During discharging of capacitor $C$ through inductance $L$, let at any instant, charge in capacitor be $q$ and current in inductance be changing at the rate of ${d} I / d t$

The emf equation of the circuit is

$\frac{Q}{C}+L \frac{d I}{d t}=0$ or $\frac{Q}{C}+L \frac{d^{2} Q}{d t^{2}}=0$

or $\frac{d^{2} Q}{d t^{2}}+\frac{Q}{L C}=0$

It is a differntial equation of second order with shows that the charge is oscillaroty

in $L-C$ circuit, i.e., $Q=Q_{0} \sin \omega t$

Comparing $(i)$ with the relation

$\frac{d^{2} Q}{d t^{2}+\omega^{2} Q=0}$

We have $\omega^{2}=\frac{1}{L C}$ or $\omega=\frac{1}{\sqrt{L V}}$

Max. energy stored in capacitor $=\frac{1}{2} \frac{Q_{0}^{2}}{C}$

Let at instant $t$, the energy be stored eqally between electric and magnetic field. Then energy stored in electric field at instant $t$ is

$\frac{1}{2} \frac{Q^{2}}{C}=\frac{1}{2}\left[\frac{1}{2} \frac{Q_{0}^{2}}{C}\right]$ or $Q^{2}=\frac{Q_{0}^{2}}{2}$ or $Q=\frac{Q_{0}}{\sqrt{2}}$

$Q_{0} \sin \omega t=\frac{Q_{0}}{\sqrt{2}}$ or $\omega t=\frac{\pi}{4}$

or $t=\frac{\pi}{4 \omega}=\frac{\pi}{4 \times(1 / \sqrt{L C})}=\frac{\pi \sqrt{L C}}{4}$