In the figure is shown a small block B of mass m resting on a smooth horizontal floor and the block is attached to an ideal spring (of force constant k). the spring is attached to vertical wall $W_1$. At a distance d from the block, right side of it, is present the vertical wall $W_2$. Now, the block is compressed by a distance $\frac{5d}{3}$ and released. It starts oscillating. If the collisions of the block with wall $W_2$ are perfectly elastic, the time period of oscillation of the block is

In the absence of wall $W_2$, the time period of block would have been $2\pi \sqrt{\frac{m}{k}}$. But due to the presence of $W_2$, it alters. But for the left part of oscillation (from the mean position shown), the period will be $T_1=\pi \sqrt{\frac{m}{k}}$.
For the right side part, $d=\sin \omega t[fromx=A\sin \omega t]$
$$\begin{gathered} \therefore d=\frac{5d}{3}\sin \omega t \Rightarrow \sin \omega t=\frac{3}{5} \\ \omega t={\sin }^{-1}\left(\frac{3}{5}\right) \Rightarrow t=\frac{T}{2\pi }{\sin }^{-1}\left(\frac{3}{5}\right) \end{gathered}$$

This is the time taken by the block to reach $W_2$ from mean position.
Collision with $W_2$ is perfectly elastic (given).
Time taken for right side part of oscillation will be
$$\begin{gathered} T_2=2t=2\left[\frac{T}{2\pi }{\sin }^{-1}\left(\frac{3}{5}\right)\right] \\ =\frac{2.2\pi \sqrt{\frac{m}{k}}}{2\pi }{\sin }^{-1}\left(\frac{3}{5}\right)=2\sqrt{\frac{m}{k}}{\sin }^{-1}\left(\frac{3}{5}\right) \end{gathered}$$

$\therefore$ Total time period is  $T=T_1+T_2$

$$\begin{gathered} =\pi \sqrt{\frac{m}{k}}+2\sqrt{\frac{m}{k}}{\sin }^{-1}\left(\frac{3}{5}\right) \\ =\sqrt{\frac{m}{k}}[\pi +2{\sin }^{-1}\left(\frac{3}{5}\right)] \end{gathered}$$