Figure shows a system consisting of a massles pulley, a spring of force constant `k` and a block of mass `m`. If the block is sligthtly displaced vertically down from its equilibrium and released, find the period of its vertical oscillation in cases (a) and (b).

Explanation

Let the block is displacement down a small displacement ment 'y'. The corresponding
extension of spring will be `(y)/(2)`. The extra force devolved in the spring is `(Ky)/(2)`. If `T` be the tension in the string, then
`2T = (Ky)/(2)` or `T = (Ky)/(4)`
The restoring force on the block
`F_(rest) =- T =(-Ky)/(4)` and `a = (F_(rest))/(m) = (K)/(4m) (-y)`
As force is proportional to the displacement form mean postion, so it represents `SHM`. compering with standard equation of `SHM, a =- omega^(2)y`, we get `omega = sqrt((K)/(4m))` and time period `T = (2pi)/(omega) = 2pi sqrt((4m)/(K))`
Note: The expression `T = 2pi sqrt((4m)/(k))` can be written as `T = 2pi sqrt((m)/(k_(eff)))`, where `k_(eff) = (k)/(4)`. The given device is equivalent to a block of mass `m` connected to a spring of force constant `k_(eff)`.