The transmitted and incident waves have the same angular frequency and tension, but the transmitted wavenumber is twice as large. The time-average string-power formula therefore gives

\frac{P _{t}}{P _{i}} = \frac{k _{2} ∣ A _{t} ∣ ^{2}}{k _{1} ∣ A _{i} ∣ ^{2}} = 2 (\frac{2}{3})^{2} = \frac{8}{9} .

The extra factor

k_{2} / k_{1}

is essential. Squaring the displacement transmission coefficient alone would give

4/9

, which is not the transmitted power fraction in strings with different linear mass densities.

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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