Question

initially mass m is held such that spring is in relaxed condition. if mass m is suddenly released, maximum elongation in spring will be

Answer 1

Here we shall use the law of conservation of energy for the spring system

so,

initial total energy = final total energy

potential energy = kinetic energy + spring elastic energy

now if x be elongation of the spring, then

mgx = (1/2)mv2 + (1/2)kx2

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)kx2

or

the maximum elongation would be

x = 2mg / k

Answer 2

I hope you help !!!!!!
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=> If mass of spring is negligible as compared to mass m then:

mg=kx

X=mg/k ,where

K is spring constant of spring and g is acc. due to gravity.