I've seen a lot of great videos demonstrating how resonance in a tube depends on some "alignment" between the wave frequency and the tube length.

Example: https://www.youtube.com/watch?v=bHdHaYNX4Tk

I also think I understand how waves get reflected once they hit the end of the tube and that the interference is responsible for the formation of standing waves.

What I don't understand is what happens when the wave frequency and the tube length don't match. In other words, what's preventing the sound from getting amplified in the experiment above?

I wrote an animation to try to understand it, but you'll see that no matter what wavelength or tube length we choose, a standing wave always gets created. Why are certain frequencies, then, special?

enter image description here

What am I misunderstanding here?

EDIT: per Mike's answer I've updated my simulation which behaves more like I would expect.



It's the phenomenon of destructive interference. When you add up the reflections, you get the sum of waves with random relative phases, and this averages to zero.

Imagine shooting a wave through a rod of length $L$ and watching it so that you can see the wave reflect off the edges. Each reflection adds to the amplitude at an individual point on the rod. After $N$ reflections, the amplitude at a point $x$ is something like:

\begin{align} A &= \sum_{n=0}^{N/2} \left[ e^{ik(x + 2nL)} + e^{ik(L-x + 2nL)}\right ] \end{align}

where $k$ is the wavenumber (the left complex exponential is for the forward propogating wave, the right term is for the backward propogation). Note that this can be simplified:

$$ A = \left[ e^{ikx} + e^{ik(L-x)}\right ] \sum_{n=0}^{N/2} e^{ik(2nL)} $$

Let's focus on the sum term on the right. If the wave lines up where the length $L$ constitutes an integer $m$ full periods, then $k$ has the form $2\pi m /L$ and

$$\sum_{n=0}^{N/2} e^{ik(2nL)} =\sum_{n=0}^{N/2} e^{i(2\pi m/L)(2nL)} = \sum_{n=0}^{N/2} e^{i4\pi mn} = N/2 $$

Since $4nm \pi$ is an integer multiple of $2\pi$. Therefore

$$ A = \frac{N}{2} \left[ e^{ikx} + e^{ik(L-x)}\right ] $$

If NOT, I will argue that this sum is 0 because then $e^{ik(2nL)}$ is just a random number on the complex unit sphere. On average, these should go to zero, see illustration.

enter image description here

  • $\begingroup$ Oh yeah, that's definitely something my simulation is missing - the wave will keep reflecting back from both ends. Let me try implementing this idea! $\endgroup$ – kunigami Apr 7 at 23:47

The tube is like a collection of harmonic oscillators (or a multi-dimensional harmonic oscillator so to say). Each standing wave represents a single harmonic oscillator that has its own so-called eigen-frequency, meaning that the harmonic oscillator will vibrate with only this frequency, if left on its own.

But, real harmonic oscillators are also damped, which means that the eigen-solution of the oscillator is not purely sinusoidal, but a sine/cosine that is damped exponentially. For example, the movement of the air in the tube is subject to viscosity, which dissipates energy. Or, for a tube that is open on one end, sound is radiated into the environment (the tube is actually coupled to the open environment), which you will hear, and this again will draw energy from it.

This exponential damping, however, means that the "left-alone" vibration of each of the oscillators is not a single "spectral line" (frequency), but actually a band of frequencies that reaches from zero to infinity, and has only a more or less pronounced resonance amplification close to the pure sinusoidal frequency part (mathematically: the real frequency part). In the spatial domain, the damped eigen-solutions will most likely resemble much the non-damped standing waves, but the details depend on where the damping occurs locally. It will be different for a radiation damped tube with an open end, compared to a closed, viscosity damped tube.

As with any other harmonic oscillator (or rather any linear vibration system, like a pendulum, or a mass-spring-damper), instead of leaving it alone, you can also excite it externally with any single frequency you like, or with an impulse, or with any other non-sinusoidal force. Due to the frequency-widened nature of the damped eigen-solutions, the tube will respond to the external excitation at all contained frequencies. Only close to the real eigen-frequencies will the response amplitudes be higher than elsewhere. The less damping the tube (or one specific oscillator) has, the higher will be the response close to the resonance compared to further away from it. But it will always be finite if damping is non-zero.

Practically this means something like, if you speak into a damped tube, you will usually hear almost all frequencies that are already contained in your voice. Some will be attenuated, however, while others will be amplified. (those close to the eigen-frequencies). What happens in the time domain, is that the response amplitude to the excitation will rise until it reaches an equilibrium more or less quickly (often much quicker than noticable), and if you stop the excitation (the talking) the amplitude will decay (exponentially) again. It is much like what you experience in an echo chamber on a larger scale. Each of the standing waves will get excited with an amplitude as a function of time depending on how your voice excites the tube in time. One could say, that your voice gets "projected" to the standing waves, which represent a whole band of frequencies each.

Now what makes things complicated, when people think about standing waves, like you did as a cause for your question, is that they neglect the damping. An undamped oscillator is unexpectedly a little more complicated than you might think. Besides exact eigen-frequencies/standing waves vibrating forever when the tube is left alone, and off-resonant frequencies in the excitation, the response to an additional sinusoidal excitation that exactly matches one of the eigen-frequencies of the tube (or more specifically a single harmonic oscillator), is going to have linearly rising amplitude, until forever, so there is no equilibrium as for the damped oscillator (because the excitation energy accumulates more and more in the resonator). This is what is commonly called resonance-catastrophe because it will eventually lead to the destruction of some part of the system. This will rarely happen with airborne sound because the goal is most often, to hear what is produced (guitar, trombone), which causes radiation damping as sketched in my second paragraph, and the coupling to the solid enclosure will be comparatively weak due to the different densities of air/solid. Only if you build the tube out of a brittle material, like glass, you could expect a resonance catastrophe under reasonable conditions.

If you talked into such an undamped tube, you would quickly hear a deafening sound of standing-wave eigen-frequencies at ever higher volume, while every frequency content that is not "eigen" would stay at a level that is insignificant compared the resonances. Hence, you would hear only the standing waves after a short period of time. Because of that, you would not be able to recognize your voice's character anymore, nor even the words that were spoken.


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