question_answer

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 [JEE MAIN Held on 07-01-2020 Morning]

A) \[L\leftrightarrow m,C\leftrightarrow \frac{1}{k},R\leftrightarrow b\]

B) \[L\leftrightarrow k,C\leftrightarrow b,R\leftrightarrow m\]

C) \[L\leftrightarrow \frac{1}{b},C\leftrightarrow \frac{1}{m},R\leftrightarrow \frac{1}{k}\]

D) \[L\leftrightarrow m,C\leftrightarrow k,R\leftrightarrow b\]

Correct Answer: A

Solution :

[a]

In damped oscillation,

\[\frac{m{{d}^{2}}x}{d{{t}^{2}}}=-kx-bv\]

\[\Rightarrow \frac{m{{d}^{2}}x}{d{{t}^{2}}}+b\frac{dx}{dt}+kx=0...(i)\]

In LCR circuit, \[\frac{-q}{C}-iR-\frac{Ldi}{dt}=0\]

\[L\frac{{{d}^{2}}q}{d{{t}^{2}}}+R\frac{dq}{dt}+\frac{q}{C}=0\]

\[\therefore m\equiv L,b\equiv R,k\equiv \frac{1}{C}\]