 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/ii/paper-2/40d/c/iv/solution

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

The preceding formulas give

\frac{c}{c _{g}} = \frac{c ^{2}}{c _{s}^{2}} = 1 + \frac{ω _{n}^{2}}{c _{s}^{2} κ ^{2}} .

Therefore

c = m c_{g}, m = 1 + (\frac{( 2 n + 1 ) π}{2 κh})^{2} > 1 .

 New to topics? Read the docs here!