$A$ stone of mass $m$ is tied to an elastic string of negligible mass and spring constant $k$. The unstretched length of the string is $L$. The other end of the string is fixed to a nail at a point $P$. Initially,the stone is held at the same level as point $P$ and is dropped vertically.
$(a)$ Find the distance $y$ from the top when the mass comes to rest for an instant,for the first time.
$(b)$ What is the maximum velocity attained by the stone in this drop?
$(c)$ What shall be the nature of the motion after the stone has reached its lowest point?

(N/A) Consider the diagram,the stone is dropped from point $P$.
$(a)$ The stone is in free fall up to length $L$. After that,the elasticity of the string exerts a restoring force. Suppose the stone is at rest at an instantaneous distance $y$ from $P$.
By the law of conservation of energy,the loss in gravitational potential energy of the stone equals the gain in elastic potential energy of the string:
$mgy = \frac{1}{2}k(y - L)^2$
$mgy = \frac{1}{2}k(y^2 - 2yL + L^2)$
$2mgy = ky^2 - 2kyL + kL^2$
$ky^2 - 2(mg + kL)y + kL^2 = 0$
Using the quadratic formula $y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:
$y = \frac{2(mg + kL) \pm \sqrt{4(mg + kL)^2 - 4k^2L^2}}{2k}$
$y = \frac{(mg + kL) + \sqrt{m^2g^2 + 2mgkL + k^2L^2 - k^2L^2}}{k}$
$y = L + \frac{mg + \sqrt{m^2g^2 + 2mgkL}}{k}$
$(b)$ The maximum velocity occurs when the acceleration is zero,i.e.,when the tension equals the weight: $k(y_{eq} - L) = mg$,so $y_{eq} = L + \frac{mg}{k}$.
Using energy conservation between the start and the equilibrium position:
$mgy_{eq} = \frac{1}{2}mv_{max}^2 + \frac{1}{2}k(y_{eq} - L)^2$
$mg(L + \frac{mg}{k}) = \frac{1}{2}mv_{max}^2 + \frac{1}{2}k(\frac{mg}{k})^2$
$mgL + \frac{m^2g^2}{k} = \frac{1}{2}mv_{max}^2 + \frac{m^2g^2}{2k}$
$v_{max} = \sqrt{2gL + \frac{m^2g^2}{mk}} = \sqrt{2gL + \frac{mg^2}{k}}$.
$(c)$ After reaching the lowest point,the stone will perform Simple Harmonic Motion $(SHM)$ about the equilibrium position $y_{eq} = L + \frac{mg}{k}$.