In the given circuit, two capacitors of capacity C & two inductors of inductance L are connected as shown. Points w, x, y, z are connected through a switch which can be connected between any two points in it. Initially C marked 1 is charged with charge Q & points x, w are connected at t = 0. If $\omega =\frac{1}{\sqrt{LC}}$, then, [q is charge of this capacitor at general time]

$$\begin{gathered} \frac{q}{C}-L\frac{di}{dt}=0,i=-\frac{dq}{dt} \\ \Rightarrow \frac{q}{C}+L\frac{d^{2}q}{d{t}^2}=0 \\ \Rightarrow \frac{d^{2}q}{d{t}^2}=-\frac{1}{LC}q \\ \mathrm{Solution} \mathrm{is} q=A\sin \left(\omega t+\varphi \right) \\ At t=0,q=Q \\ Q=A\sin \varphi ...\left(1\right) \\ i=A\omega \cos \left(\omega t+\varphi \right) \\ At t=0,i=0 \\ 0=A\omega \cos \varphi ...\left(2\right) \\ From \left(1\right) and \left(2\right), \\ \varphi =\frac{\pi }{2},A=Q \\ \Rightarrow q=Q\cos \omega t \mathbf{ }\mathit{O}\mathit{p}\mathit{t}\mathit{i}\mathit{o}\mathit{n}\mathbf{ }\mathbf{\left(}\mathit{a}\mathbf{\right)}\mathbf{ }\mathit{i}\mathit{s}\mathbf{ }\mathit{w}\mathit{r}\mathit{o}\mathit{n}\mathit{g}\mathbf{.} \end{gathered}$$

At time $\frac{2\pi }{\omega }$ , charge on 1 is $q=Q\cos \omega \frac{2\pi }{\omega }=Q$ 
For (b): Just after this instant, we have two capacitors in series and this combination is connected to an inductor

$-\frac{q}{C}+\frac{Q-q}{C}-L\frac{di}{dt}=0,\frac{Q}{C}-\frac{2q}{C}-L\frac{d^{2}q}{d{t}^2}=0$

$\frac{d^{2}q}{d{t}^2}=-\frac{2}{LC}\left(q-\frac{Q}{2}\right)$

Let $q\text{'}=q-\frac{Q}{2}$

$\Rightarrow \frac{dq\text{'}}{dt}=\frac{dq}{dt}and\frac{d^{2}q\text{'}}{d{t}^2}=\frac{d^{2}q}{d{t}^2}$

$\Rightarrow \frac{d^{2}q\text{'}}{d{t}^2}=-\frac{2}{LC}q\text{'}$

$\Rightarrow q\text{'}=q-\frac{Q}{2}=A\sin \left(\sqrt{\frac{2}{LC}}t+\varphi \right)$

At  $t=0,q=0,$

$\Rightarrow 0-\frac{Q}{2}=A\sin \left(\varphi \right)$                            ...(3)

$i=\frac{dq}{dt}=A\sqrt{\frac{2}{LC}}\cos \left(\sqrt{\frac{2}{LC}}t+\varphi \right)$

$At t=0,i=0$

$\Rightarrow 0=A\sqrt{\frac{2}{LC}}\cos \varphi$                       ...(4)

From (3) and (4)

$\varphi =\frac{3\pi }{2},A=\frac{Q}{2}$

$\Rightarrow q=\frac{Q}{2}+A\sin \left(\sqrt{\frac{2}{LC}}t+\frac{3\pi }{2}\right)$

$\Rightarrow q=\frac{Q}{2}\left[1-\cos \sqrt{\frac{2}{LC}}t\right]$

This is charge on right plate of capacitor 2. Charge on left plate of capacitor 1 is,

$Q-q=\frac{Q}{2}\left[1+\cos \sqrt{\frac{2}{LC}}t\right]$

Option (b) is wrong.
For (c): We have a charged capacitor of capacitance connected to an inductor of inductance 2L
Charge on the capacitor varies as $q=Q\cos \omega \text{'}t$ where $\omega \text{'}=\frac{1}{\sqrt{\left(2L\right)C}}=\frac{\omega }{\sqrt{2}}$

$q=Q\cos \left(\frac{\omega t}{\sqrt{2}}\right)$                 Option (c) is correct, option (d) is wrong.