Standing waves: How are the waves not respecting the boundary conditions "cancelled" from the normal mode?

Question

I'm asking this question because I did not find any explanation in textbooks.

Consider a string fixed at both ends with an impulse provided to it (like in a guitar). The impulse can be described using Fourier integral and contains all the possible frequencies (with their weights). Nevertheless only the frequencies that satisfy the condition $$f=n \frac{v}{2L}$$ will "survive" and all the other components will disappear in a short time, after some reflections.

I would like to have an approximate, but quite clear, idea of how the wave components not satisfying to $(1)$ will disappear after being reflected some times.

Consider the situation in the picture, which is the case of $n=2$ standing wave (in black) and the red one and green one are the incident and reflected wave.

Because of $(1)$, in this case the incident and reflected wave add up to zero in both ends of the rope at any time $t$ (this is indeed the condition from which $(1)$ is derived).

Anyway, if this does not hold, i.e. the incident and reflected wave do not add up to zero in one of the ends of the rope, what happens to the wave?

In my view these two things must hold anyway:

1. The end of the rope is physically fixed so it won't move in any case
2. Since the end does not move, the reflected wave is always reflected upside down. So also the components not obeying $(1)$ will be reflected up-side down

But this does not explain why the components not satisfying $(1)$ eventually disappear. And it also looks like point 2. This is quite absurd since, if the waves are reflect upside down (like in picture), but independently from the fact that they satisfy condition $(1)$ or not, it seems to me that they should add up to zero at the ends of the rope in any case.

But this is surely incorrect and I'm quite confused on what happens to the components of the wave not obeying $(1)$, so any explanation is highly appreciated.

Answer 1

you are right in stating that if we ignore damping all types of waves will keep reflecting forever.

However, if we add damping, the situation changes. The amount of damping depends on the frequency of the oscillation. Only the low frequency oscillations will survive.

Typical excitations of a guitar string, do not have a single frequency however: they are a mixture of different frequencies. The only excitations that do have only a single frequency are those having a sinsoidal shape and obeying the boundary conditions.

Coincidentally, all other excitations can be described as a sum of these basis components using a mathematical trick called the Fourrier series. Thus whatever the shape of your string is, it can always be described as a sum of sinusoidal excitations obeying the boundary conditions.

Now since all the high frequency components of this sum are strongly damped, you are left with the low frequency components, whatever your original excitation might have been.

Answer 2

The reason you don't get a continuous frequency spectrum is because of the boundary conditions at the ends of the string.

Suppose you start the string moving with an arbitrary displacement and velocity at time $t = 0$.

Suppose the string has a finite length $L$ and call the displacement of the string $u(x,t)$ for $0 <= x <= L$ and $0 <= t$. If the ends of the string are fixed, $u(0,t) = u(L,t) = 0$ for all $t$.

Since the length is finite, you can describe any initial displacements and velocities with a discrete Fourier series with terms $$ u(x,0) = \sum_{k=1}^\infty a_k \sin \frac{\pi kx}{L}$$

The motion continues as a sum of forwards and backwards travelling waves, all moving at the same speed $c$ (the d'Alembert solution to the wave equation).

The individual terms in the above Fourier series produce standing waves whose frequencies are all multiples of the fundamental frequency $f_0$, i.e. $f_k = kf_0$ for $k = 1, 2, \dots$.

Answer 3

You are assuming that all conceivable frequencies are present at the beginning (when you pluck the string), and then over time misfitting frequencies vanish. Your assumption is incorrect since ends of the string are held fixed right from the beginning, and any frequency that would not give zero displacement at string ends (equivalently that which do not satisfy the condition you have mentioned) simply will not be present, right from beginning.