(a) Total energy is initially in the form of electric field within the plates of charged capacitor. $U_e$ = $\frac{Q^2}{2C}$ = $\frac{(10 \times 10^{-3})^2}{2 \times 50 \times 10^{-6}}$ = 1 J If we neglect the losses due to resistance of connecting wires, the total energy remain consumed during LC oscillations. (b) Natural frequency of the circuit f = $\frac{1}{2\pi \sqrt{LC}}$ = $\frac{1}{2\pi \times \sqrt{20 \times 10^{-3} \times 50 \times 10^{-6}}}$ = $\frac{500}{\pi}$ Hz = 159 Hz (c) Instantaneous electrical energy $U_e$ = $\frac{q_0^2 \cos^2 \omega t}{2C}$ At ωt = 0, π, 2π, 3π… the energy is completely electrical. t = $\frac{n\pi}{2\pi f}$ = $\frac{n}{2f}$ = $\frac{n\pi}{1000}$ s , n = 0 , 1 , 2 , 3 , 4 or t = T/2 , T , 3T/2 ... Instantaneous magnetic energy $U_B$ = $\frac{1}{2} L q_0^2 \omega^2 \sin^2 \omega t$ or $U_B$ = $\frac{q_0^2}{2C} \sin^2 \omega t$ so at ωt = π/2, 3π/2, 5π/2… The energy is completely magnetic t = $\frac{(2n+1)\pi}{2(2\pi f)}$ = $\frac{2n+1}{4f}$ = $\frac{(2n+1)\pi}{2000}$ s where n = 0, 1, 2, 3, 4 … or t = T/4, 3T/4, 5T/4… (d) timings for energy shared equally between inductor and capacitor. $U_B$ = $U_E$ $\frac{q_0^2}{2C} \sin^2 \omega t$ = $\frac{q_0^2}{2C} \cos^2 \omega t$ $\tan^2 \omega t$ = 1 or tan ωt = tan π/4 t = $\frac{\pi}{4\omega}, \frac{3\pi}{4\omega}, \frac{5\pi}{4\omega}$ ... or t = $\frac{T}{8}, \frac{3T}{8}, \frac{5T}{8}$ ... (e) When a resistor is inserted in the circuit, eventually all the energy will be lost as heat across resistance. The oscillation will be damped.