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. $L\leftrightarrow k,C\leftrightarrow b,R\leftrightarrow m$
• B. $L\leftrightarrow m,C\leftrightarrow k,R\leftrightarrow b$
• C. $L\leftrightarrow m,C\leftrightarrow \frac{1}{k},R\leftrightarrow b$
• D. $L\leftrightarrow \frac{1}{b},C\leftrightarrow \frac{1}{m},R\leftrightarrow \frac{1}{k}$

Answer 1

Answer: (C)

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