 Solution

ID: past-exam-of-the-mathematical-methods-ii-course-of-the-university-of-oxford/2015/11/d/ii/solution

The reflected wave remains on the incident string, so its wavenumber magnitude is still

k_{1}

. Its outward power magnitude is

\frac{∣ P _{r} ∣}{P _{i}} = \frac{∣ A _{r} ∣ ^{2}}{∣ A _{i} ∣ ^{2}} = \frac{1}{9} .

Thus

∣ P_{r} ∣ + P_{t} = (1/9 + 8/9) P_{i} = P_{i}

, verifying energy conservation. With signed positive-

x

P_{r} = - P_{i} /9

and the same statement is

P_{i} + P_{r} = P_{t}

. The minus sign of the reflected displacement is a phase reversal, whereas the minus sign of its flux records travel in the opposite direction; these are different uses of a sign.

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