The question states that the pulley is frictionless and *fixed* but not that it is *massless*. I think the latter (*fixed*) must be a mistake : the pulley must be free to rotate, in order to get the result that the blocks have the same period. We must also assume that the pulley is *massless*. Then tension in the string and springs is always the same on both sides of the pulley, as is the case when there is string but no springs.

The masses are equal and the tensions acting on them are equal, so the acceleration must also be equal : $m\ddot x=mg-T$ (taking down as +ve for both masses). Both masses have the same equation of motion, so the oscillation frequency and period is the same.

It is tempting to write $T=kx_1=2kx_2$ where $x_1, x_2$ are the position co-ordinates of the masses. Then we seem to have different equations of motion and different frequencies :

$\ddot x_1+\frac{k}{m}x_1=g$

$\ddot x_2+\frac{2k}{m}x_2=g$.

However, this is incorrect. It assumes that the upper ends of the springs are fixed in position. They are not. The pulley turns freely. Then some of the larger extension of spring 1 is transferred smoothly and instantaneously to the other side, compensating for the smaller extension of spring 2. Therefore the extension of spring 1 is not equal to the displacement of mass 1, and likewise for spring 2/mass 2.

The masses are released at the same time, and the above mechanism keeps the lower ends of the springs always level, where the masses are attached. So the masses oscillate *in phase*.

It is a little more difficult to anticipate what happens if the masses are not released at the same time...

Do you think periods will be the same or different for the 2 masses? Why?

edited Feb 12, 2017 by sammy gerbil