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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
$$L\leftrightarrow m, C\leftrightarrow k, R\leftrightarrow b$$
B
$$L\leftrightarrow \dfrac{1}{b}, C\leftrightarrow \dfrac{1}{m}, R\leftrightarrow \dfrac{1}{k}$$
C
$$L\leftrightarrow k, C\leftrightarrow b, R\leftrightarrow m$$
D
$$L\leftrightarrow m, C\leftrightarrow \dfrac{1}{k}, R\leftrightarrow b$$
Solution
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#### 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 circuituse 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.

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