Past exam of the physics A course of the University of Cambridge / 2018 / Paper 2 / Section B / B6 / d

Solution

q_{0} = \frac{V _{0} / L}{1/ ( L C ) - ω ^{2} + i ( R / L ) ω} .

∣ q_{0} ∣

minimizes

D (ω) = (ω_{0}^{2} - ω^{2})^{2} + γ^{2} ω^{2}, ω_{0}^{2} = \frac{1}{L C}, γ = \frac{R}{L} .

Differentiate:

\frac{d D}{d ω} = 2 ω [- 2 (ω_{0}^{2} - ω^{2}) + γ^{2}] .

The positive-frequency minimum therefore has

ω_{res} = \frac{1}{L C} - \frac{R ^{2}}{2 L ^{2}}, R^{2} < \frac{2 L}{C} .

This is the resonance of charge, or capacitor voltage. If the stated inequality fails, their maximum is at zero frequency, and no positive displacement-resonance peak exists. The electric current has the additional factor

ω

, and instead peaks exactly at

ω_{0} = 1/ L C

for every

R > 0

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