Question 131 of 190

A $LCR$ circuit behaves like a clamped harmonic oscillator. Comparing it with a physical spring-mass damped oscillator having damping constant $'{b}^{'},$ the correct equivalence would be:

A$L↔m,C↔k,R↔b$B$L↔\dfrac{1}{b},C↔\dfrac{1}{m},R↔\dfrac{1}{k}$C$L↔k,C↔b,R↔m$D$L↔m,C↔\dfrac{1}{k},R↔b$

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Answer: (D) $L↔m,C↔\dfrac{1}{k},R↔b$

In damped oscillation $ma+bv+kx=0$

$m\dfrac{{d}^{2}x}{d{t}^{2}}+b\dfrac{dx}{dt}+kx=0\ldots ..\left(i\right)$
In the circuit
$-iR-L\dfrac{di}{dt}-\dfrac{q}{c}=0$
$L\dfrac{{d}^{2}q}{d{t}^{2}}+R\dfrac{dq}{dt}+\dfrac{1}{c}.q=0..\left(ii\right)$
Comparing equation (i) and (ii)
$m=L,b=R,k=\dfrac{1}{c}$