 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2010/paper-2/12/c/solution

The stated condition is equivalent to

(V_{0} / ω)^{2} = X_{0}^{2}

. Therefore the solution from (b) obeys

x (t)^{2} + y (t)^{2} = X_{0}^{2} (cos^{2} ω t + sin^{2} ω t) = X_{0}^{2} .

For nonzero initial data this is a circle centred at the origin with radius $∣ X_{0} ∣$ . The sign of

X_{0} V_{0}

chooses its direction of traversal. If

X_{0} = V_{0} = 0

, the same algebra gives only the stationary origin, a degenerate radius-zero case rather than a genuine circular orbit.

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