Q116 / 120 · JEE Advanced 2006 · Physics

Paragraph: The capacitor of capacitance $C$ can be charged (with the help of a resistance $R$ ) by a voltage source $V$, by closing switch $S_1$ while keeping switch $S_2$ open. The capacitor can be connected in series with an inductor $L$ by closing switch $S_2$ and opening $S_1$.Figure
Question: If the total charge stored in the $L C$ circuit is $Q_0$, then for $t \geq 0$

  • A
    the charge on the capacitor is $Q=Q_0 \cos \left(\frac{\pi}{2}+\frac{t}{\sqrt{L C}}\right)$
  • B
    the charge on the capacitor is $Q=Q_0 \cos \left(\frac{\pi}{2}-\frac{t}{\sqrt{L C}}\right)$
  • Correct answer
    C
    the charge on the capacitor is $Q=-L C \frac{d^2 Q}{d t^2}$
  • D
    the charge on the capacitor is $Q=-\frac{1}{\sqrt{L C}} \frac{d^2 Q}{d t^2}$

Solution

Comparing the LC oscillations with normal SHM, we get $$ \frac{d^2 Q}{d t^2}=-\omega^2 Q $$ Here, $$ \begin{aligned} \omega^2 & =\frac{1}{L C} \\ Q & =-L C \frac{d^2 Q}{d t^2} \end{aligned} $$ $$ \therefore \quad Q=-L C \frac{d^2 Q}{d t^2} $$

Hence, the correct answer is option: 'C'.

Relevant Questions