(a) As we know that, when we have given the support by hand to net force $=0$ and when it is released, all of its PE will convert into $\mathrm{KE}$ and to mass $m$ then the body oscillates about a mean position.


Let the mass reaches at its new position $x$ unit. $x$ is the maximum displacement in the spring when it reaches the lowest point in oscillation.
Then PE of spring $=$ gravitational PE lost by man. loss in PE of the block $=m g x$ where, $m=$ mass of the block
But, Elastic potential energy of the spring is $=\frac{1}{2} k x^2$
Equating both (i) and (ii) by using conserving the mechanical energy,
we get, $m g x=\frac{1}{2} k x^2$ or $x=\frac{2 m g}{k}$
The mean position spring oscillation will be, when the block is balanced by the spring.

If $x$ is the extension in that case, then
$$
F=+k x_0
$$
We know that, $F=m g$
$$
\therefore m g=+k x_0
$$
or $x_0=\frac{m g}{k}$
From Eq. (iii) and Eq. (iv),
$$
\begin{aligned}
&\frac{x}{x_0}=\frac{2 m g}{k} / \frac{m g}{k}=2 \\
&x=2 x_0
\end{aligned}
$$
As given that $x=4 \mathrm{~cm}$ (maximum extension from the unstretched position)
So, $2 x_0=4$
$$
\Rightarrow x_0=\frac{4}{2}=2 \mathrm{~cm}
$$
But the displacement of mass from the mean position to the position when spring attains its natural length is equal to amplitude of the motion. It is the maximum distance from mean position.
So, $A=x-x_0=4-2=2 \mathrm{~cm}$.
where, $A=$ amplitude of the motion.
(b) Time period of the oscillating system depends on mass spring constant is
$$
T=2 \pi \sqrt{\frac{m}{k}}
$$
which does not depend on amplitude.
But from Eq. (iii),
$$
\frac{2 m g}{k}=x,\left(\frac{m}{k}=\frac{x}{2 g}\right) \text { (maximum extension) }
$$
As given that, $x=4 \mathrm{~cm}=4 \times 10^{-2} \mathrm{~m}$
$$
\begin{aligned}
\frac{m}{k} &=\frac{4 \times 10^{-2}}{2 g}=\frac{2 \times 10^{-2}}{g} \\
\frac{k}{m} &=\frac{g}{2 \times 10^{-2}}
\end{aligned}
$$
As we know that,
$$
v(\text { frequency })=\frac{1}{T}=\frac{1}{2 \pi} \sqrt{\frac{k}{m}}
$$
$$
\begin{aligned}
\therefore v &=\frac{1}{2 \times 3.14} \sqrt{\frac{g}{2 \times 10^{-2}}} \\
&=\frac{1}{2 \times 3.14} \sqrt{\frac{4.9}{10^{-2}}}=\frac{1}{6.28} \times \sqrt{4.9 \times 100} \\
&=\frac{10}{6.28} \times 2.21=3.52 \mathrm{~Hz} .
\end{aligned}
$$
The block during oscillation cannot rise above the position from where it was released because total extension in spring is $4 \mathrm{~cm}$ when released and amplitude is $2 \mathrm{~cm}$.
So, it oscillates below the released position.