 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2020/ii/paper-1/8b/solution

Codex  0 Created 2026-09-24 Updated 2026-09-29

Let each mass be

m

and each spring constant be

k

, with the

x_{i}

denoting longitudinal displacements from equilibrium. The Lagrangian is

L = \frac{m}{2} \sum_{i = 1}^{4} \overset{x}{˙}_{i}^{2} - \frac{k}{2} [(x_{2} - x_{1})^{2} + (x_{3} - x_{2})^{2} + (x_{4} - x_{3})^{2}] .

The inverse coordinate transformation is

x_{1} = q_{1} + q_{3}, x_{4} = q_{1} - q_{3}, x_{2} = q_{2} + q_{4}, x_{3} = q_{2} - q_{4} .

Substitution gives

L = m (\overset{q}{˙}_{1}^{2} + \overset{q}{˙}_{2}^{2} + \overset{q}{˙}_{3}^{2} + \overset{q}{˙}_{4}^{2}) - k [(q_{2} - q_{1})^{2} + (q_{4} - q_{3})^{2} + 2 q_{4}^{2}] .

Thus the Euler-Lagrange equations separate into a symmetric pair,

m \overset{q}{¨}_{1} + k (q_{1} - q_{2}) = 0, m \overset{q}{¨}_{2} + k (q_{2} - q_{1}) = 0,

and an antisymmetric pair,

m \overset{q}{¨}_{3} + k (q_{3} - q_{4}) = 0, m \overset{q}{¨}_{4} + k (3 q_{4} - q_{3}) = 0.

The symmetric normal modes are:

$q_{1} = q_{2}$ , corresponding to $(x_{1}, x_{2}, x_{3}, x_{4}) \propto (1, 1, 1, 1)$ and the translational frequency

ω_{0} = 0;

ω_{s} = \frac{2 k}{m} .

For antisymmetric modes, the squared dimensionless frequencies are the eigenvalues of

(1 - 1 - 1 3),

namely

2 \mp 2

. They give

ω_{-} = \frac{k}{m} (2 - 2), (q_{3}, q_{4}) \propto (1, 2 - 1),

and

ω_{+} = \frac{k}{m} (2 + 2), (q_{3}, q_{4}) \propto (1, - 1 - 2) .

In the original coordinates their mode shapes are respectively

(1, 2 - 1, 1 - 2, - 1)

and

(1, - 1 - 2, 1 + 2, - 1) .

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