A LCR circuit behaves like a damped harmonic oscillator. Comparing it with a physical spring-mass damped oscillator having damping constant ' $b$ ', the connect equivalence would be:

Option 1 - <math> <mi mathvariant="normal">L</mi> <mo>↔</mo> <mfrac> <mrow> <mrow> <mn>1</mn> </mrow> </mrow> <mrow> <mrow> <mtext> </mtext> <mi mathvariant="normal">b</mi> </mrow> </mrow> </mfrac> <mo>,</mo> <mi mathvariant="normal">C</mi> <mo>↔</mo> <mfrac> <mrow> <mrow> <mn>1</mn> </mrow> </mrow> <mrow> <mrow> <mtext> </mtext> <mi mathvariant="normal">m</mi> </mrow> </mrow> </mfrac> <mo>,</mo> <mi mathvariant="normal">R</mi> <mo>↔</mo> <mfrac> <mrow> <mrow> <mn>1</mn> </mrow> </mrow> <mrow> <mrow> <mi mathvariant="normal">k</mi> </mrow> </mrow> </mfrac> </math>

Option 2 - <math> <mi mathvariant="normal">L</mi> <mo>↔</mo> <mi mathvariant="normal">k</mi> <mo>,</mo> <mi mathvariant="normal">C</mi> <mo>↔</mo> <mi mathvariant="normal">b</mi> <mo>,</mo> <mi mathvariant="normal">R</mi> <mo>↔</mo> <mi mathvariant="normal">m</mi> </math>

Option 3 - <math> <mi mathvariant="normal">L</mi> <mo>↔</mo> <mi mathvariant="normal">m</mi> <mo>,</mo> <mi mathvariant="normal">C</mi> <mo>↔</mo> <mfrac> <mrow> <mrow> <mn>1</mn> </mrow> </mrow> <mrow> <mrow> <mi mathvariant="normal">k</mi> </mrow> </mrow> </mfrac> <mo>,</mo> <mi mathvariant="normal">R</mi> <mo>↔</mo> <mi mathvariant="normal">b</mi> </math>

Option 4 - <math> <mi mathvariant="normal">L</mi> <mo>↔</mo> <mi mathvariant="normal">m</mi> <mo>,</mo> <mi mathvariant="normal">C</mi> <mo>↔</mo> <mi mathvariant="normal">k</mi> <mo>,</mo> <mi mathvariant="normal">R</mi> <mo>↔</mo> <mi mathvariant="normal">b</mi> </math>

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Vishal Baghel | Contributor-Level 10

5 months ago

Correct Option - 3

Detailed Solution:

In damped oscillation

$∴ v_{2} = \frac{1}{8} * \frac{1}{T} = \frac{1}{8} * \frac{10^{16}}{1.6} = 7.8 * 10^{14}$

$m a + b v + k x = 0$

In the circuit

$m \frac{d^{2} x}{{d t^{2}}^{2}} + b \frac{d x}{d t} + k x = 0$

$- i R - L \frac{d i}{d t} - \frac{q}{C} = 0$

Comparing equation (i) and (ii)

$L \frac{d^{2} q}{d t^{2}} + R \frac{d q}{d t} + \frac{1}{c} q = 0$

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A LCR circuit behaves like a damped harmonic oscillator. Comparing it with a physical spring-mass damped oscillator having damping constant ' b ', the connect equivalence would be:

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