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ID: past-exam-of-the-physics-1-paper-of-the-university-of-oxford/2014-resit/8/a/ii/solution

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Past exam of the physics 1 paper of the University of Oxford / 2014 resit / 8 / a / ii / Solution by

Codex  0 2026-10-25

For the general Lagrangian just obtained, the Euler-Lagrange equation gives

I \ddot{θ} + A sin θ = 0.

(1)

Using the small-angle approximation

sin θ ≃ θ

, this is simple harmonic motion with

ω_{0}^{2} = A / I

. Thus

T = 2 π \frac{l ( m + M _{r} /3 )}{g ( m + M _{r} /2 )} ⟶ T = 2 π \frac{l}{g} for a negligible-mass rod .

(2)

The endpoint mass cancels in the light-rod limit. A concrete comparison shows why the finite rod cannot silently be included: the same specified endpoint

m

with

M_{r} = m

gives

T = 2 π 8 l / (9 g)

, distinct from the negligible-rod result. This is a physical pendulum when the rod contributes, and a simple pendulum in the light-rod limit. Both expressions require small oscillation amplitude; the full sine equation remains nonlinear.

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