A capacitor is connected across an inductor. At time t = 0 charge on the capacitor is equal to `1/sqrt2q_"max"`, where `q_"max"` is the maximum charge on the capacitor. The time t, at which the energy stored in the capacitor is equal to the energy stored in the inductor is (The inductance of the inductor is L and capacitance of the capacitor is C. Resistance of the circuit is zero)

Options

A. `2pisqrt(LC)`
B. `1/(2pi)sqrt(LC)`
C. `(pisqrt(LC))/2`
D. `1/(2pisqrt(LC))`

Answer

To solve the problem, we need to find the time \( t \) at which the energy stored in the capacitor is equal to the energy stored in the inductor. Let's break down the solution step by step.

### Step 1: Understanding the initial conditions
At time \( t = 0 \), the charge on the capacitor is given as:
\[ q(0) = \frac{1}{\sqrt{2}} q_{\text{max}} \]
where \( q_{\text{max}} \) is the maximum charge on the capacitor.

### Step 2: Relating charge to time
The charge \( q(t) \) on the capacitor at any time \( t \) in an LC circuit is given by:
\[ q(t) = q_{\text{max}} \sin(\omega t + \phi) \]
where \( \omega = \frac{1}{\sqrt{LC}} \) and \( \phi \) is the phase constant.

### Step 3: Finding the phase constant \( \phi \)
At \( t = 0 \):
\[ q(0) = q_{\text{max}} \sin(\phi) = \frac{1}{\sqrt{2}} q_{\text{max}} \]
This implies:
\[ \sin(\phi) = \frac{1}{\sqrt{2}} \implies \phi = \frac{\pi}{4} \]

### Step 4: Current in the circuit
The current \( I(t) \) in the circuit can be found by differentiating \( q(t) \):
\[ I(t) = \frac{dq}{dt} = \omega q_{\text{max}} \cos(\omega t + \phi) \]
Substituting \( \phi = \frac{\pi}{4} \):
\[ I(t) = \omega q_{\text{max}} \cos\left(\omega t + \frac{\pi}{4}\right) \]

### Step 5: Energy stored in the capacitor and inductor
The energy stored in the capacitor \( U_C \) is given by:
\[ U_C = \frac{1}{2} \frac{q^2}{C} \]
The energy stored in the inductor \( U_L \) is given by:
\[ U_L = \frac{1}{2} L I^2 \]

### Step 6: Setting the energies equal
We need to find \( t \) such that:
\[ U_C = U_L \]
Substituting the expressions for \( U_C \) and \( U_L \):
\[ \frac{1}{2} \frac{q^2}{C} = \frac{1}{2} L I^2 \]
Substituting \( I(t) \):
\[ \frac{q^2}{C} = L \left(\omega q_{\text{max}} \cos\left(\omega t + \frac{\pi}{4}\right)\right)^2 \]

### Step 7: Substituting \( q(t) \)
Substituting \( q(t) = q_{\text{max}} \sin\left(\omega t + \frac{\pi}{4}\right) \):
\[ \frac{(q_{\text{max}} \sin(\omega t + \frac{\pi}{4}))^2}{C} = L \left(\omega q_{\text{max}} \cos\left(\omega t + \frac{\pi}{4}\right)\right)^2 \]

### Step 8: Simplifying the equation
Cancelling \( q_{\text{max}}^2 \) from both sides:
\[ \frac{\sin^2(\omega t + \frac{\pi}{4})}{C} = L \omega^2 \cos^2(\omega t + \frac{\pi}{4}) \]
Substituting \( \omega^2 = \frac{1}{LC} \):
\[ \frac{\sin^2(\omega t + \frac{\pi}{4})}{C} = \frac{L}{LC} \cos^2(\omega t + \frac{\pi}{4}) \]
This simplifies to:
\[ \sin^2(\omega t + \frac{\pi}{4}) = \cos^2(\omega t + \frac{\pi}{4}) \]

### Step 9: Using trigonometric identities
This implies:
\[ \tan^2(\omega t + \frac{\pi}{4}) = 1 \implies \tan(\omega t + \frac{\pi}{4}) = 1 \]
Thus:
\[ \omega t + \frac{\pi}{4} = \frac{\pi}{4} + n\pi \quad (n \in \mathbb{Z}) \]
Taking \( n = 1 \):
\[ \omega t + \frac{\pi}{4} = \frac{5\pi}{4} \implies \omega t = \pi \implies t = \frac{\pi}{\omega} \]

### Step 10: Final expression for time
Substituting \( \omega = \frac{1}{\sqrt{LC}} \):
\[ t = \frac{\pi}{\frac{1}{\sqrt{LC}}} = \pi \sqrt{LC} \]

### Final Answer
Thus, the time \( t \) at which the energy stored in the capacitor is equal to the energy stored in the inductor is:
\[ t = \frac{\pi}{2} \sqrt{LC} \]