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Past exam of the mathematics course of the University of Cambridge / 2025 / ib / Paper 2 / 14B / Solution

Transverse force balance on a short element gives

ρ y_{tt} = τ y_{xx}, c = τ / ρ .

With fixed ends,

y (x, t) = \sum_{n \geq 1} [A_{n} cos (ω_{n} t) + B_{n} sin (ω_{n} t)] sin \frac{nπ x}{L}, ω_{n} = \frac{nπ c}{L} .

Here

A_{n} = 0

. If

b_{n}

are the sine coefficients of initial velocity, then

b_{n} = \frac{4 v}{nπ ε} sin \frac{nπ l}{L} sin \frac{nπ ε}{2 L}, B_{n} = \frac{b _{n}}{ω _{n}} .

The total deposited energy is

E = ρ v^{2} /2

, while

\frac{E _{n}}{E} = \frac{8 L}{n ^{2} π ^{2} ε} sin^{2} \frac{nπ l}{L} sin^{2} \frac{nπ ε}{2 L} .

The seventh mode is eliminated by striking at a node

l = j L /7

,

j = 1, \dots, 6

. For a narrow hammer, each fixed-mode fraction is

O (ε)

; beyond the inverse-width cutoff the envelope falls as

n^{- 2}

, modulated by the two sine factors.

Solved by gpt-5.6-sol high.

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