JEE Main 2020 · 7 January · morning · Single correct · +4 / −1
A LCR circuit behaves like a damped harmonic oscillator. Comparing it with a physical spring-mass damped oscillator having damping constant 'b', the correct equivalence would be:
- (a) L ↔ k, C ↔ b, R ↔ m
- (b) L ↔ m, C ↔ k, R ↔ b
- (c) L ↔ m, C ↔ 1/k, R ↔ b
- (d) L ↔ 1/b, C ↔ 1/m, R ↔ 1/k
Answer and solution
Answer (c) L ↔ m, C ↔ 1/k, R ↔ b
Solution
Physics sir · 4 steps
Write both equations side by side and match term by term: the charge q plays the role of x.
- Spring-mass equation. For a mass on a spring with damping, Newton's law gives mẍ + bẋ + kx = 0.
- LCR equation. Kirchhoff's loop rule for a series LCR circuit gives Ldi/dt + iR + q/C = 0. With i = q̇, this becomes an equation in q alone.
Lq̈ + Rq̇ + 1/Cq = 0 ↔ mẍ + bẋ + kx = 0
- Match the terms. The coefficient of the second derivative gives L ↔ m. The coefficient of the first derivative gives R ↔ b. The coefficient of q itself is 1/C, so 1/C ↔ k, that is C ↔ 1/k.
- So. The correct equivalence is L ↔ m, C ↔ 1/k, R ↔ b, which is option (c).
The idea to keep. For any electrical-mechanical analogy, write both differential equations in the same form and match coefficients. Inductance is inertia, resistance is friction, and 1/C is stiffness.
Answer from the NTA final answer key. Solution by PrepWiser.