 Solution

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

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

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

The first mode remains the solution from part (b). The applied force excites only the third mode, so

y (x, t) = A t e^{- k t} sin \frac{π x}{l} + q_{3} (t) sin \frac{3 π x}{l},

where, using

π c / l = k

,

q_{3}^{''} + 2 k q_{3}^{'} + 9 k^{2} q_{3} = α cos k t, q_{3} (0) = q_{3}^{'} (0) = 0.

A particular solution is

q_{3, p} (t) = \frac{α}{34 k ^{2}} (4 cos k t + sin k t) .

Adding the decaying complementary solution and imposing the initial conditions gives

q_{3} (t) = \frac{α}{34 k ^{2}} [4 cos k t + sin k t - e^{- k t} (4 cos (22 k t) + \frac{5}{2} sin (22 k t))] .

t \to \infty

, all transients decay but the driven third-mode oscillation remains. The string approaches a periodic steady-state response, rather than coming to rest.

With gravity omitted, its mechanical energy is

E = \frac{μ}{2} \int_{0}^{l} (y_{t}^{2} + c^{2} y_{x}^{2}) d x .

s = α / (34 k^{2})

and letting

t \to \infty

gives

E_{\infty} (t) = \frac{μ l α ^{2}}{4624 k ^{2}} [25 sin^{2} k t + 145 cos^{2} k t + 64 sin k t cos k t] .

The limiting energy is periodic because the external force continually supplies the energy dissipated by drag.

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