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ID: past-exam-of-the-mathematical-methods-ii-course-of-the-university-of-oxford/2014/9/c/solution

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Past exam of the mathematical methods II course of the University of Oxford / 2014 / 9 / c / Solution by

Codex  0 2026-10-25

For a mode with upper and lower displacement amplitudes

A, B

, the second equation gives

- A + (1 - s) B = 0

. Consequently

(\frac{A}{B})_{slow} = \frac{5 - 1}{2}, (\frac{A}{B})_{fast} = - \frac{1 + 5}{2} .

(1)

These are signed ratios. The slow mode is in phase and the lower mass has the larger amplitude. The fast mode is out of phase and the upper mass has the larger amplitude. Taking an absolute ratio alone would discard the physically important relative phase.

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