 Solution

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/a/solution

Put

ω_{0}^{2} = 1/ (L C)

q (t) = Q e^{- γ t} (cos ω_{d} t + \frac{γ}{ω _{d}} sin ω_{d} t), V_{C} (t) = V_{0} e^{- γ t} (cos ω_{d} t + \frac{γ}{ω _{d}} sin ω_{d} t) .

The printed electric current arrow points toward the positively labelled capacitor plate, so its signed electric current is

I = \overset{q}{˙}

, not the positive discharge magnitude. Hence

I (t) = - \frac{V _{0}}{L ω _{d}} e^{- γ t} sin ω_{d} t .

γ = 25 s^{- 1}, ω_{0} ≃ 447.21 s^{- 1}, ω_{d} ≃ 446.51 s^{- 1}, T_{d} ≃ 14.07 ms .

The voltage starts at $6 V$ with zero slope; the signed electric current starts at zero and initially becomes negative. Its initial slope is

\dot{I} (0) = - V_{0} / L = - 6 A s^{- 1}

. Both oscillate with an

e^{- 25 t}

envelope and a

40 ms

amplitude time constant. The phase difference is close to a quarter period, with the small damping correction displayed explicitly in

V_{C}

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