Question:
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 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:
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Compare the equation of the damped mechanical oscillator, \(m\ddot x+b\dot x+kx=0\), with that of the series LCR circuit, \(L\ddot q+R\dot q+q/C=0\).
Updated On: Oct 31, 2026
- L \( \leftrightarrow \) k, C \( \leftrightarrow \) b, R \( \leftrightarrow \) m
- L \( \leftrightarrow \) m, C \( \leftrightarrow \) k, R \( \leftrightarrow \) b
- L \( \leftrightarrow \) m, C \( \leftrightarrow \) \( {1 \over k} \), R \( \leftrightarrow \) b
- L \( \leftrightarrow \) \( {1 \over b} \), C \( \leftrightarrow \) \( {1 \over m} \), R \( \leftrightarrow \) \( {1 \over k} \)
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The Correct Option is C
Solution and Explanation
Spring-mass with damping: \(m\dfrac{d^2x}{dt^2}+b\dfrac{dx}{dt}+kx=0\).
Series LCR circuit: \(L\dfrac{d^2q}{dt^2}+R\dfrac{dq}{dt}+\dfrac{q}{C}=0\).
Comparing term by term: \(x\leftrightarrow q\), \(m\leftrightarrow L\), \(b\leftrightarrow R\), \(k\leftrightarrow\dfrac1C\), i.e. \(C\leftrightarrow\dfrac1k\).
Answer: (C) L \(\leftrightarrow\) m, C \(\leftrightarrow\) 1/k, R \(\leftrightarrow\) b
Series LCR circuit: \(L\dfrac{d^2q}{dt^2}+R\dfrac{dq}{dt}+\dfrac{q}{C}=0\).
Comparing term by term: \(x\leftrightarrow q\), \(m\leftrightarrow L\), \(b\leftrightarrow R\), \(k\leftrightarrow\dfrac1C\), i.e. \(C\leftrightarrow\dfrac1k\).
Answer: (C) L \(\leftrightarrow\) m, C \(\leftrightarrow\) 1/k, R \(\leftrightarrow\) b
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