Assume we have a frictionless pendulum of length $l$ with mass $m$. This pendulum hangs from some weightless contraption, which is itself bolted to a platform. This platform can move horizontally in the direction of the swing of the pendulum. There are no other forces than gravity.
If the pendulum is set in motion, as it swings on one direction, the whole platform and the pendulum move to the direction of swing and when it swings to the other direction, again the whole system moves to that direction.
Question 1
At time $t$, what is the angle $\theta(t)$ of swing of the pendulum away from a perpendicular line through the point the pendulum swings?
Assume we have another pendulum, identical to the other one, and it too is then bolted to the same platform as the other one. When set in motion, the pendulums have same frequency. Assume that pendulum 1 starts at angle $\theta_1(0)$ and the other at $\theta_2(0).$
Question 2
What are $\theta_1(t)$ and $\theta_2(t)$?
Now, if $\theta_1(0) = \theta_2(0)$, then it would seem intuitive that $\theta_1(t) = \theta_2(t).$
Also, if $\theta_1(0) = -\theta_2(0)$ it would seem that $\theta_1(t) = -\theta_2(t).$ Now the movement of the whole system, platform included, is $0$, as the movements cancel each other out.
The motivation for this question is the video showing a large number of metronomes, out of sync, on a moving platform, synchronizing over time. This was demonstrated on recent episode of Mythbusters. They used metronomes, I should think pendulums are identical when it comes to this property.
Best Answer
This is actually an interesting problem in classical mechanics, dating back to Huygens. We'll work with the three variables you define in the question, namely $(x,\theta_1, \theta_2)$. Also we'll set your $m = l = g = 1$ for simplicity.
Kinetic Energy
The position vector of the first pendulum bob is
$$\mathbb{r}_1 = (x+\sin\theta_1, \cos\theta_1)$$
whence we deduce its kinetic energy to be
$$2T_1 = \dot{x}^2 + 2\dot{x}\dot{\theta_1}\cos\theta_1 + \dot{\theta_1}^2$$
We can similarly find the kinetic energy of the second bob.
Finally we must take into account the kinetic energy of the support with mass $M$.
$$2T_3 = M\dot{x}^2$$
It's interesting to keep $M\neq 0$ since different values of $M$ give different behaviour.
Potential Energy
We assume gravity acts on the bobs as usual, producing potential energy terms of form $\cos\theta$. We also assume an elastic potential $kx^2$ pulling the table back to equilibrium. Overall we have
$$V = kx^2 - \cos\theta_1-\cos\theta_2$$
Equations of Motion
One can easily write down the Lagrangian $L=T-V$ and from it deduce the equations of motion
$$\ddot{\theta}+\ddot{x}\cos\theta+\sin\theta-\dot{x}\dot{\theta}\sin{\theta} = 0$$
$$(M+2)\ddot{x}+\ddot{\theta_1}\cos\theta_1+\ddot{\theta_2}\cos\theta_2 - \dot{\theta_1}^2\sin\theta_1-\dot{\theta_2}^2\sin{\theta_2} + kx = 0$$
Interestingly when I put these into Mathematica, there was no synchronization! It turns out the missing ingredient is damping.
Damping
Intuitively the phase difference between the pendulums must drift in a periodic way in the absence of any dissipative effects. Indeed that's what you see when numerically solving the above equations with Mathematica.
Recall that damping is usually modelled as an additive term proportional to the velocity. Adding in such terms for $\theta_1$, $\theta_2$ and $x$ now does produce the desired synchronization behaviour. For my initial conditions and choice of constants we get antiphase locking.
Summary of the Physics
Momentum transfer through a connecting medium coupled with dissipative effects leads to synchronization.
Better Models
To fully model the video mentioned you'd need a forcing term from the escapement mechanism of the metronomes. You can read about such an approach here. See also this Wolfram demonstration and the papers it references.
Towards Chaos
Evidently this setup is nonlinear and so generically displays chaotic behaviour. The study of such systems is particularly important in chemistry and biology. Here is a good introduction.
If you want to play around with this behaviour yourself, here's my rudimentary Mathematica code. Try playing with the constants and initial conditions.
sol = NDSolve[{30 x''[t] + y''[t] Cos[y[t]] + z''[t] Cos[z[t]] - y'[t]^2 Sin[y[t]] - z'[t]^2 Sin[z[t]] + 30 x[t] + 2 x'[t] == 0, y''[t] + x''[t] Cos[y[t]] + Sin[y[t]] - x'[t] y'[t] Sin[y[t]] + 0.02 y'[t] == 0, z''[t] + x''[t] Cos[z[t]] + Sin[z[t]] - x'[t] z'[t] Sin[z[t]] + 0.02 z'[t] == 0, x[0] == 0, x'[0] == 0, y[0] == Pi/10, y'[0] == 0, z[0.5] == 1, z'[2] == 0}, {x, y, z}, {t, 0, 1000}] Plot[{Evaluate[y[t] /. sol], Evaluate[z[t] /. sol]}, {t, 0, 250}, PlotRange -> All]