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Question:

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

1L

1C

LC

1LC

Solution:

For an LC circuit, applying Kirchoff's Voltage Law gives, qC + L(di/dt) = 0 => qC + L(d²q/dt²) = 0 => d²q/dt² = -1/LC q
In an SHM due to spring, F = d²x/dt² = -kx
Thus, by comparing both equations, we get the mechanical equivalent of k in LC circuit is 1/LC.
Hence option D is correct.