Consider the diagram, the stone is dropped from point $P$.

$(a)$ Stone is in free fall upto length L. After that elasticity of string exerted force for $SHM$. Suppose, stone is at rest at instantaneous distance ' $\mathrm{y}$ '.

Loss in potential energy of stone $=$ Gain in elastic potential energy in string.

$m g y=\frac{1}{2} k(y-\mathrm{L})^{2}$

$\therefore m g y=\frac{1}{2} k y^{2}-k y \mathrm{~L}+\frac{1}{2} k \mathrm{~L}^{2}$

$\Rightarrow \frac{1}{2} k y^{2}-(k \mathrm{~L}+m g) y+\frac{1}{2} k \mathrm{~L}^{2}=0$

$y=\frac{(k \mathrm{~L}+m g) \pm \sqrt{(k \mathrm{~L}+m g)^{2}-k^{2} \mathrm{~L}^{2}}}{k}$

$\therefore y=\frac{(k \mathrm{~L}+m g) \pm \sqrt{2 m g k \mathrm{~L}+m^{2} g^{2}}}{k}$

$\therefore y=\frac{(k \mathrm{~L}+m g)+\sqrt{2 m g k \mathrm{~L}+m^{2} g^{2}}}{k}$