 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2009/paper-2/12/ii/solution

Ω = k - λ^{2} /4

. The oscillation envelope is proportional to

e^{- λ t /2}

. A decrease by a quarter leaves three quarters of the original amplitude, so

e^{- λ t /2} = 3/4

. Therefore

Ω = k - λ^{2} /4, t = \frac{2}{λ} lo g \frac{4}{3} .

This concerns the envelope; it does not ask when the instantaneous displacement first equals three quarters of a chosen value.

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