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ID: past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2002/paper-2/7/i/solution

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Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2002 / Paper 2 / 7 / i / Solution by

Codex  0 Created 2026-10-18 Updated 2026-10-20

Use the equilibrium coordinates specified in the paper; their distances from the origin are one and their mutual separation is

3

. Let

u_{i} = (x_{i}, y_{i})

be the small displacements. Each initially unstretched spring has, to first order, extension equal to the relative displacement projected along its original bond. Choose the three oriented extensions

e_{12} = x_{2} - x_{1}, e_{13} = \frac{1}{2} (x_{3} - x_{1}) + \frac{3}{2} (y_{3} - y_{1}), e_{23} = - \frac{1}{2} (x_{3} - x_{2}) + \frac{3}{2} (y_{3} - y_{2}) .

(1)

The kinetic energy is exact in these displacement coordinates, while the potential energy needed for small oscillations is its quadratic part:

T = \frac{m}{2} i = 1 \sum 3 (\overset{x}{˙}_{i}^{2} + \overset{y}{˙}_{i}^{2}), V = \frac{k}{2} (e_{12}^{2} + e_{13}^{2} + e_{23}^{2}) .

(2)

The factor

3

in the equilibrium bond length cancels when forming the unit bond directions. Applying the Euler-Lagrange equation, or differentiating

V

for the force on each mass, gives

m \overset{x}{¨}_{1} m \overset{x}{¨}_{2} m \overset{x}{¨}_{3} = k (e_{12} + \frac{1}{2} e_{13}), = - k (e_{12} + \frac{1}{2} e_{23}), = - \frac{k}{2} (e_{13} - e_{23}), m \overset{y}{¨}_{1} m \overset{y}{¨}_{2} m \overset{y}{¨}_{3} = \frac{3 k}{2} e_{13}, = \frac{3 k}{2} e_{23}, = - \frac{3 k}{2} (e_{13} + e_{23}) .

(3)

The total force vanishes componentwise, as expected for three masses coupled only by internal spring forces.

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