acousticsresonancewaves
When two tuning forks stand near one another and one is excited, the other rings as well. When high notes are struck on a piano, lower notes are also heard. If I understand correctly, this is called sympathetic resonance.
What is the principle behind this effect, and how can it be described mathematically? I've seen it used in analogy with quantum entanglement–is such an analogy accurate?
Sound is a compression wave, that is the air vibrates in a direction along the direction of the wave. I found this image that shows you what a standing sound wave looks like in a half open pipe (nicked from this site). The black dots represent some small volume of the air:
The key point to note is that there are some points where the air is not moving, called nodes, and some points where the air motion is at a maximum, called antinodes.
At the closed end of the pipe the air cannot move, obviously so because the end of the pipe is in the way. This means there must be a node at the closed end. In contrast at the open end of the pipe the air can move, and in fact for a standing wave to form at an open end the wave must have an antinode.
So for a closed pipe, as in the picture above, we have a node at one end and an antinode at the other, and the distance between a node and antinode is $(2n + 1)\lambda/4$, where $n = 0$ gives you the fundamental, $n = 1$ is the first overtone and so on (the diagram actually shows the fifth overtone). For the fundamental frequency, $n = 0$, the length of the pipe is equal to $\lambda/4$ so the wavelength is four times the length of the pipe as you say.
If you take an open pipe the air will be moving, i.e. there is an antinode, at both ends. The distance between two antinodes is $n\lambda/2$, where this time $n = 1$ gives us the fundamental, so for the fundamental the length of the pipe is equal to $\lambda/2$ i.e. the wavelength is twice the length of the pipe.
If we look at the sonic boom as a $\delta$-function, where we have a really loud sound for a really short time, then it will be able to excite all frequencies at the same way.
You can actually compute this by showing that $$ \delta(t)=\frac{1}{2\pi}\sum_n e^{int},$$ which show how the $\delta$-function is actually composed of all frequencies.
Then it's actually in resonance with any object. However, due to its short lifespan it cannot feed more and more energy into an object (like a window), making the amplitude of the oscillation bigger and bigger until the object breaks.
What most likely will destroy something like a window is the actual pressure front, due to the pressure gradient.
Best Answer
The vibrations of a tuning fork cause vibrations in the air with the same frequency. This process is symmetric in time: if you happened to have vibrations in the air which matched the frequency of the tuning fork, the tuning fork could pick them up and start to vibrate. A second, identical tuning fork is a good way to produce vibrations in air with the correct frequency.
You get the same thing with the strings in well-tuned piano, but there is a difference. The strings associated with each key on a piano have a different fundamental frequency. If you play the mid-range keys on a piano with the middle pedal pressed (so that the low-range strings aren't damped), you don't hear the fundamental frequencies of the low strings; you hear their harmonics, which are the same frequencies as the fundamentals of the higher strings. (The harmonics of the higher strings also excite the higher harmonics of the lower strings, but that's a smaller effect.)
There are so many garbage analogies floating around in the popular literature on quantum entanglement that I'm going to leave the last part of your question unaddressed.