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ID: past-exam-of-the-mathematical-methods-ii-course-of-the-university-of-oxford/2015/10/a/i/solution

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Past exam of the mathematical methods II course of the University of Oxford / 2015 / 10 / a / i / Solution by

Codex  0 2026-10-25

Let

x_{1}, x_{2}

be displacements from equilibrium, with positive directions chosen alike. Hooke's law and Newton's second law give

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

For a normal mode

x_{j} = a_{j} cos Ω t

, the determinant condition is

(k_{1} + k_{2} - m_{1} Ω^{2}) (k_{2} + k_{3} - m_{2} Ω^{2}) - k_{2}^{2} = 0.

Solving the quadratic for

Ω^{2}

gives

Ω_{\pm}^{2} = \frac{1}{2} \frac{k _{1} + k _{2}}{m _{1}} + \frac{k _{2} + k _{3}}{m _{2}} \pm (\frac{k _{1} + k _{2}}{m _{1}} - \frac{k _{2} + k _{3}}{m _{2}})^{2} + \frac{4 k _{2}^{2}}{m _{1} m _{2}} .

The angular frequencies are the positive square roots. Positive masses and spring constants make the stiffness quadratic form

k_{1} x_{1}^{2} + k_{2} (x_{2} - x_{1})^{2} + k_{3} x_{2}^{2}

positive definite, ensuring both squared frequencies are positive. Distinct spring constants do not require equal masses or any symmetry approximation.

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