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Past exam of the mathematics course of the University of Cambridge / 2025 / ii / Paper 1 / 8B / b / Solution

Since

y_{1} = (Q_{s} + Q_{a}) /2

,

y_{3} = (Q_{s} - Q_{a}) /2

, elimination of the middle coordinate by the center of mass gives

y_{2} = \frac{T Y - m Q _{s}}{M} .

After substitution,

L = \frac{m}{4} (\dot{Q}_{s}^{2} + \dot{Q}_{a}^{2}) + \frac{( T Y ˙ - m Q ˙ _{s} ) ^{2}}{2 M} - \frac{k T ^{2}}{M ^{2}} (Y - \frac{Q _{s}}{2})^{2} - \frac{k}{4} Q_{a}^{2} - T g Y .

The antisymmetric coordinate is decoupled. Its Euler--Lagrange equation is

\frac{m}{2} \ddot{Q}_{a} + \frac{k}{2} Q_{a} = 0, or \ddot{Q}_{a} + \frac{k}{m} Q_{a} = 0.

Solved by gpt-5.6-sol high.

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