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Physics · Ch 4 — Electromagnetic Induction and Alternating Current

Analogies Between LC Oscillations and Simple Harmonic Oscillations

4.9.3

Analogies Between LC Oscillations and Simple Harmonic Oscillations

Qualitative comparison. The electromagnetic oscillations of an LC circuit can be directly compared with the mechanical oscillations of a spring-mass system already studied in Class 11. Both systems involve energy oscillating between exactly two forms: in the LC circuit, between the capacitor's electrical energy and the inductor's magnetic energy; in the spring-mass system, between the spring's potential energy and the mass's kinetic energy (Table 4.2 sets these two energy pairs directly side by side).

Building the full analogy. Extending this pairing quantity by quantity (Table 4.3): electrical charge q corresponds to mechanical displacement x; current i=dq/dti=dq/dt corresponds to velocity v=dx/dtv=dx/dt; inductance L corresponds to mass m (both are measures of a system's inertia -- its resistance to a change in current, or in velocity, respectively); and the reciprocal of capacitance, 1/C1/C, corresponds to the spring's force constant k. The electrical energy 12(1/C)q2\tfrac12(1/C)q^2 then matches the spring's potential energy 12kx2\tfrac12kx^2 term-by-term, and the magnetic energy 12Li2\tfrac12Li^2 matches the kinetic energy 12mv2\tfrac12mv^2 term-by-term, so the total electromagnetic energy U=12(1/C)q2+12Li2U=\tfrac12(1/C)q^2+\tfrac12Li^2 corresponds exactly to the total mechanical energy E=12kx2+12mv2E=\tfrac12kx^2+\tfrac12mv^2.

Deriving the LC angular frequency by analogy. The spring-mass system's angular frequency is already known (from Class 11, unit 10) to be ω=k/m\omega=\sqrt{k/m}. Applying the substitutions k→1/Ck\to1/C and m→Lm\to L established above, this becomes directly

ω=1LC(4.58)\omega = \dfrac{1}{\sqrt{LC}} \qquad (4.58) …

Table 4.2Energy in the two oscillatory systems
LC oscillator ElementLC oscillator EnergySpring-mass ElementSpring-mass Energy
CapacitorElectrical Energy =12q2C=\dfrac{1}{2}\dfrac{q^2}{C}SpringPotential energy =12kx2=\dfrac{1}{2}kx^2
Table 4.3Analogies between electrical and mechanical quantities
Electrical systemMechanical system
Charge qqDisplacement xx
Current i=dq/dti=dq/dtVelocity v=dx/dtv=dx/dt
Inductance LLMass mm
Reciprocal of capacitance 1/C1/CForce constant kk
Electrical energy =121Cq2=\dfrac{1}{2}\dfrac{1}{C}q^2Potential energy =12kx2=\dfrac{1}{2}kx^2
Magnetic energy =12Li2=\dfrac{1}{2}Li^2Kinetic energy =12mv2=\dfrac{1}{2}mv^2