Pendulum Waves - Mathematical Description

Everyone likes bling, so let's start this post off with some eye-candy (skip to 20s if you've got very little patience):

That was pretty damn cool. Even if you're not admitting it, I know you're thinking it.

Lately, that video has been posted and reposted all over the internet. Now, don't get me wrong, I love spreading the joy of physics to no end. For those unsatisfied with simply seeing the video however, this post will be dedicated to presenting the nitty-gritty mathematical details of this marvelous phenomenon. If you would like to skip the math and just get some specifications to making your own pendulum wave machine, proceed here.

This is how the apparatus works in a nutshell:

The period of one complete cycle of the dance is 60 seconds. The length of the longest pendulum has been adjusted so that it executes 51 oscillations in this 60 second period. The length of each successive shorter pendulum is carefully adjusted so that it executes one additional oscillation in this period. Thus, the 15th pendulum (shortest) undergoes 65 (51 oscillations + 14 pendulums from the first) oscillations. When all 15 pendulums are started together, they quickly fall out of sync—their relative phases continuously change because of their different periods of oscillation. However, after 60 seconds they will all have executed an integral number of oscillations and be back in sync again at that instant, ready to repeat the dance. [1]

Let's begin. In a typical undergraduate physics course, you have most likely come across the (somewhat useless) travelling wave equation:

\[y(x,t)=A\cos\left(kx+\omega t+\phi\right)\]
where $A$ is the amplitude, $k=\frac{2\pi}{\lambda}$ is the wave number, $\omega=\frac{2\pi}{T}$ is the angular frequency, and $\phi$ is the phase shift. By convention, $\lambda$ and $T$ are used to denote the wavelength and period of the wave respectively. Lets also call the number of oscillations the longest pendulum goes thru $N$ (51 in this case) and the total time for the entire "dance" $\Gamma$ (60s in this case).

From the video, we can see that at $t=0$ and for any $x$ (we pick $x=0$), we have the maximum amplitude, $A$:

\[y\left(0,0\right)=A\cos\left(k\cdot0+\omega\cdot0+\phi\right)\]\[A=A\cos\left(\phi\right)\implies\phi=0.\]
Another thing we can see is that because the waves change shape in time, the wavelength must be a function of time and so the wave number (since it depends on $\lambda$) must also be a function of time. So instead of a fixed $k$, we have $k(t)$. Another thing to note is that since each individual pendulum is adjusted to give one more oscillation than their longer neighbour, the period and therefore the angular frequency must be a function of location, so we have $\omega(x)$.

Putting it all together, we have the equation:

\[y(x,t)=A\cos\left(k(t)x+\omega(x)t\right).\]
Now lets think for a moment. If $k(t)$ and $\omega(x)$ were independent of each other, the combination of some random $k$ and $\omega$ probably would not create the pretty patterns we see in the video. Aha! We have to conclude that $k(t)$ and $\omega(x)$ are some how working together - that these two functions are not independent of each other. More specifically, the cause to $k$ changing with time is due to constructing the machine so that the oscillations of each adjacent pendulum differs by 1 (i.e. due to tuning $w$ with x). With a bit of handwaving, we can choose to express one function in terms of the other and essentially throw away one of them. Our wave equation can now be modified to either one of:

\[y(x,t)=A\cos\left(k_{0}x+\omega(x)t\right)\]\[y(x,t)=A\cos\left(k(t)x+\omega_{0}t\right).\]
The constants $k_{0}$ and $\omega_{0}$ are used to describe the wave at $t=0$ and $x=0$ respectively.

Now the two expressions above are really the same, since they're both very general functions of $x$ and $t$. This means we can pick our choice of which equation to use.

Since I don't like overly long, dry, and boring posts, I'll end this post right here and continue the derivation in my next post. If you feel like doing the derivation yourself, go for it! It's a great way to learn.