What I tried to do was first consider this system at equilibrium, assuming the pulley to be a disc of radius $'r'$ and I calculated the extensions of the 2 springs as $e(=mg/k)$ for lower spring and $3e$ for upper spring.
Then I decided to write the total energy of the system and equate it to a constant since energy is conserved in SHM.
$$\frac{1}{2}mv^2+\frac{1}{2}m\big(\frac{v}{2}\big)^2+\frac{1}{2}I\omega^2+\frac{1}{2}k\big(\frac{x}{2}+3e\big)^2+\frac{1}{2}k\big(x+e\big)^2-mgx-mg\frac{x}{2}=k$$
Where,
$m$= mass of pulley and block
$k$= spring constant
$v$= speed of mass
$I$= moment of inertia of pulley about centre
$\omega$= angular speed of pulley
Then I differentiate this energy and equate it to $0$ to obtain an expression of the form $$\frac{d^2x}{dt^2}=-\omega^2x$$
But I am facing 2 issues: 1) The gravitational potential energy terms are not cancelling out. Does this mean that our motion is occurring wrt to some other origin?
2) Is the constraint of pulley and mass used correct? I thought that when the block is shifted down by $x$, the pulley shifts down by $\frac{x}{2}$ but my friend argued that this would mean that the lower spring does not extend? I can't quite see why and how spring should affect the constraint.
Any help would be appreciated.
