For the arrangement shown in the figure, find the period of oscillation

When the block displaced down by x, the spring will stretch by $\frac{x}{2}$  

From the free body diagram of the pulley, $\frac{kx}{2} =2T$  or $T=\frac{kx}{4}$

The net restoring force on the block is T. Using the Second Law of motion, we get $m\frac{d^{2}x}{d{t}^2}=-T=-\frac{kx}{4}$

$$\begin{gathered} \frac{d^{2}x}{d{t}^2}=-\frac{kx}{4m}={\omega }^{2}x \\ \omega =\sqrt{\frac{k}{4m}} T=2\pi \sqrt{\frac{4m}{k}} \end{gathered}$$