# "Standing Waves" at Arbitrary Frequencies

(Although this is a similar question as Standing waves arbitary frequency and Standing waves on string, but I believe that the answers do not do enough justice to the questions, so here is a more in-depth elaboration on the question, and hopefully a more in-depth response could be received)

## Introduction:

This question started from my confusion on EM waveguides, on how come EM waves get reflected if the width of the waveguide is too narrow (I might ask this question at some point in the future, but this will not be the main point of this one). As I went down the rabbit hole, it became obvious that the answer lies in the nature of standing waves.

If you have studied waveguides, you should know that since the wavelength of the EM wave is too short, it is not enough to form a standing wave in the waveguide, which leads to the main point of this question:

What happens if we produce waves at arbitrary frequencies in a system with fixed/closed boundaries on both sides (ψ(0,t) = ψ(L,t) = 0)?

To simplify the problem, let us only consider a string wave. Now I know a lot of you, upon looking at the question, would just go down to the answer section and say, "resonance only happens at quantised frequencies, and any other frequencies would produce a jumbled mess", but that is exactly what I want to know, what does the "jumbled mess" look like?

## Derivation by Reflection of Waves:

Before jumping into the more professional way of solving this problem using Fourier Analysis, let me start with the more simple way to solve it.

The simplest way to explain the formation of standing waves is by visualising a travelling wave travelling towards a fixed boundary, and then getting reflected. The reflected wave is π radians out of phase with the incident wave and travelling in the opposite direction, superpositioning/interfering with the incident wave, so as to satisfy the condition ψ(L,t) = 0.

As the reflected wave reaches the other boundary, the condition ψ(0,t) = 0 is also satisfied and so no more reflection is present if it is vibrating at their resonating frequencies.

So let us consider now a string wave not vibrating at its resonating frequencies. If we continue from the reflected wave reaching the other fixed boundary as described above, we would see that the condition ψ(0, t) = 0 is not satisfied and therefore another reflection is needed.

And so here comes the problem, the second reflected wave simply makes the wave function become discontinuous, making the solution infeasible. A demonstration of the above can be seen here in this desmos graph (three instead of only two reflected waves are simulated in this graph).

So why does this method not work? And is there a way to actually compute what happens?

## Derivation by Fourier Analysis:

Now let us move on to the more formal way of solving these kinds of problems: Fourier Analysis. The question in this section is actually pretty simple, as the method of using the fact that every wave pattern can be modelled with the sum of standing waves vibrating at its own frequencies given the initial condition is well-established.

The question is simply this: What should the initial condition be for this problem?

For a normal problem like plucking a guitar string, it can be well-defined that the boundaries are fixed initially, but for this case, it is hard to do so, as the problem technically starts with one end of the string not having a fixed boundary, while after a while, that boundary should become fixed. Even if we try to set the initial instant as the instant when that boundary becomes fixed, we would have no idea what the function looks like at that instant. And so, we have no idea what the initial condition should be.

## Summary

To summarise, my main questions in this post are:

1. Why does the derivation by reflection result in a discontinuity and stop working?
2. What is the initial condition for the derivation by fourier analysis method?
3. If none of the above methods work, what other successful ways can we model this phenomenon?

Hopefully, someone has the time to read through this long-winded question and has the knowledge to answer it.

• In your paragraph starting "The simplest way", how did you produce a travelling wave in one direction without also producing a travelling wave in the other direction? What mechanism are we using to introduce energy to the system here? Commented Nov 17, 2021 at 6:06
• The energy is introduced from one of the sides (let’s say it’s the left side with reference to the Desmos graph) using a vibrator, and the vibrator stops when the reflected wave reaches the vibrator. Commented Nov 17, 2021 at 8:05
• The travelling wave travelling in the opposite direction should only form when the wave travelling to the right hits the fixed boundary on the right. You can refer to the Desmos graph for reference. Commented Nov 17, 2021 at 8:27
• Consider that when you generated the wave, the "leading edge" of the wave must have been a point with displacement zero. Then the reflection of that point would also have displacement of zero. Then there can be no discontinuity as shown in your graph. Commented Nov 17, 2021 at 16:06
• I see what you mean, but I’m afraid that the problem is that what you mentioned only applies to the first reflection. The reason why the second reflection does not work is because although the first reflected wave has zero displacement in its “leading edge” when it reaches the left boundary, the wave travelling to the right (the incident wave) does not have zero displacement at the left boundary when the first reflected wave arrives, therefore, the second reflected wave does not start with zero displacement in order to satisfy the condition ψ(0, t) = 0. Commented Nov 19, 2021 at 2:40