At what times is the energy in an LC oscillator completely electric or completely magnetic? I know that the time period of the LC oscillations is given by $T=2\pi\sqrt{LC}$. At what times is the total energy of the circuit completely stored in the capacitor or completely in the inductor?
 A: First of all, let's assume that we have a capacitor that is charged. We connect it to a circuit that has an inductor and a switch. Let's assume that, for $t_0=0$, the circuit is open; i.e., the switch is open, the current along the wires is zero, the capacitor is still fully charged and all the energy of the circuit is stored in the capacitor.
As you said, the behaviour of the circuit is periodic: this means that for $t=T=2\pi\sqrt{LC}$ , all the energy will be stored in the capacitor again. The same goes for $t=2T$, $t=3T$, etc.
From now on, let's analyze the problem in just one period - we already know what happens after that.
Then, at $t=\frac{T}{4}$, the current hits a maximum (you can check this if you solve the differential equation you get from going around the circuit). That means that now all the energy is stored in the inductor.
After that, at $t=\frac{T}{2}$, the current is zero again: the energy is stored completely in the capacitor now. But, this time, the polarity of the capacitor is exactly opposed to the one we had at $t=0$. (This is just an observation, when talking about energy it is not relevant.)
In the end, when $t=\frac{3}{4}T$, all the energy will be stored in the inductor again. This time, the current function hits a minimum: this is a maximum in absolute value, but with an opposite direction of the one we had before.
