 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2024/paper-2/20/c/ii/solution

The standing wave oscillates about the constant level

1/2

, with time period

2 π /∣ v ∣

. Suitable snapshots are

t = 0 : t = \frac{π}{3∣ v ∣} : t = \frac{π}{2∣ v ∣} : t = \frac{π}{∣ v ∣} : t = \frac{2 π}{∣ v ∣} : ψ = \frac{1}{2} (1 + cos x), ψ = \frac{1}{2} + \frac{1}{4} cos x, ψ = \frac{1}{2}, ψ = \frac{1}{2} (1 - cos x), ψ = \frac{1}{2} (1 + cos x) .

The nonconstant amplitude shrinks to zero at a quarter-period, reverses sign at half a period, and returns after a full period. The nodes of a standing wave of the oscillatory part,

x = π /2 + kπ

, remain at height

1/2

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