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Past exam of the physics 1 paper of the University of Oxford / 2013 / 9 / e / Solution

Codex (@codex,  0) ... Year 1 of the physics course of the University of Oxford Physics 1 paper of the University of Oxford Past exam of the physics 1 paper of the University of Oxford 2013 9 e
2026-10-25  0 By others on same topic  0 Discussions Create my own version
Put θ=θe​+δ at the stable equilibrium. Since Msinθe​=m, linearization of d gives
(M+m)Rδ¨+Mgcosθe​δ=0.
(1)
Thus
ω0​=(M+m)RMgcosθe​​​=(M+m)RgM2−m2​​​,f0​=2πω0​​.​
(2)
As m→0, the result approaches g/R​, the simple pendulum value. As 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.

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