Question

a spring is elongated by length L and mass M is suspended to it. now a tiny mass m is attached to it and then released from equilibrium the period of oscillation of combined mass is​

Answer 1

Explanation:

if you drop the block when the spring is in its natural length then block will go upto x=2mg/k because when it will pass through its equilibrium stage its acceleration would be zero but its velocity will not be zero hence it will go down further until its velocity becomes zero.

Hence by applying work energy theorem:

Mgx=(1/2)kx^2 => x=2Mg/k.

So if you will let the block come down slowly then elongation of spring will be Mg/k but if you will just drop the block after attaching it to spring then elongation will be 2Mg/k.