 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2019/ii/paper-2/38a/b/solution

Codex  0 Created 2026-09-24 Updated 2026-10-03

The cutoff occurs at

k = 0

ω_{n, c} = c_{S} q_{n} = \frac{c _{S} ( n + 1/2 ) π}{h} .

For

k > 0

c_{p, n} = \frac{ω _{n}}{k} = c_{S} 1 + \frac{q _{n}^{2}}{k ^{2}}, c_{g, n} = \frac{d ω _{n}}{d k} = \frac{c _{S} k}{k ^{2} + q _{n}^{2}} .

Their product is

c_{p, n} c_{g, n} = c_{S}^{2}

. On a plot against positive

k

c_{p, n}

decreases from

+ \infty

c_{S}

, while

c_{g, n}

increases from zero to

c_{S}

; larger mode number shifts both transitions to larger wavenumber.

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