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

Options

A. `2pi sqrt((m)/(K))`
B. `pi sqrt((m)/(4K))`
C. `pi sqrt((m)/(K))`
D. `4pi sqrt((m)/(K))`

Answer

D

Explanation

Let us assume thet in equilibrium condition spring is `x_(0)` elongate from its natural length.

In equilibrium `T_(0) = mg`
and `kx_(0) = 2T_(0)`
`rArr kx_(0) = 2mg` (1)
If the mass `m` distance `x` from its equilibrium position then pulley will move down by `(x)/(2)` so the extra force then spring will be `(kx)/(2)`. From figure
`F_("net") = mg - T = mg - (k)/(2) (x_(0) + (x)/(2))`
`F_("net") = mg - (kx_(0))/(2) - (kx)/(4)`
From eq. (1) we get
`F_("net") = (-kx)/(4)` ...(2)
Now compare eq. (2) with `F = - K_(SHM) x`
then `K_(SHM) = (K)/(4)`
`rArr T = 2 pi sqrt((m)/(K_(SHM))) = 2pi sqrt((4m)/(K))`