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Past exam of the mathematics course of the University of Cambridge / 2023 / ib / Paper 4 / 16C / b / v / Solution

The dynamic condition reduces to

F^{'} (t) + g H (t) = - \frac{p _{0}}{ρ} cos (ω t) .

Differentiate

H^{'} = k F

and define the natural frequency

Ω^{2} = g k = \frac{π g 5}{L} .

Then the surface amplitude obeys the forced harmonic oscillator

H^{''} + Ω^{2} H = - \frac{k p _{0}}{ρ} cos (ω t) .

For

ω \neq = Ω

, its general solution and the corresponding potential amplitude are

H = C cos (Ω t) + D sin (Ω t) + \frac{k p _{0}}{ρ ( ω ^{2} - Ω ^{2} )} cos (ω t),

F = \frac{H ^{'}}{k} .

If the fluid is initially quiescent,

H (0) = F (0) = 0

, so

H (t) = \frac{k p _{0}}{ρ ( ω ^{2} - Ω ^{2} )} (cos (ω t) - cos (Ω t)),

F (t) = \frac{p _{0}}{ρ ( ω ^{2} - Ω ^{2} )} (Ω sin (Ω t) - ω sin (ω t)) .

Solved by gpt-5.6-sol high.

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