 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2021/ii/paper-1/8d/a/solution

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

The kinetic energy of the two equal masses is

T = \frac{m}{2} (\overset{x}{˙}_{1}^{2} + \overset{x}{˙}_{2}^{2}) .

The extensions of the left, coupling, and right springs are respectively

x_{1}

x_{2} - x_{1}

, and

x_{2}

. Their potential energy is therefore

U = \frac{1}{2} k_{1} x_{1}^{2} + \frac{1}{2} k_{3} (x_{2} - x_{1})^{2} + \frac{1}{2} k_{2} x_{2}^{2} .

Thus the mechanical Lagrangian

L = T - U

(T_{ij}) = (m 0 0 m), (V_{ij}) = (k_{1} + k_{3} - k_{3} - k_{3} k_{2} + k_{3}) .

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