Are the accelerations the same at either end of a moveable pully? Given a moveable pulley with a fixed pulley on either side,

Is the acceleration of the left weight (m1) the same as that of the right weight (m2)?
Intuitively, I would imagine it to be, since if m1 drops by 10 metres, then m3 would rise by 5 metres, and thus m2 would drop by 10 metres. But mathematically the accelerations don't appear to be the same... Perhaps there is a mistake in my reasoning.

Assume the system is released from rest, an inextensible and massless string is used for the pulleys, and there is no friction forces acting on the system.
 A: Hint: The tension in the string is the same throughout.
Hope this helps.
A: While it is certainly possible that $x_1 = - \frac{1}{2} x_3$ and that $x_2 = - \frac{1}{2} x_3$, as your intuition suggests, it is not necessarily the case.  (We define $x_1$, $x_2$, and $x_3$ to be the displacements from the original positions of the masses.)  If both of the equations were automatically true, then rope would never move relative to $m_3$, since the same amount of rope would be going over the side pulleys at all times.  But hopefully it seems intuitive that you could come up with some highly unbalanced set of masses (with $m_1 \gg m_2$, for example) where the rope should slide over $m_3$.
Instead, you should look to find one equation that relates $x_1$, $x_2$, and $x_3$ all together.  This equation will be a mathematical statement that the overall length of the rope does not change.  Once you have this equation, you can differentiate it to get a parallel statement concerning the accelerations of each mass.
