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what i am talking about is something along the lines of this video, http://www.youtube.com/watch?v=5v5eBf2KwF8 where 30 metronomes sync themselves on a table. Will the same happen with double or triple pendulums?

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As this is a rather complicated system (in particular, if you want to take into account friction), only sophisticated simulations or experiments can give you the definite answer, but here are the three cases to consider and what would be required to observe them:

Synchronous chaotic motion

Let’s first assume that we have no friction. Furthermore, let’s tackle this one backwards and look at the completely synchronous chaotic state, i.e., each pendulum performs the same chaotic motion. This is a solution of the equations of motion of the system because:

  • the chaotic motion is a solution of the single (double or triple) pendulum;
  • the coupling (through the platform) is small and does not strongly affect the solution;
  • all pendulums interact with the platform in exactly the same way, so there is no complicated interaction between the pendulums. From a single pendulum’s point of view, it’s as its effect on the platform would be amplified (by the other pendulums doing exactly the same thing).

So, if I set up all pendulums in exactly the same initial conditions that lead to chaos for the single pendulum (and the slight interaction with the platform does not destroy the chaos), I would observe synchrony.

Now, is this state stable, i.e., would I observe it in reality? For the frictionless case, we have energy conservation and hence time symmetry. Therefore for every solution leading to the stable state, there also exists one leading away from it (the time inverse). In the language of dynamical systems: No attractors in conservative systems. Note that the same reasoning would apply to the setup with single pendulums, so it’s not surprising that we need friction.

In a real setup with small friction, only the state without any motion would be really stable, so let’s look at the case where the friction affects the dynamics on a larger time scale than the internal dynamics of the internal dynamics of the pendulums, and let’s ignore the slow decay due to friction when we talk about stability. Again, the same thing applies to the single pendulums.

In this case, the completely synchronous state would be stable for the same reason it is for single pendulums: Let’s consider a slight perturbation of this state, more specifically, one pendulum being slightly out of step. Then the interaction through the platform would bring it back to synchrony with the other pendulums. (To be precise, the dynamics of the other pendulums would also be slightly affected and shifted towards that of the perturbed one. The final dynamics would be in between with a strong tendency towards the bigger group of pendulums3)

So, if I set up all pendulums in exactly the same initial conditions that lead to chaos for the single pendulum and I make some inevitable errors, I would still observe a synchronous chaotic state.

There may be other solutions to the equations of motion, e.g., half of the pendulums swinging in exactly the opposite way than the other half, but these states should not be stable, and if they are, they should be considerably less likely (i.e., have a smaller basin of attraction) than the synchronous state. Once more, the same thing applies to the single pendulums. We can thus assume that the synchronous state is the attractor for most initial conditions.

So, for random initial conditions, this globally synchronous state should be eventually assumed, but due to the more complex dynamics, this may take considerably longer than for the single pendulums. Most importantly, it may take longer than it takes for friction to take its toll. But, for a benign setup (friction, mass of the platform, …), you should observe a completely synchronous chaotic state.

Synchronous periodic motion

For small amplitudes, the double pendulum tends to perform periodic motions (behaving more or less like single pendulums), which are more easy to synchronise. Thus, if the effect of friction is too strong or you start with small amplitudes, the pendulums will become periodic before they can synchronise, but then they will synchronise rather quickly, because periodic motions / single pendulums are easier to synchronise.

No synchrony whatsoever

For this to happen, your friction needs to be so strong that it eats up all kinetic energy before any synchronisation. Just once more, this also applies to single pendulums.

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  • $\begingroup$ +1 a far, far better analysis than my effort. that's a answer I would proud of. $\endgroup$ – user108787 Sep 29 '16 at 22:21
  • $\begingroup$ I'm not convinced by your argument for synchronized chaotic motion. 1. Isn't the essence of chaos that small differences will quickly produce distinctly different motions? No matter how small the coupling, differences will emerge. 2. You seem to be assuming that the platform can affect the pendulums but not vice versa. But this contradicts the condition in the title (they are on a platform they can affect). The pendulums cannot all affect the platform identically, because their effect depends on their positions on it. $\endgroup$ – sammy gerbil Sep 29 '16 at 22:51
  • $\begingroup$ @sammygerbil: 1. Yes, but that’s small differences between states of the entire system. Small differences between subsystems can still be subject to synchrony. So, if you consider the completely synchronised chaotic state, and slightly perturb one pendulum, then the dynamics of the entire system will converge back to the completely synchronised state, but that state will evolve considerably differently from the case where you haven’t perturbed a single pendulum, i.e., both will exhibit a chaotic synchronous motion, but it will be drastically different (after a while). $\endgroup$ – Wrzlprmft Sep 30 '16 at 0:50
  • $\begingroup$ […] To convince yourself that chaotic systems can synchronise, just take a bunch of instances of any chaotic toy model and couple them diffusively. $\endgroup$ – Wrzlprmft Sep 30 '16 at 0:51
  • $\begingroup$ @sammygerbil: 2. You seem to be assuming that the platform can affect the pendulums but not vice versa. – I clarified my answer a bit in that manner. — The pendulums cannot all affect the platform identically, because their effect depends on their positions on it. – I would consider that effect negligible. I am not completely sure about the setup in the video linked in the question, but for a setup like this one (plank on tins), I think it’s a valid assumption (due to an almost perfectly rigid platform). $\endgroup$ – Wrzlprmft Sep 30 '16 at 1:14
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As the comments above indicate, the probability of synchronised motion of the proposed system is virtually nil.

Taking the double pendulum as an example, I include two illustrations that may persuade you that inherent and unavoidable chaotic motion assures us that synchronised motion will not result and/or be maintained.

enter image description here

Motion of the double compound pendulum (from numerical integration of the equations of motion.

enter image description here

Long exposure of double pendulum exhibiting chaotic motion.

And this is just for a double pendulum!!

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    $\begingroup$ For synchronised motion to occur, the surface that supports the double pendulums must convey a periodic "beat" between the pendulums, […] – That’s not true. Identical chaotic systems can adopt a synchronous chaotic dynamics that does not involve any periodicity and does not need periodic beats to synchronise. — which seems very unlikely given the very low probability that both systems will ever be in any kind of linked repetitive motion. – That’s also not true: The double pendulum does exhibit periodic orbits for some certain starting conditions (which do not have zero measure). $\endgroup$ – Wrzlprmft Sep 28 '16 at 6:11
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    $\begingroup$ You've gone to town on the equations, but I don't think they answer the question. $\endgroup$ – sammy gerbil Sep 29 '16 at 20:56
  • $\begingroup$ @sammygerbil . You are perfectly correct, the math is not needed at all here at all and is just a distraction. I will edit it out. Thanks for the comment. $\endgroup$ – user108787 Sep 29 '16 at 22:25

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