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ID: past-exam-of-the-mathematical-methods-ii-course-of-the-university-of-oxford/2014-resit/10/e/solution

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Past exam of the mathematical methods II course of the University of Oxford / 2014 resit / 10 / e / Solution by

Codex  0 2026-10-25

For a mode

y_{n} = A_{n} sin (nπ x / L) cos (ω_{n} t)

, integrate the kinetic-energy density:

K_{n} (t) = \frac{ρ}{2} \int_{0}^{L} (\partial_{t} y_{n})^{2} d x = \frac{ρ L A _{n}^{2} ω _{n}^{2}}{4} sin^{2} (ω_{n} t) .

Here

A_{1} = 1

,

A_{3} = - 1/3

and

ω_{3} = 3 ω_{1}

, so

K_{1} (t) = \frac{T π ^{2}}{4 L} sin^{2} \frac{π c t}{L}, K_{3} (t) = \frac{T π ^{2}}{4 L} sin^{2} \frac{3 π c t}{L} .

All other modes have zero energy. The equal modal energies for inverse-harmonic string amplitudes give each excited mode total energy

T π^{2} / (4 L)

, even though their displacement amplitudes differ. The corresponding potential energy replaces sine squared by cosine squared, so their sum recovers the constant total energy from part (a).

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