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 will be

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

Answer 1

The correct option is D $L\leftrightarrow m$, $C\leftrightarrow \frac{1}{k}$, $R\leftrightarrow b$

For damped oscillator
by Newton's second law
$$\begin{array}{cc}& -kx-bv=ma\\ & kx+bv+ma=0\\ & kx+b\frac{dx}{dt}+m\frac{d^{2}x}{d{t}^2}=0\\ & \text{For LCR circuit by KVL}\\ & -IR-L\frac{dI}{dt}-\frac{q}{c}=0\\ & ⟹IR+L\frac{dI}{dt}+\frac{q}{c}=0\\ & ⟹\frac{q}{c}+R\frac{dq}{dt}+L\frac{d^{2}q}{d{t}^2}=0\\ & \text{By comparing}\\ & R⟹b\\ & c⟹\frac{1}{k}\\ & m⟹L\end{array}$$