 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/ib/paper-3/16c/b/v/solution

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

Equating the two linearized expressions for

p (a, t) - p_{\infty}

from parts (iii) and (iv) gives

ρ a_{0}^{2} lo g \frac{b _{0}}{a _{0}} \overset{ϵ}{¨} + 2 p_{\infty} ϵ = 0.

ω = (\frac{2 p _{\infty}}{ρ a _{0}^{2} lo g ( b _{0} / a _{0} )})^{1/2} .

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