As shown in the figure, the spring is unstretched when a man begins...

Let $x$ be the displacement of the block from its initial unstretched position.
The equation of motion for the block is given by:
$M \ddot{x} = F - kx$

This can be rewritten as:
$M \ddot{x} = -k\left(x - \dfrac{F}{k}\right)$

This equation represents Simple Harmonic Motion (SHM) with a shifted equilibrium position. The new equilibrium position is where the net force is zero, i.e., $x_0 = \dfrac{F}{k}$.

Since the block starts from rest at the unstretched position ($x = 0$), this initial position is one of the extreme positions of the SHM. The amplitude $A$ of the motion is the distance between the extreme position and the equilibrium position:
$A = \left| x_0 - 0 \right| = \dfrac{F}{k}$

The time period of a spring-mass system depends only on the mass $M$ and the spring constant $k$, and is independent of any constant external force applied to it. Therefore, the time period is:
$T = 2\pi \sqrt{\dfrac{M}{k}}$