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JEE Mains · Physics · STD 11 - Oscillations
  • JEE Mains 2020
  • Hard
  • MCQ

A \(LCR\) circuit behaves like a damped harmonic oscillator. Comparing it with a physical springmass damped oscillator having damping constant \(\mathrm{b}\), the correct equivalence would be:

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Answer & Solution

Correct Answer

(A) \(\mathrm{L} \leftrightarrow \mathrm{m}, \mathrm{C} \leftrightarrow \frac{1}{\mathrm{k}}, \mathrm{R} \leftrightarrow \mathrm{b}\)

Step-by-step Solution

Detailed explanation

By \(kVL\)

\(-\mathrm{L} \frac{\mathrm{di}}{\mathrm{dt}}-\frac{\mathrm{q}}{\mathrm{C}}-\mathrm{iR}=0\)

\(\mathrm{L} \frac{\mathrm{d}^{2} \mathrm{q}}{\mathrm{dt}^{2}}+\frac{1}{\mathrm{C}} \mathrm{q}+\mathrm{R} \frac{\mathrm{dq}}{\mathrm{dt}}=0\)

for damped oscillator

net force \(=-\mathrm{kx}-\mathrm{bv}=\mathrm{ma}\)

\(\frac{m d^{2} x}{d t^{2}}+k x+\frac{b d x}{d t}=0\)

by comparing : Equivalence is

\(\mathrm{L} \rightarrow \mathrm{m}: \mathrm{C} \rightarrow \frac{1}{\mathrm{K}} ; \mathrm{R} \rightarrow \mathrm{b}\)

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