Question

What is the mechanical equivalent of spring constant k in LC oscillating circuit?

• A. $\frac{1}{L}$
• B. $\frac{1}{C}$
• C. $\frac{L}{C}$
• D. $\frac{1}{LC}$

Answer 1

The correct option is D $\frac{1}{LC}$

For an LC circuit, applying Kirchoff's Voltage Law gives,

$\frac{q}{C}+L\frac{di}{dt}=0$

$⟹\frac{q}{C}+L\frac{d^{2}q}{d{t}^2}=0$

$⟹\frac{d^{2}q}{d{t}^2}=-\frac{1}{LC}q$

In an SHM due to spring, $F=\frac{d^{2}x}{d{t}^2}=-kx$

Thus, by compairing both equations, we get the mechanical equivalent of $k$ in LC circuit is $\frac{1}{LC}$.

Hence option D is correct.