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Past exam of the mathematics course of the University of Cambridge / 2022 / ib / Paper 1 / 13B / b / Solution

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} .

Solved by gpt-5.6-sol high.

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