A system shown in fig. (5) consists of mass less pulley,
 

a spring of force constant k and a block of mass m. If block is just slightly displaced vertically down from its equilibrium position and released. The time period of vertical oscillation is
 

If m moves downward through a distance y from its equilibrium position, then pulley will also move through a distance. So, the stretching of string will
be2y. 

Tension

 $$\begin{gathered} \mathrm{T}=\mathrm{F}=\mathrm{k}\left(2\mathrm{y}\right) \\ \mathrm{Also} \mathrm{mg}=2\mathrm{T} \\ \therefore \text{ Restoring force, }{\mathrm{F}}^{\mathrm{\prime }}=2\mathrm{T}=4\mathrm{ky} \\ \text{ Now }\mathrm{ma}=-4ky\text{ or }\mathrm{a}=-4ky/\mathrm{m} \\ \frac{{\mathrm{d}}^2\mathrm{y}}{{\mathrm{dt}}^2}=-\left(\frac{4\mathrm{k}}{\mathrm{m}}\right)\mathrm{y}\text{ or }\frac{{\mathrm{d}}^2\mathrm{y}}{{\mathrm{dt}}^2}+\left(\frac{4\mathrm{k}}{\mathrm{m}}\right)\mathrm{y}=0 \\ \text{ or }\frac{{\mathrm{d}}^2\mathrm{y}}{{\mathrm{dt}}^2}+{\mathrm{\omega }}^2\mathrm{y}=0\text{ where }\mathrm{\omega }=\sqrt{\left(\frac{4\mathrm{k}}{\mathrm{m}}\right)} \\ \mathrm{This} \mathrm{represents} \mathrm{S}.\mathrm{H}.\mathrm{M}. \mathrm{The} \mathrm{time} \mathrm{period} \mathrm{is} \mathrm{given} \mathrm{by} \\ \mathrm{T}=\frac{2\mathrm{\pi }}{\mathrm{\omega }}=2\mathrm{\pi }\sqrt{\left(\frac{\mathrm{m}}{4\mathrm{k}}\right)} \end{gathered}$$