In the figure shown, the spring deflects by δ to

In the figure shown, the spring deflects by δ to position A (the equilibrium position) when a mass m is kept on it. During free vibration, the mass is at position B at some instant. The change in potential energy of the spring mass system from position A to position B is

Potential energy at A = mg (l – δ)

Total energy at B = mg [l – (δ + x)] +	1	kx²
	2

∴ Change in energy = mgl – mg(δ + x) +	1	kx² - mgl + mgδ
	2

=	1	kx² - mgx . δ
	2

1	k(x + δ)
2