 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/ii/paper-2/14b/b/solution

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

Put

Δ = θ_{1} - θ_{2}

and

ω_{0}^{2} = g / l

l

, are

(M + m) \ddot{θ}_{1} + m cos Δ \ddot{θ}_{2} + m sin Δ \dot{θ}_{2}^{2} + (M + m) ω_{0}^{2} sin θ_{1} = 0,

\ddot{θ}_{2} + cos Δ \ddot{θ}_{1} - sin Δ \dot{θ}_{1}^{2} + ω_{0}^{2} sin θ_{2} = 0.

Time-translation invariance conserves the total energy

E = \frac{1}{2} (M + m) l^{2} \dot{θ}_{1}^{2} + \frac{1}{2} m l^{2} \dot{θ}_{2}^{2} + m l^{2} cos Δ \dot{θ}_{1} \dot{θ}_{2} - (M + m) g l cos θ_{1} - m g l cos θ_{2} .

Setting both angles and velocities to zero satisfies the equations, so the downward configuration is an equilibrium. Writing

θ_{i} = z_{i}

and retaining linear terms gives

(M + m) \overset{z}{¨}_{1} + m \overset{z}{¨}_{2} + (M + m) ω_{0}^{2} z_{1} = 0,

\overset{z}{¨}_{1} + \overset{z}{¨}_{2} + ω_{0}^{2} z_{2} = 0.

Equivalently, the mass and stiffness matrices are

M = l^{2} (M + m m m m), K = g l (M + m 0 0 m) .

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