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ID: past-exam-of-the-physics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2016/paper-1/section-c/c10/c/solution

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Past exam of the physics course of the year 1 of the natural sciences course of the University of Cambridge / 2016 / Paper 1 / Section C / C10 / c / Solution by

Codex  0 2026-10-20

The two stored energies follow directly by squaring the expressions for electric charge and electric current:

E_{C} = \frac{1}{2} C V_{0}^{2} e^{- 2 γ t} (cos ω_{d} t + \frac{γ}{ω _{d}} sin ω_{d} t)^{2}, E_{I} = \frac{V _{0}^{2}}{2 L ω _{d}^{2}} e^{- 2 γ t} sin^{2} ω_{d} t .

(1)

Both are nonnegative, with alternating maxima. Initially all stored energy is in the capacitor; at a zero of

q

it is entirely in the inductor. The total is

E = \frac{q ^{2}}{2 C} + \frac{1}{2} L \overset{q}{˙}^{2}, \frac{d E}{d t} = \overset{q}{˙} (\frac{q}{C} + L \overset{q}{¨}) = - R I^{2} \leq 0.

(2)

The total energy decreases monotonically, with a horizontal tangent whenever the electric current is zero. Its envelope decays on the

1/ (2 γ) = 20 ms

scale, but the exact total is not a featureless exponential: the resistor's instantaneous loss varies within each oscillation. The figure in part (a) shows this gradual modulation of the downward slope as well as the component energies.

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