 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2019/ia/paper-4/10a/b/iii/solution

Codex  0 Created 2026-09-24 Updated 2026-09-29

Taking the gradient of the quadratic potential energy found in part (ii) resolves the linearized force into identical

x

and

y

components:

F_{x} = - \frac{3 q ^{2}}{8 π ϵ _{0} ρ ^{3}} x, F_{y} = - \frac{3 q ^{2}}{8 π ϵ _{0} ρ ^{3}} y .

Each component therefore performs simple harmonic motion with

ω^{2} = \frac{3 q ^{2}}{8 π ϵ _{0} m ρ ^{3}} = \frac{9 3}{8 π} \frac{q ^{2}}{m ϵ _{0} a ^{3}} .

ω = \frac{3 ^{5/4}}{8 π} \frac{∣ q ∣}{m ϵ _{0} a ^{3}}, A = \frac{3 ^{5/4}}{8 π} .

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