 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2015/ii/paper-4/7d/a/solution

Codex  0 Created 2026-09-24 Updated 2026-10-06

Let

x_{1}, x_{2}, x_{3}

be longitudinal displacements from equilibrium. The original diagram has masses

m, 2 m, m

, and no attachments to fixed walls. The Lagrangian is

L = \frac{1}{2} m (\overset{x}{˙}_{1}^{2} + 2 \overset{x}{˙}_{2}^{2} + \overset{x}{˙}_{3}^{2}) - \frac{1}{2} k [(x_{2} - x_{1})^{2} + (x_{3} - x_{2})^{2}] .

m \overset{x}{¨}_{1} = k (x_{2} - x_{1}), 2 m \overset{x}{¨}_{2} = k (x_{1} - 2 x_{2} + x_{3}), m \overset{x}{¨}_{3} = k (x_{2} - x_{3}) .

The forces sum to zero, consistent with momentum conservation for an isolated molecule.

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