One end of an un-stretched vertical spring is attached to the ceiling and an object attached to the other end is slowly lowered to its equilibrium position. If S is the gain in spring energy and G is the loss in gravitational potential energy in the process, then : 

S = G/2

As : \(kx = mg\)

At equilibrium position : \(x = \frac{mg}{k}\)

\(U_{spring} = \frac{1}{2} k x^2\)

  = \(\frac{1}{2} k (\frac{mg}{k})^2\)

  = \(\frac{mgx}{2}\)

  = \(\frac{1}{2} \text{loss in GPE}\)

\(S = \frac{1}{2}G\)