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 :

Answer 1

Correct option is D. $$L\leftrightarrow m, C\leftrightarrow \dfrac{1}{k}, R\leftrightarrow b$$
For a spring-mass damped oscillator

$$\dfrac{d^2x}{dt^2}+\left(\dfrac{2x}{2\sqrt{mk}}\sqrt{\dfrac{k}{m}}\right)\dfrac{dx}{dt}+\left(\sqrt{\dfrac{k}{m}}\right)^2x=0$$

$$\Rightarrow \boxed{\dfrac{d^2x}{dt^2}+2\tau w_0\dfrac{dx}{dt}+w_0^2x=0}$$

$$\tau=\dfrac{c}{2\sqrt{mk}}\Rightarrow $$ damping ratio $$c=b\Rightarrow$$ damping constant

$$W_0=\sqrt{\dfrac{k}{m}}\Rightarrow$$ angular frequency

$$\Rightarrow \boxed{\dfrac{d^2x}{dt^2}+(\dfrac{b}{m})\dfrac{dx}{dt}+(\dfrac{k}{m})x=0}$$..........(1)

For LCR circuit

use KVL:-

$$V-L\dfrac{dI}{dt}-IR-\dfrac{q}{c}=0$$

$$\Rightarrow \boxed{\dfrac{d^2q}{dt^2}+\dfrac{R}{L}\dfrac{dq}{dt}+(\dfrac{1}{LC})q=\dfrac{V}{L}}$$...........(2) $$I=\dfrac{dq}{dt}$$

By comparing (1) and (2):=

$$L\rightarrow m,\ \ \ R\rightarrow b,\ \ \ \ \ LC\rightarrow \dfrac{m}{k}$$

$$\Rightarrow C\rightarrow \dfrac{1}{k}$$

option (D) is correct.