As shown in the figure, the left block travels at a speed v towards...

Let the cycle begin when the left block moves from the left wall towards the right block with velocity $v$.

The distance to the right block is $L$. The time taken by the left block to reach the right block is:
$t_1 = \dfrac{L}{v}$

Since the collision is perfectly elastic and both blocks have the same mass $m$, they exchange their velocities upon impact. The left block comes to rest, and the right block acquires the velocity $v$ towards the right.

The right block then compresses the spring, comes to a momentary halt, and returns to its equilibrium position. This motion constitutes exactly half of a simple harmonic oscillation. The time taken for this half-cycle is:
$t_2 = \dfrac{1}{2} \left( 2\pi \sqrt{\dfrac{m}{k}} \right) = \pi \sqrt{\dfrac{m}{k}}$

When the right block returns to its equilibrium position, it has a velocity $v$ directed towards the left. It then collides elastically with the stationary left block. Velocities are exchanged once again: the right block comes to rest, and the left block acquires a velocity $v$ towards the left.

The left block travels the distance $L$ back to the left wall. The time taken for this return journey is:
$t_3 = \dfrac{L}{v}$

Upon reaching the left wall, the left block collides elastically with it and rebounds with velocity $v$ towards the right. This brings the system back to its exact initial state, completing one full cycle.

The total time period $T$ of the periodic motion is the sum of these time intervals:
$T = t_1 + t_2 + t_3$

$T = \dfrac{L}{v} + \pi \sqrt{\dfrac{m}{k}} + \dfrac{L}{v}$

$T = \pi \sqrt{\dfrac{m}{k}} + \dfrac{2L}{v}$