 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/ia/paper-4/9c/b/solution

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

a + x

, its tension is

k x

, and the orbit radius is

r = L + (a + x) sin ϕ_{2} .

Vertical balance first gives

k x cos ϕ_{2} = m g, x = \frac{m g}{k cos ϕ _{2}} .

Radial balance is

k x sin ϕ_{2} = m ω^{2} r .

Using the vertical equation,

ω = \frac{g tan ϕ _{2}}{L + ( a + \frac{m g}{k cos ϕ _{2}} ) sin ϕ _{2}}^{1/2} .

Unlike the rigid-rod result, this depends on

m

: a larger mass stretches the spring farther and changes the radius of the orbit.

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