Vibrating string: why exactly these harmonics?

Question

I read the following (informal) description of how harmonics arise when a string is plucked:

How can you be sure that your guitar string is indeed just one continuous string, rather than two half strings, each half as long as the original one, but seamlessly joined. (...) [E]ach of these half-strings weighs half as much and is twice as stiff as the whole string, and therefore each half-string will have a resonance frequency that is twice as high as that of the whole string. When you pluck your guitar string, you make it vibrate and play its note, and the string must decide whether it is to vibrate as one whole or as two halves; if it chooses the latter option, the frequency at which it vibrates, and the sound frequency it emits, will double! And the problem doesn’t end there. If we can think of a string as two half-strings, then we can just as easily think of it as three thirds, or four quarters, and so forth. (Schnupp, Nelken, King, Auditory Neuroscience (2012), p. 8-9)

This sounds like a nice intuitive explanation (for someone like me who has not been exposed to a formal treatment of this problem), but it raises one obvious question: why should the string "split up" only in equal size segments? It could also split up in, say, two strings of length 3/7 and 4/7. (The 3/7 part would be strongly excited if the string is plucked at 3/14 of its length.)

So why then do exactly these harmonics arise? Are the "non-equal splits" initially present in the transient but then die out due to destructive interference, while the known harmonics are sustained by constructive interference?

Answer 1

Trying to explain this at the same level as the excerpt you've provided:

Think about a junction point where two portions of the string meet. It's constantly "feeling" the influence of each side, pulling it up and down as each side vibrates. If the two portions of the string were oscillating at different frequencies (which would be the case if the portions were different lengths), then there would be a net force on the junction point at most moments of time.

But this junction point is just an infinitesimal point, and it doesn't have a mass of its own. Since the net force on this point will be mass times acceleration, and mass is zero, there cannot be a net force on the junction point. The only way for this to happen is for each side of the string to be pulling in opposite directions on the junction point at all times. And this implies that both sides of the string are vibrating at the same frequency, and therefore that the portions on each side of the junction have equal length.

Answer 2

The string divides itself in this way because the endpoints are fixed, they cannot move. These are the boundary conditions of the problem. Say you divided the string into $3/4$ and $1/4$. That means that there's a node at a point $3L/4$ of the string. But the only way that this node can exist while having two nodes at the ends of the string is for the wavelength to be different between the two regions, and this would imply a different vibratory frequency for each point in the string. However, all the points of a standing wave oscillate in phase and with the same frequency, so this cannot happen.

For a more mathematical explanation, start acknowledging that a standing wave is the superposition of two "equal" waves propagating in different directions. Therefore

$$\eta_1 = \frac{A}{2}e^{i(kx-\omega t)} \ ; \ \eta_2=-\frac{A}{2}e^{-i(kx+\omega t)} \to \eta = \eta_1+\eta_2 = A \sin(kx)\sin(\omega t). $$

The boundary conditions imply $\sin(k_nL)=0$, which can only be true for $k_n=\frac{n\pi}{L}$. Since $k=2\pi/\lambda$, this implies $\lambda_n=2L/n$. The wavelength of each harmonic is $2L, \ L, \ 2L/3$, etc. In other words, you are dividing the string into one, two, three, four...