In fig. (6), the spring has a force constant k The pulley
 

s light and smooth, the spring and string are light. The suspended block has a mass m. If the block is slightly displaced from its equilibrium position and then released, the period of its vertical oscillation is

see fig (12)

Let the tension in spring at equilibrium position for block be${\mathrm{T}}_0\text{. Then }\mathrm{mg}={\mathrm{T}}_0$ From force diagram of pulley
$$\begin{gathered} 2{\mathrm{T}}_0={\mathrm{ky}}_0\text{ or }2\mathrm{mg}={\mathrm{ky}}_0 \\ \mathrm{When} \mathrm{the} \mathrm{block} \mathrm{is} \mathrm{displaced} \mathrm{through} \mathrm{a} \mathrm{small} \mathrm{distanceyin} \\ \mathrm{downward} \mathrm{direction} \mathrm{from} \mathrm{equilibrium} \mathrm{position}, \mathrm{let} \mathrm{the} \\ \mathrm{tension} \mathrm{in} \mathrm{string} \mathrm{be} \mathrm{T}. \mathrm{The} \mathrm{stretching} \mathrm{of} \mathrm{string} \mathrm{will} \mathrm{be} \\ (\mathrm{y}/2)\text{. Therefore } \\ \mathrm{and} 2\mathrm{T}=\mathrm{k}\left({\mathrm{y}}_0+\frac{{\mathrm{y}}_0}{2}\right)={\mathrm{ky}}_0+\frac{{\mathrm{ky}}_0}{2} \\ \mathrm{From} \mathrm{eqs}. \left(1\right) \mathrm{and} 2), \mathrm{we} \mathrm{get} \\ 2\mathrm{T}=2\mathrm{mg}+\frac{{\mathrm{ky}}_0}{2}\text{ or }\mathrm{T}=\mathrm{mg}+\left(\frac{{\mathrm{ky}}_0}{4}\right) \\ \mathrm{Now} \mathrm{mg}-\left(\mathrm{mg}+\frac{{\mathrm{ky}}_0}{4}\right)=\mathrm{ma} \\ \mathrm{or} \mathrm{ma}=-\frac{{\mathrm{ky}}_0}{4}\text{ or }\mathrm{a}=-\left(\frac{\mathrm{k}}{4\mathrm{m}}\right){\mathrm{y}}_0 \\ \mathrm{therofore} \frac{{\mathrm{d}}^2{\mathrm{y}}_0}{{\mathrm{dt}}^2}=-\left(\frac{\mathrm{k}}{4\mathrm{m}}\right){\mathrm{y}}_0\text{ or }\frac{{\mathrm{d}}^2{\mathrm{y}}_0}{{\mathrm{dt}}^2}+\left(\frac{\mathrm{k}}{4\mathrm{m}}\right){\mathrm{y}}_0=0 \\ \mathrm{T}=\frac{2\mathrm{\pi }}{\mathrm{\omega }}=2\mathrm{\pi }\sqrt{\left(\frac{4\mathrm{m}}{\mathrm{R}}\right)} \end{gathered}$$