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

In damped oscillation ∴ v 2 = 1 8 × 1 T = 1 8 × 10 16 1.6 = 7.8 × 10 14 m a + b v + k x = 0 In the circuit m d 2 x d t 2 2 + b d x d t + k x = 0 - i R - L d i d t - q C = 0 Comparing equation (i) and (ii) L d 2 q d t 2 + R d q d t + 1 c q = 0

A LCR circuit behaves like a damped harmonic oscillator. Comparing it with a physical spring-mass damped oscillator having damping constant '&nbsp;<math></math> <math> <mi>b</mi> </math> ', the connect equivalence would be:

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Alternating Current Overview Physics Alternating Current Physics Class 12th

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 -

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

Option 2 -

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

Option 3 -

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

Option 4 -

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

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1 Answer

Answered by

Vishal Baghel | Contributor-Level 10

a month ago

Correct Option - 3

Detailed Solution:

In damped oscillation

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

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

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)

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

In damped oscillation <math> <mo>∴</mo> <mrow> <mi mathvariant="normal"> </mi> </mrow> <msub> <mrow> <mrow> <mi>v</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msub> <mo>=</mo> <mfrac> <mrow> <mrow> <mn>1</mn> </mrow> </mrow> <mrow> <mrow> <mn>8</mn> </mrow> </mrow> </mfrac> <mo>×</mo> <mfrac> <mrow> <mrow> <mn>1</mn> </mrow> </mrow> <mrow> <mrow> <mtext> </mtext> <mi mathvariant="normal">T</mi> </mrow> </mrow> </mfrac> <mo>=</mo> <mfrac> <mrow> <mrow> <mn>1</mn> </mrow> </mrow> <mrow> <mrow> <mn>8</mn> </mrow> </mrow> </mfrac> <mo>×</mo> <mfrac> <mrow> <mrow> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mn>16</mn> </mrow> </mrow> </msup> </mrow> </mrow> <mrow> <mrow> <mn>1.6</mn> </mrow> </mrow> </mfrac> <mo>=</mo> <mn>7.8</mn> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mn>14</mn> </mrow> </mrow> </msup> </math> <math> <mi>m</mi> <mi>a</mi> <mo>+</mo> <mi>b</mi> <mi>v</mi> <mo>+</mo> <mi>k</mi> <mi>x</mi> <mo>=</mo> <mn>0</mn> </math> In the circuit <math> <mi>m</mi> <mfrac> <mrow> <mrow> <msup> <mrow> <mrow> <mi>d</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msup> <mi>x</mi> </mrow> </mrow> <mrow> <mrow> <msup> <mrow> <mrow> <mi>d</mi> <msup> <mrow> <mrow> <mi>t</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msup> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msup> </mrow> </mrow> </mfrac> <mo>+</mo> <mi>b</mi> <mfrac> <mrow> <mrow> <mi>d</mi> <mi>x</mi> </mrow> </mrow> <mrow> <mrow> <mi>d</mi> <mi>t</mi> </mrow> </mrow> </mfrac> <mo>+</mo> <mi>k</mi> <mi>x</mi> <mo>=</mo> <mn>0</mn> </math> <math> <mo>-</mo> <mi mathvariant="normal">i</mi> <mi mathvariant="normal">R</mi> <mo>-</mo> <mi mathvariant="normal">L</mi> <mfrac> <mrow> <mrow> <mi mathvariant="normal">d</mi> <mi mathvariant="normal">i</mi> </mrow> </mrow> <mrow> <mrow> <mi mathvariant="normal">d</mi> <mi mathvariant="normal">t</mi> </mrow> </mrow> </mfrac> <mo>-</mo> <mfrac> <mrow> <mrow> <mi mathvariant="normal">q</mi> </mrow> </mrow> <mrow> <mrow> <mi mathvariant="normal">C</mi> </mrow> </mrow> </mfrac> <mo>=</mo> <mn>0</mn> </math> Comparing equation (i) and (ii)<img > <math> <mi>L</mi> <mfrac> <mrow> <mrow> <msup> <mrow> <mrow> <mi>d</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msup> <mi>q</mi> </mrow> </mrow> <mrow> <mrow> <mi>d</mi> <msup> <mrow> <mrow> <mi>t</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msup> </mrow> </mrow> </mfrac> <mo>+</mo> <mi>R</mi> <mfrac> <mrow> <mrow> <mi>d</mi> <mi>q</mi> </mrow> </mrow> <mrow> <mrow> <mi>d</mi> <mi>t</mi> </mrow> </mrow> </mfrac> <mo>+</mo> <mfrac> <mrow> <mrow> <mn>1</mn> </mrow> </mrow> <mrow> <mrow> <mi>c</mi> </mrow> </mrow> </mfrac> <mi>q</mi> <mo>=</mo> <mn>0</mn> </math>

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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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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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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: