A LCR circuit be haves like a damped harmonic oscillator. The discharge of capacitor C, through an `C^(square)` inductor L having resistance R is oscillatory. Then which of following is correct _____.

Options

A. `1/(LC)lt(R^(2))/(4L^(2))`
B. `1/(LC)gt(R^(2))/(4L^(2))`
C. `1/(LC)gt(R^(3))/(4L^(4))`
D. `1/(sqrt(LC))gt(R^(3))/8`

Explanation

To solve the problem regarding the behavior of an LCR circuit as a damped harmonic oscillator, we will derive the necessary equations and conditions step by step.

### Step-by-Step Solution:

1. **Understanding the LCR Circuit**:
- An LCR circuit consists of an inductor (L), a capacitor (C), and a resistor (R) connected in series. When the capacitor discharges through the inductor, the energy oscillates between the capacitor and the inductor.

2. **Writing the Voltage Equation**:
- According to Kirchhoff's loop rule, the sum of potential differences around the circuit must equal zero:
\[ V_L + V_R + V_C = 0 \]
- Where:
- \( V_L = L \frac{dI}{dt} \) (voltage across the inductor)
- \( V_R = IR \) (voltage across the resistor)
- \( V_C = \frac{Q}{C} \) (voltage across the capacitor)

3. **Substituting Current**:
- The current \( I \) is the rate of flow of charge \( Q \), so \( I = \frac{dQ}{dt} \).
- Substituting this into the equation gives:
\[ L \frac{d^2Q}{dt^2} + R \frac{dQ}{dt} + \frac{Q}{C} = 0 \]

4. **Forming the Differential Equation**:
- Rearranging the terms leads to the second-order differential equation:
\[ L \frac{d^2Q}{dt^2} + R \frac{dQ}{dt} + \frac{Q}{C} = 0 \]

5. **Identifying the Damped Oscillation**:
- The general solution to this equation indicates that the charge \( Q \) oscillates and can be expressed as:
\[ Q(t) = Q_0 e^{-\frac{R}{2L}t} \cos(\omega t) \]
- Here, \( \omega \) is the angular frequency of oscillation.

6. **Finding the Angular Frequency**:
- The angular frequency \( \omega \) is given by:
\[ \omega = \sqrt{\frac{1}{LC} - \left(\frac{R}{2L}\right)^2} \]

7. **Condition for Oscillation**:
- For oscillation to occur, the term under the square root must be positive:
\[ \frac{1}{LC} - \left(\frac{R}{2L}\right)^2 > 0 \]
- Rearranging this gives:
\[ \frac{1}{LC} > \frac{R^2}{4L^2} \]

8. **Conclusion**:
- Therefore, the correct condition that must be satisfied for the LCR circuit to behave as a damped harmonic oscillator is:
\[ \frac{1}{LC} > \frac{R^2}{4L^2} \]
- This implies that option (b) is correct.