$A$ body of mass $m$ is attached to one end of a massless spring which is suspended vertically from a fixed point. The mass is held in hand,so that the spring is neither stretched nor compressed. Suddenly,the support of the hand is removed. The lowest position attained by the mass during oscillation is $4 \, cm$ below the point where it was held in hand.
$(a)$ What is the amplitude of oscillation?
$(b)$ Find the frequency of oscillation.

(N/A) Let the spring constant be $k$. When the hand is removed,the mass $m$ starts oscillating. The lowest point reached is $x_{max} = 4 \, cm = 0.04 \, m$. At this point,the velocity is zero,so by the work-energy theorem,the loss in gravitational potential energy equals the gain in elastic potential energy:
$mgx_{max} = \frac{1}{2} k x_{max}^2$
$mg = \frac{1}{2} k x_{max} \implies x_{max} = \frac{2mg}{k}$.
The equilibrium position (mean position) is at $x_{eq} = \frac{mg}{k}$.
The amplitude $A$ is the distance from the mean position to the extreme position:
$A = x_{max} - x_{eq} = \frac{2mg}{k} - \frac{mg}{k} = \frac{mg}{k}$.
Since $x_{max} = 4 \, cm$,then $\frac{2mg}{k} = 4 \, cm$,which means $\frac{mg}{k} = 2 \, cm$.
Thus,the amplitude $A = 2 \, cm = 0.02 \, m$.
$(b)$ The frequency of oscillation $f$ is given by $f = \frac{1}{2\pi} \sqrt{\frac{k}{m}}$.
From $\frac{mg}{k} = 0.02 \, m$,we have $\frac{k}{m} = \frac{g}{0.02} = \frac{9.8}{0.02} = 490 \, s^{-2}$.
$f = \frac{1}{2\pi} \sqrt{490} \approx \frac{22.136}{6.28} \approx 3.52 \, Hz$.