If a spring extends by x on loading ,then energy stored by the spring is(if T is the tension in the spring and k is the spring constant)

$\mathrm{work} = \mathrm{Force}.\mathrm{displacement}$

$\mathrm{dw} =\mathrm{F}.\mathrm{dx}---\left(1\right)$

$\mathrm{in} \mathrm{a} \mathrm{spring} \mathrm{F}=-\mathrm{kx}$

$\Rightarrow \mathrm{dF}=-\mathrm{kdx}$

$\Rightarrow -\frac{\mathrm{dF}}{\mathrm{k}}=\mathrm{dx}---\left(2\right)$

$\mathrm{substitute} \mathrm{eqn}\left(2\right) \mathrm{in} \left(1\right), \mathrm{we} \mathrm{get}:$

$\mathrm{dw} =\mathrm{F}(-\frac{\mathrm{dF}}{\mathrm{k}})$

$$\begin{gathered} \mathrm{integrating} \\ \int \mathrm{dw} =\int \mathrm{F}(-\frac{\mathrm{dF}}{\mathrm{k}}) \end{gathered}$$

$\Rightarrow \mathrm{w}=-\frac{{\mathrm{F}}^2}{2\mathrm{k}}$

$$\begin{gathered} \mathrm{here} \mathrm{work} \mathrm{w} \mathrm{is} \mathrm{energy} \mathrm{stored} \\ \mathrm{F} \mathrm{is} \mathrm{tension} \mathrm{in} \mathrm{the} \mathrm{spring} \mathrm{T} \\ \mathrm{negative} \mathrm{is} \mathrm{for} \mathrm{restoring} \mathrm{force} \mathrm{of} \mathrm{spring}. \end{gathered}$$

$\therefore \mathrm{energy} \mathrm{stored} =\frac{{\mathrm{T}}^2}{2\mathrm{k}}$