(a) Till the stone drops through a length L it will be in free fall. After that the elasticity of the string will force it to a SHM. Let the stone come to rest instantaneously at y.

The loss in P.E. of the stone is the P.E. stored in the stretched string.

(b) The maximum velocity is attained when the body passes, through the “equilibrium, position” i.e. when the instantaneous acceleration is zero. That is mg - kx = 0 where x is the extension from L:

(c) Consider the particle at an instantaneous position y. Then

Make a transformation of variables: z = k/m(y - L) - g

Thus the stone performs SHM with angular frequency *ω* about the point