A spring of negligible mass having a force constant k extends by an amount y when a mass m is hung from it. The mass is pulled down a little and released. The system begins to execute simple harmonic motion of amplitude A and angular frequency $\omega$. The total energy of the mass-spring system will be

Let L be the relaxed length of the spring and y the extension produced in it due to force mg so that
$k y=m g$                        (i)
The displacement of the mass during oscillation is given by
$x=A\sin (\omega t+\varphi )$              (ii)
At the instant when the displacement is x
$\mathrm{KE}$ of mass $=\frac{1}{2}m{V}^2=\frac{1}{2}m{\left(\frac{dx}{dt}\right)}^2$$=\frac{1}{2}m{A}^2{\omega }^2{\cos }^2(\omega t+\varphi )$             (iii)
PE of spring $=\frac{1}{2}k(y+x{)}^2=\frac{1}{2}k\left(y^2+2yx+x^2\right)$$=\frac{1}{2}k{y}^2+kyx+\frac{1}{2}k{x}^2$
Using (i) and (ii) and $\omega =\sqrt{\frac{k}{m}}$, we have
PE of spring $=\frac{1}{2}k{y}^2+mgx$$+\frac{1}{2}·m{\omega }^2{A}^2{\sin }^2(\omega t+\varphi )$                        (iv)
Taking gravitational PE at the mean position to be zero,
Gravitational PE at $x=-m g x$                                             (v)
Adding (iii), (iv) and (v), we get
Total energy of mass-spring system
$$\begin{gathered} =\frac{1}{2}m{A}^2{\omega }^2{\cos }^2(\omega t+\varphi )+\frac{1}{2}k{y}^2+mgx \\ +\frac{1}{2}m{A}^2{\omega }^2{\sin }^2(\omega t+\varphi )-mgx \\ =\frac{1}{2}m{A}^2{\omega }^2+\frac{1}{2}k{y}^2 \end{gathered}$$