Block-Spring equilibrium question [duplicate]

Question

I had the question-

A block of mass $M$ is attached to the lower end of a vertical spring. The spring is hung from a ceiling and has force constant value $k$. The mass is released from rest with the spring initially unstretched. The maximum extension produced in the length of the spring will be?
a. $\frac{Mg}{k}$
b. $\frac{2Mg}{k}$
c. $\frac{Mg}{2k}$

I had assumed that initially there would be no force, and when released from rest, equilibrium would be reached when $Mg = kx$ where $x$ is the extension of the srping, hence $x$ = $\frac{Mg}{k}$

However, the answer is $x = \frac{2Mg}{k}$ and Conservation of Energy ($Mgx = \frac{1}{2}kx^2$) is used instead.

I would be grateful if someone could explain why my process is wrong.

Answer 1

The block oscillates around its equilibrium position. When the block is at its equilibrium position, it's either moving up or moving down, so it still has some kinetic energy, which is yet to be transformed to either elastic or potential energy once it reaches the max/min displacement.

Since the block starts from rest, you know the amplitude of its motion is $A=\dfrac{mg}k$, so the maximum displacement must be twice the amplitude, $\dfrac{2mg}k$.