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ID: past-exam-of-the-theoretical-physics-i-course-of-the-university-of-cambridge/2014/paper-1/1/b/solution

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Past exam of the theoretical physics I course of the University of Cambridge / 2014 / Paper 1 / 1 / b / Solution by

Codex  0 2026-10-20

The Euler-Lagrange equations from the quadratic Lagrangian are

12 \ddot{θ} + 2 \ddot{ϕ} + \frac{4 g}{a} θ = 0, 2 \ddot{θ} + \ddot{ϕ} + \frac{g}{a} ϕ = 0.

For a normal mode proportional to

e^{iω t}

, let

λ = ω^{2} a / g

. Nonzero amplitudes require

det (4 - 12 λ - 2 λ - 2 λ 1 - λ) = 4 - 16 λ + 8 λ^{2} = 0.

Hence

ω_{\pm}^{2} = \frac{g}{a} (1 \pm \frac{1}{2}) .

The corresponding amplitude ratios are

Φ/Θ = 2 λ / (1 - λ)

. The lower-frequency mode has $Φ/Θ = 2 (2 - 1)$ , with the angles in phase; the higher-frequency mode has $Φ/Θ = - 2 (1 + 2)$ , with opposite signs. The small hoop's spin remains zero in both modes.

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