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 position and then released, the period of its vertical oscillation is

Let us assume that in equilibrium condition spring is ${\mathrm{x}}_0$ elongate from its natural length

In equilibrium ${\mathrm{T}}_{0 }= \mathrm{mg}$ and ${\mathrm{kx}}_0 = 2{\mathrm{T}}_0$

$\Rightarrow {\mathrm{kx}}_0 = 2\mathrm{mg}$-------------(1)

If the mass m moves down a distance x from its equilibrium position then pulley will move down by $\frac{\mathrm{x}}{2}$. So the extra force in spring will be $\frac{\mathrm{kx}}{2}$. From figure

${\mathrm{F}}_{\mathrm{net}}= \mathrm{mg}-\mathrm{T} = \mathrm{mg}-\frac{\mathrm{k}}{2}({\mathrm{x}}_0+\frac{\mathrm{x}}{2})$

${\mathrm{F}}_{\mathrm{net}} = \mathrm{mg}-\frac{{\mathrm{kx}}_0}{2}-\frac{\mathrm{kx}}{4}$

from eq.(1), we get

${\mathrm{F}}_{\mathrm{net}} = \frac{-\mathrm{kx}}{4}$-----------------(2)

Now compare eq.(2) with F = $-{\mathrm{K}}_{\mathrm{SHM}}\mathrm{x}$

then ${\mathrm{K}}_{\mathrm{SHM}} = \frac{\mathrm{K}}{4}$

$\Rightarrow T = 2\mathrm{\pi }\sqrt{\frac{\mathrm{m}}{{\mathrm{K}}_{\mathrm{SHM}}}} = 2\mathrm{\pi }\sqrt{\frac{4\mathrm{m}}{\mathrm{K}}} = 4\mathrm{\pi }\sqrt{\frac{\mathrm{m}}{\mathrm{K}}}$