In the figure shown, the spring is light and has a force constant k. The pulley is light and smooth and the string is light. The suspended block has a mass m. On giving a slight displacement vertically to the block in the downward direction from its equilibrium position the block executes S.H.M. on being released with time period T. Then

Let the extension in the spring be x₀ at equilibrium. If F₀ be the tension in the string then F₀ = kx₀ . Further if T₀ is the tension in the thread then T₀ = mg and 2T₀ = kx₀ . 

Let the mass m be displaced through a slight displacement x downwards. Let the the new tension in the string and spring be T and F respectively.

         $$\begin{gathered} \Rightarrow \mathrm{F}=k\left({\mathrm{x}}_0+\frac{\mathrm{x}}{2}\right) \mathrm{and} \mathrm{F}=2\mathrm{T} \\ \Rightarrow 2\mathrm{T}=\mathrm{k}\left({\mathrm{x}}_0+\frac{\mathrm{x}}{2}\right) \\ \Rightarrow \left(\mathrm{T}-{\mathrm{T}}_0\right)=\frac{\mathrm{kx}}{4} \\ \Rightarrow \mathrm{Time} \mathrm{period}=2\mathrm{\pi }\sqrt{\frac{\mathrm{m}}{{\displaystyle \frac{\mathrm{k}}{4}}}}=4\mathrm{\pi }\sqrt{\frac{\mathrm{m}}{\mathrm{k}}} \end{gathered}$$