 09 '23 at 15:14
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    @silver When you pluck a string, you displace it at a point, which roughly has the shape of the fundamental. In other words if you decompose the initial displacement into Fourier components, the low-order terms will be dominant in the spectrum. See: https://physics.stackexchange.com/a/111916/11645 – llama Oct 09 '23 at 22:56
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    The physics part is correct, but the "well-designed instrument", not so much. String instrument making goes to great lengths to support or suppress particular harmonics. Harmonics such as 1/2, 1/3, 1/4, 1/6 or 1/8 are welcome companions of the main tone (up to a point). Also notice that the "preferred" place to pluck at is rarely (except for the harp) in the middle of the string, which is where you'd go if the harmonics were considered harmful. – Jirka Hanika Oct 10 '23 at 12:18
  • @silver If you actually have a guitar, try for yourself that the tone contains relatively more of the lower overtones the more your initial displacement of the string looks like the fundamental, for example if you plug the string right in the middle, where the initial displacement looks like an isosceles triangle which looks "closer" to a pure sine, compared to plugging it close to the bridge, where the triangle has widely different side lengths i.e. more overtones in the Fourier decomposition, which leads to the tone being very bright. – Koschi Oct 10 '23 at 12:50
  • Actually plugging all the tones exactly at half of the respective string length can lead to so little overtones it may does not really sound guitar-like as everyone is used to anymore. This leads back to the last sentence of John's answer that a well-designed instrument does NOT mean we only need the fundamental note, as @JirkaHanika pointed out. But besides that, great answer!! – Koschi Oct 10 '23 at 12:53
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A string vibrating at exactly one frequency can still have finite energy. This can be derived from considering the kinetic energy of the moving string.

In a real string, it doesn't matter what basis you use to decompose the frequencies, each frequency carries a finite piece of the total energy. The more frequencies you have, the smaller each piece is.

Once you approach a continuum of frequencies, there's still no problem in a real system. Frequencies that aren't exact harmonics don't die instantly, so there is always some $\Delta f$ you could integrate over to obtain your energy. Look at the Fourier spectrum of any stringed instrument: the peaks are sharp, but not infinitesimally so. There is energy in some width around each harmonic.

HiddenBabel
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When you pluck a guitar string, initially the vibration is chaotic and complex, but the components of the vibration that aren't eigenmodes die out over time due destructive interference

This is a misconception. The vibration is initially only as complex (or non-complex) as later during the tone†. The string moves the whole time according to some partial differential equation; it so happens that eigenmodes provide an efficient way of solving this PDE (because the PDE is to good approximation linear), but you can just as well look at it completely in time-domain. The string shape starts out as an asymmetric triangle (cf. Why do harmonics occur when you pluck a string?), then wobbles its way to the other side, then back. In the ideal case (infinitely thin perfectly elastic no air infinite-impedance end stops), it will then return exactly to the original shape. Rinse and repeat.

The string shape can at all times be described as a superposition of the harmonic eigenmodes, because these form a basis of the space of all possible shapes the string can have. But that doesn't really mean it "is" the superposition of these harmonics, you could also expand the shape in any other basis, such as Legendre- or Chebyshev polynomials. Only, these aren't eigenbaseis, and therefore you couldn't easily predict how the expansion evolves over time. The sense in which the shape "is" the expansion in sinusoidal harmonics in particular is that these are the overwhelmingly most convenient basis, so we choose it. Once we've chosen it, any talk about "non-harmonics" is meaningless – the harmonics are the only functions there are in our basis.‡

This post contains some animations that may illustrate the point.

†Actually it is true that the vibration gets simpler over time, but this has nothing to do with non-harmonics. Instead, it is because higher-frequency harmonics tend to get damped more quickly than low-frequency ones, so at the start you have many more harmonics contributing to the movement than later on. But they're still all harmonics.

‡If you added more functions to it, the system wouldn't be linearly independent anymore and therefore the expansion ambiguous. This has nothing to do with the physics of strings, it's just a matter of how the mathematics behind "expanding an element of a vector space in some basis" works.

leftaroundabout
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