Question
If a LC circuit is considered analogous to a harmonically oscillating spring block system, which energy of the LC circuit would be analogous to potential energy and which one analogous to kinetic energy?
✓
Answer
When a charged capacitor C having an initial charge $q_0$ is discharged through an inductance L, the charge and current in the circuit start oscillating simple harmonically. If the resistance of the circuit is zero, no energy is dissipated as heat. We also assume an idealized situation in which energy is not radiated away from the circuit. The total energy associated with the circuit is constant.
The oscillation of the LC circuit are an electromagnetic analog to the mechanical oscillation of a block-spring system.
The total energy of the system remains conserved.
$\frac{1}{2}\text{CV}^2+\frac{1}{2}\text{LI}^2=\text{Constant}=\frac{1}{2}\text{CV}^2_0=\frac{1}{2}\text{LI}^2_0$
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Mass spring systerm
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v/s
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LC circuit
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Displacement (x)
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Charge (q)
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Velocity (v)
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Current (i)
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Acceleration (a)
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Rate of change of current $\Big(\frac{\text{di}}{\text{dt}}\Big)$
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Mass (m) [Inertia]
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Inductance (L) [Inertia of electricity]
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Momentum (p = mv)
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Magnetic flux $(\phi=\text{Li})$
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Retarding force $\Big(-\text{m}\frac{\text{dv}}{\text{dt}}\Big)$
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Self induced emf $\Big(-\text{L}\frac{\text{di}}{\text{dt}}\Big)$
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Equation of free oscillations:
$\frac{\text{d}^2\text{x}}{\text{dt}^2}=-\omega^2\text{x};$
where $\omega=\sqrt{\frac{\text{K}}{\text{m}}}$
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Equation of free oscillations:
$\frac{\text{d}^2\text{q}}{\text{dt}^2}=-\Big(\frac{1}{\text{LC}}\Big).\text{q};\text{where }\omega^2=\frac{1}{\text{LC}}$
$\Rightarrow\ \omega=\frac{1}{\sqrt{\text{LC}}}$
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Force constant K
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Capacitance C
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Kinetic energy $=\frac{1}{2}\text{mv}^2$
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Magnetic energy $=\frac{1}{2}\text{Li}^2$
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Elastic potential energy $=\frac{1}{2}\text{Kx}^2$
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Elecrical potential energy $=\frac{1}{2}\frac{\text{q}^2}{\text{C}}$
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