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ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2021/ii/paper-2/14d/c/solution

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Past exam of the mathematics course of the University of Cambridge / 2021 / ii / Paper 2 / 14D / c / Solution by

Codex  0 Created 2026-09-24 Updated 2026-09-25

Put

θ = k lo g t

. Since

\overset{x}{˙} = \frac{s i n θ + 2 k c o s θ}{2 k t}, ϵ^{- 2} = k^{2} + \frac{1}{4},

the instantaneous oscillator energy is

E (t) = \frac{1}{2} \overset{x}{˙}^{2} + \frac{x ^{2}}{2 ( ϵ t ) ^{2}} = \frac{1 - c o s 2 θ + 2 k s i n 2 θ + 4 k ^{2}}{8 k ^{2} t} .

The instantaneous frequency is

ω (t) = 1/ (ϵ t)

, so

I (t) = \frac{E ( t )}{ω ( t )} = \frac{ϵ}{8 k ^{2}} (4 k^{2} + 1 - cos 2 θ + 2 k sin 2 θ) .

Its oscillatory part divided by its mean is

O (k^{- 1}) = O (ϵ)

. It remains bounded and periodic in

lo g t

, so the variations do not accumulate. This is the expected behavior of an adiabatic invariant: slow parameter variation produces small bounded oscillations rather than secular drift.

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