 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2022/ib/paper-1/13b/b/solution

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

With gravity omitted, the initial data contain only the first fixed-end normal mode, so write

y (x, t) = q (t) sin \frac{π x}{l} .

Then

q^{''} + 2 k q^{'} + (\frac{π c}{l})^{2} q = 0, q (0) = 0, q^{'} (0) = A .

Because

k = π c / l

, this oscillator is critically damped, and

q (t) = A t e^{- k t} .

Therefore

y (x, t) = A t e^{- k t} sin \frac{π x}{l} .

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