The question asked about guitar strings and resonant cavities. I don’t feel like doing much math for this – the wave equation which approximately describes the motion of the guitar string is covered in any freshman physics book, it’s good to know, but I don’t have much to say about it.

Instead, we’ll simply use the result that the pitch made by a guitar string depends on a few things:

• the thickness of the string: A thicker string has more inertia, and hence vibrates more slowly, making a lower pitch.
• the tension on the string: Higher tension means it takes more energy to deform the string, and so the string also has higher kinetic energy, and moves faster. Higher tension will result in a higher pitch
• the length of the string: The speed of a wave propagating down a string is an intrinsic property, meaning it depends only on what the string is like at some particular location, not on how big the string is as a whole. That’s because when a wave is propagating, it doesn’t even know how long the string is. With that in mind, a long string can support a long wavelength. It takes more time for a long wave to pass, and so on a long string the frequency goes down.

With this in mind, we can use a dimensional analysis to guess that the frequency of a guitar string is proportional to $\sqrt{\frac{T}{L M}}$.

This will let us find the interval between two strings of the guitar. We’ll take whichever string is lower and go down the frets until the two strings have the same pitch. Now measure the length of the lower string when your finger is pinching it, and its natural length. The ratio of the pitches between the two strings is $\sqrt{\frac{L_{pinch}}{L_{natural}}}$.

A string half as long vibrates with twice the frequency, because its length is half as much and so is its mass, and both these figures are inverse-square-root proportional to the frequency. So by lightly putting your finger down on the middle of the guitar string, it will act as two independent strings half as long, and the pitch will jump an octave.

It is also clear now why the lower-pitched strings are thicker. They must be thicker, longer, or have less tension (or some combination). Thicker is a good way to go.

Next we’ll consider the cement tube. If it’s closed at both ends, it acts very much like a guitar string. The waves are of pressure rather than displacement, but they are still well approximated by the same wave equation.

The boundary condition is that the derivative of the pressure at the wall be zero. The pressure can be as high as you want – the wall can take it. However, if there is some pressure gradient, the air next to the wall would bounce off it and readjust until it killed off that gradient. So, if we imagine the wall as stretching from $l=0$ to $l = 2\pi$, then instead of the sine functions used to model the displacement of a guitar string, we’d use cosine functions to model the pressure in the air.

We could measure the length of the tube by finding a resonant frequency (a note which sounds very loudly), then increasing the pitch until we found another one. The resonant frequencies are given by $\frac{nv}{l}$, with $n$ some integer, $v$ the speed of sound, and $l$ the length of the tube. With two adjacent resonant frequencies we know
$\begin{array}{rcl}f_1 & = & \frac{nv}{L} \\ f_2 & = & \frac{(n+1)v}{L} \\ f_2 - f_1 & = & \frac{v}{L} \\ L & = & \frac{v}{f_2-f_1}\end{array}$

If an end of the tube is open, the boundary condition is that the pressure does not deviate from equilibrium there, because the pressure will be the same as in the open air outside. So we’d describe the pressure by sine waves again. However, this wouldn’t make a difference to what the resonant frequencies are. It would make a difference at where the nodes are, so that the same pitch might resonate better at a different spot along the length of the tube.

If the tube had one open and one closed end, we’d be mixing the two boundary conditions. This would let us fit a quarter-wavelength wave in the tube, as opposed to the half-wavelength minimum when the boundary conditions are identical. The next higher frequency would fit 3/4 a wavelength in the tube. This is why an instrument like a clarinet, which has one end closed and the other open, “jumps a twelfth”, meaning that when you play a harmonic on it, the frequency is three times as high.

As for the metronome, someone suggested to me measuring the length of a cavity by shouting and timing the echo. For short tubes, I thought it would be better to adjust a metronome until the echo fell exactly with the next tick, as this would be more accurate than starting and stopping a watch by hand. However, yesterday afternoon I spent some quality time crawling around in a sewer on campus in hopes of finding a good resonant cavity. I discovered that in general the echoes from the metronome app on my iphone were much too weak to pull this procedure off.

Finally, for the ear, it is really fantastic. You basically have a little biological fourier-analyzer inside you. If you haven’t heard about how it works, you should definitely look it up online.