Pendulum Wave Period

Question

Recently I've seen various videos showing the pendulum wave effect. All of the videos which I have found have a pattern which repeats every $60\mathrm{s}$.

I am trying to work out the relationship between the period of the overall pattern and the difference in length between each of the pendulums.

From the small-angle approximation for the period of a simple pendulum, for each pendulum with period $T$ we have:

$$T\approx2\pi\sqrt{\frac{L}{g}},$$

Where $L$ is the length of the pendulum and $g$ is the acceleration due to gravity.

If we take (say) $n$ pendulums, with lengths $L,L+d,\dots,L+(n-1)d$, then the pattern repeats at a period $t$ when all of $\frac{t}{L}\in\mathbb{Z}^{*}$, $\frac{t}{L+d}\in\mathbb{Z}^{*}$, $\dots$, $\frac{t}{L+(n-1)d}\in\mathbb{Z}^{*}$. where $\mathbb{Z}^{*}=\mathbb{N}\cup\{0\}$.

However, I'm not sure how to develop a direct relationship between $t$, $d$ and $L$.

Answer 1

Taking equally spaced lengths will not reproduce the pendulum wave effect, simply because it's the periods that you want to fulfil commesurability conditions, and lengths and periods are not linearly related.

To get a pendulum wave, say you have $N$ pendula with lengths $l_n$. Then, as you know, the $n$th pendulum will have a period $$T_n=2\pi\sqrt{\frac{l_n}{g}}.$$ For the pendulum wave to work you need to prearrange a time $T$ at which all the pendulums come back into step. (Most demonstrations choose $T=60\text{ s}$, but this is not necessary.) For this to happen, $T$ needs to be an integer number of periods for each pendulum, so therefore $$\boxed{nT_n=T.}$$ (This presupposes that we're using the number of periods in time $T$ to index the pendula. This is not strictly necessary but simplifies things, and people usually choose some relatively narrow range in $n$.)

Given the above relationship between $T_n$ and $l_n$, the commesurability condition reads $$ l_n=g\left(\frac{T}{2\pi n}\right)^2 $$ in terms of the lengths $l_n$. Here $n$ should be an integer with $n\geq1$, but most demonstrations take $10\lesssim n\lesssim 30$ or so. Using very slow pendula, with low $n$, would mean the ratio of lengths between the first and the last pendulum would come out rather large.

Answer 2

The question is equivalent to this: given the physical state of the system at time $t_0$, $S(t_0)$, for what $T$ will $S(t_0)=S(t_0+T)$? That is, for what $T$ will $$S_0(t)=S_0(t+T),$$$$ S_1(t)=S_1(t+T)$$$$...$$ hold, where $S_i(t)$ is the substate (namely, $\theta$) of the $i$th pendulum? Note that, letting $k=\frac{2 \pi}{\sqrt{g}}$, it is known that

$$S_0(t)=S_0(t+k \sqrt{L})$$ $$S_1(t)=S_1(t+k \sqrt{L+d})$$ $$S_2(t)=S_2(t+k \sqrt{L+2d})$$ $$...$$

When $S(t_0)=S(t_0+T)$, all the substates $S_n(t)=S_n(t+T)$ are pairwise equal.

Evidently, $k \sqrt{L+id}|T,(i \in [0,n])$ follows from this. Or, letting $w=\frac{T}{k}, \sqrt{L+id}|w,$ $(i \in [0,n])$ .

There is not necessarily a solution for $w$, for example in the case where $L=5, i=0,1, d=2,$ $w=n \sqrt{5}, w=m \sqrt{7}$, which would give $\frac{n}{m}=\sqrt{\frac{7}{5}}$,$\text{ rational = irrational}$, a contradiction.

This isn't such a strange result. Imagine two non-interacting planets orbiting the sun with periods $\pi$ and $e$ years. When $t=0$, let them align so that $\theta_1=\theta_2=0$. Let them go, and they will never again align so that $\theta_1=\theta_2=0$. However, later they will approximately align at $0$ degrees.