In the following arrangements, bock is slightly displaced vertically down from its equilibrium position and released. Find time period of vertical oscillations. The pulley is light.

Explanation

Let block is displaced down by 'y'. The corresponding extension of the spring will be `2y`m, as string is inextensible. The extra force developed in the spring is `2ky` if `T` is the tension in the spring, then `T = 2Ky`The restroing force on the block
`F_(rest) = 2T=-2(2Ky) =- 4Ky` and the acceleration of the block `a = (F_(rest))/(m) = (4K)/(m) (-y)` As force acting on block is proporional to the displacement from mean position, so it it will represent `SHM`. compering this with standard equation of `SHM`,
`a =- omega^(2) y`,
we get `omega = sqrt((4K)/(m))` and `T = (2pi)/(omega) = 2pi sqrt((m)/(4K))`
Note: The expression `T = 2pi sqrt((m)/(4K))` can be written as, `T = 2pi sqrt((m)/(k_(eff)))`, where `k_(eff) = 4k`.
Thus the given device is equivalent to a block of mass `m` connected to a spring of force constant `k_(eff)`.