Find the time period of mass M when displaced from its equilibrium position and then released for the system shown in figure.

Explanation

Let spring constant of the spring be k, When the mass M moves down by, l, then the spring will extend 2l (because each side will expand by length l).
So restoring force
`F =-2k(2l) =-4kl`
In equilibrium position.
`Mg = 4kl`
On pulling the mass down by y, the restoring force F will be
`F -2k (2l + 2y)`
Net restoring force,
`f =F. = F =-4kl -4ky -(-kl)`
or `f =-4ky`
So the motion will be S.H.M.
And its time period, `T = 2pi sqrt((m)/(k)) = 2pi sqrt((M)/(4k))`.