Question

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

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

Answer 1

The correct option is B

Step 1. Given data,

A $LCR$circuit behaves like a damped harmonic oscillator

Step 2. Damped oscillator by Newton’s second law,

For damped oscillator by Newton’s second law,

If the velocity of the spring-mass system is $v$, acceleration is $a$, the spring constant is $k$, displacement from the mean position is $x$ and damping constant $b$ then,

$$\begin{gathered} -kx-bv=ma \\ kx+bv+ma=0 \\ kx+b\left(\frac{dx}{dt}\right)+m\left(\frac{d^{2}x}{d{t}^2}\right)=0\dots \left(1\right)\left[As,v=\frac{dx}{dt}anda=\frac{dv}{dt}\right] \end{gathered}$$

Step 3. On applying KVL in series LCR circuit,

For $LCR$ circuit by$KVL$,

If the current through the circuit is $I$, the value of resistance is$R$, the amount of charge is$q$, inductance is $L$, the capacitor is $c$ then

$$\begin{gathered} -IR-L\left(\frac{dl}{dt}\right)–\frac{q}{c}=0 \\ IR+L\left(\frac{dl}{dt}\right)+\frac{q}{c}=0 \\ \frac{q}{c}+R\left(\frac{dq}{dt}\right)+L\left(\frac{d^{2}q}{d{t}^2}\right)=0\dots \left(2\right)\left[As,I=\left(\frac{dq}{dt}\right)\right] \end{gathered}$$

By comparing equation $\left(1\right)$ and equation $\left(2\right)$,

$$\begin{gathered} R\Rightarrow b \\ c\Rightarrow \frac{1}{k} \\ m\Rightarrow L \end{gathered}$$

Hence, option $\left(B\right)$ is correct.