Why do fifteen unconnected pendulums snake, split into rows and line up again?

For small swings, the period of a pendulum depends only on its length $L$:

For pendulum $k$ (counting from 0) to swing exactly $51+k$ times every 60 seconds, its frequency is $f_k=\frac{51+k}{60}$ hertz, and the length follows:

The longest is about 34.4 cm, the shortest about 21.2 cm. Once the fifteen lengths are set, every pattern that follows is set too; no pendulum "knows" what the others are doing.

All the pendulums are let go from the same side together, so pendulum $k$ sits at

The phase gap between neighbours is the same for every pair:

that is, 6° a second. Pendulum $k$ runs $k\Delta$ ahead of the first, so at any moment the fifteen positions are one cosine wave $A\cos(\varphi_0+s\Delta)$ read at the fifteen points $s=0,1,\dots,14$. The colours on the stage show this relative phase: same colour, same step.

At $t=\frac{p}{q}\times60$ seconds (with $\frac{p}{q}$ in lowest terms), pendulum $k$ is ahead by

which depends only on the remainder of $k$ divided by $q$, so the pendulums fall into $q$ groups, each in perfect step: 4 groups at 15 s ($\frac14$), 3 at 20 s ($\frac13$), 2 at 30 s ($\frac12$), 5 at 12 s ($\frac15$).

At 60 seconds pendulum $k$ has made exactly $51+k$ swings, all are back where they started, and the show begins again.

For whole numbers $k$, adding or removing a full turn from the gap makes no difference:

At 55 seconds $\Delta=330^\circ$, which lands on the same fifteen points as $-30^\circ$. The eye joins the dots with the gentlest wave, so it sees a slow snake 30° behind, swimming back, while the real wave is far denser.

That is why the pattern at $60-t$ seconds is the one at $t$ played backwards: $\Delta(60-t)=360^\circ-\Delta(t)$, the same as $-\Delta(t)$.

This is what the sampling theorem says: a wave is only seen correctly with at least two samples per wavelength ($|\Delta|\le180^\circ$). Wagon wheels turning backwards on film and the stripes when you photograph a screen are the same aliasing.