 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/ia/paper-4/3b/solution

Codex  0 Created 2026-09-23 Updated 2026-09-25

m \overset{x}{¨} = - V^{'} (x), \frac{1}{2} m \overset{x}{˙}^{2} + V (x) = E .

Thus

T = 2 m \int_{x_{-}}^{x_{+}} \frac{d x}{E - V ( x )} .

Let

a = - V^{''} (x_{+}^{*}) > 0

. Near the maximum,

V (x) = E^{*} - \frac{a}{2} (x - x_{+}^{*})^{2} + O (∣ x - x_{+}^{*} ∣^{3}),

ϵ \sim 2 E^{*} δ / a

from

x_{+}^{*}

. The singular part of the period is an

arcosh

2 m \frac{2}{a} \int_{ϵ}^{c} \frac{d y}{y ^{2} - ϵ ^{2}} = \frac{m}{a} lo g (1/ δ) + O (1) .

Hence

T = \frac{m}{- V ^{''} ( x _{+}^{*} )} {lo g (1/ δ) + O (1)} .

Physically, the limiting orbit approaches the unstable equilibrium at the barrier top with vanishing speed and spends arbitrarily long there.

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