Question 215: AC fundamentals

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:

Accepted Answer

For spring mass damped oscillator
ma = - kx - bv
$$ \Rightarrow $$ ma + kx + bv = 0
$$ \Rightarrow $$ $$m{{{d^2}x} \over {d{t^2}}}$$ + b$${{dx} \over {dt}}$$ + kx = 0 ....(1)
For LCR circuit
L$${{di} \over {dt}}$$ + iR + $${q \over C}$$ = 0
$$ \Rightarrow $$ L$${{{d^2}q} \over {d{t^2}}}$$ + R$${{dq} \over {dt}}$$ + $${q \over C}$$ = 0 .....(2)
Comparing (1) and (2), we get
L $$ \leftrightarrow $$ m, C $$ \leftrightarrow $$ $${1 \over k}$$, R $$ \leftrightarrow $$ b

Answer

L $$ \leftrightarrow $$ k, C $$ \leftrightarrow $$ b, R $$ \leftrightarrow $$ m

Answer

L $$ \leftrightarrow $$ m, C $$ \leftrightarrow $$ k, R $$ \leftrightarrow $$ b

Answer

L $$ \leftrightarrow $$ m, C $$ \leftrightarrow $$ $${1 \over k}$$, R $$ \leftrightarrow $$ b

Answer

L $$ \leftrightarrow $$ $${1 \over b}$$, C $$ \leftrightarrow $$ $${1 \over m}$$, R $$ \leftrightarrow $$ $${1 \over k}$$

AC fundamentals

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:

Choose the correct answer: