Question

Initially mass m is held such that spring is in relaxed condition. If mass m is suddenly released maximum elongation in spring

Answer 1

Net external force is zero at equilibrium. So to calculate extension at that point

mg = kx which gives x = mg / k

The block still has some velocity in this position. So it moves further down till the velocity becomes zero. This maximum extension is given by

1/2 k * x (max) * x (max) = mg * x (max) [energy conservation]

it gives x (max) = 2 mg / k

So extension at equilibrium is mg / k

Answer 2

Answer:

it is same explaination, but explained a little clearer.

Explanation:

Using the law of conservation of energy for the spring system

initial total energy = final total energy

potential energy = kinetic energy + spring elastic energy

now if x be elongation of the spring, then

At the lowest point where spring is stretched to its maximum.

mgx = (1/2)mv^2 + (1/2)kx^2

now, maximum elongation corresponds to point where the velocity becomes zero and the spring is about to turn backwards.

So, by putting v = 0 in the above relation we get

mgx = (1/2)kx^2

=>

x = 2mg / k