The electrical analog of a spring constant k is

Options

A. L
B. R
C. C
D. `(1)/(C)`

Answer

To find the electrical analog of a spring constant \( k \), we can analyze the behavior of an LC circuit (which consists of an inductor \( L \) and a capacitor \( C \)) and compare it to the mechanics of a spring.

### Step-by-Step Solution:

1. **Understanding the LC Circuit**:
- In an LC circuit, we have an inductor (L) and a capacitor (C) connected in parallel. The capacitor stores electrical energy in the form of an electric field, while the inductor stores energy in the form of a magnetic field.

2. **Setting Up the Equations**:
- Let's denote the charge on the capacitor as \( Q \) and the current through the inductor as \( I \). The potential across the capacitor is given by:
\[ V_C = \frac{Q}{C} \]
- The back EMF (electromotive force) across the inductor is given by:
\[ V_L = -L \frac{dI}{dt} \]

3. **Relating Charge and Current**:
- Since the capacitor is discharging, we can express the current \( I \) as:
\[ I = -\frac{dQ}{dt} \]
- Substituting this into the equation for the inductor gives:
\[ V_L = -L \frac{d}{dt}\left(-\frac{dQ}{dt}\right) = L \frac{d^2Q}{dt^2} \]

4. **Equating the Potentials**:
- Since the capacitor and inductor are in parallel, their potentials are equal:
\[ \frac{Q}{C} = L \frac{d^2Q}{dt^2} \]

5. **Rearranging the Equation**:
- Rearranging gives us:
\[ L \frac{d^2Q}{dt^2} + \frac{Q}{C} = 0 \]
- This resembles the equation of motion for simple harmonic motion (SHM), which is:
\[ m \frac{d^2x}{dt^2} + kx = 0 \]

6. **Identifying Analogies**:
- In SHM, \( m \) is the mass and \( k \) is the spring constant. In our electrical analogy:
- The electrical analog of mass \( m \) is the inductor \( L \).
- The electrical analog of the spring constant \( k \) is \( \frac{1}{C} \).

7. **Conclusion**:
- Therefore, the electrical analog of the spring constant \( k \) is:
\[ k = \frac{1}{C} \]

### Final Answer:
The electrical analog of a spring constant \( k \) is \( \frac{1}{C} \).