 Solution

ID: past-exam-of-the-mathematical-methods-ii-course-of-the-university-of-oxford/2014-resit/8/b/solution

r_{1} = (2 l sin θ_{1}, - 2 l cos θ_{1})

and

r_{2} = r_{1} + (l sin θ_{2}, - l cos θ_{2})

. Differentiating and summing their kinetic energies gives the exact expression

K = \frac{m l ^{2}}{2} [8 \dot{θ}_{1}^{2} + \dot{θ}_{2}^{2} + 4 \dot{θ}_{1} \dot{θ}_{2} cos (θ_{1} - θ_{2})] .

In the linear small-oscillation approximation, set the cosine to one in this quadratic velocity expression:

K ≃ \frac{m l ^{2}}{2} (8 \dot{θ}_{1}^{2} + 4 \dot{θ}_{1} \dot{θ}_{2} + \dot{θ}_{2}^{2}) .

The cross term arises because the lower mass participates in motion generated by both rods.

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