 Solution

ID: past-exam-of-the-physics-1-paper-of-the-university-of-oxford/2013/9/e/solution

Put

θ = θ_{e} + δ

at the stable equilibrium. Since

M sin θ_{e} = m

(M + m) R \ddot{δ} + M g cos θ_{e} δ = 0.

Thus

ω_{0} = \frac{M g cos θ _{e}}{( M + m ) R} = \frac{g M ^{2} - m ^{2}}{( M + m ) R}, f_{0} = \frac{ω _{0}}{2 π} .

m \to 0

, the result approaches

g / R

m ↑ M

, the restoring curvature tends to zero and this small-amplitude frequency tends to zero; the endpoint itself is not a stable quadratic minimum.

 New to topics? Read the docs here!