= Solution
\b[<time average of harmonic power>.] Expand the two <real parts>:
$$
\langle\operatorname{Re}F\operatorname{Re}u\rangle
=\frac14\langle Fu+Fu^*+F^*u+F^*u^*\rangle
=\boxed{\frac12\operatorname{Re}(Fu^*)}.
$$
The terms proportional to $e^{\pm2i\omega t}$ average to zero over complete periods, while $Fu^*$ is time-independent. For real peak amplitudes this is $F_0u_0\cos\phi/2$.
\b[String dynamics.] A short element of length $dx$ experiences transverse <force> $T[\psi_x(x+dx)-\psi_x(x)]$. <Newton's second law> therefore gives $\rho\,dx\,\psi_{tt}=T\,dx\,\psi_{xx}$, or
$$
\boxed{\psi_{xx}=\frac1{v^2}\psi_{tt},\qquad v=\sqrt{T/\rho}}.
$$
The approximation is a flexible, lossless string with small slope and constant <tension>; $\rho$ is its <linear mass density>.
\b[<energy> transport.] Across a cut toward increasing $x$, the transverse <force> is $F_y=-T\psi_x$ and the transverse velocity is $u=\psi_t$. For a rightgoing wave these have <complex amplitudes> $ikTA$ and $i\omega A$. Hence
$$
\boxed{Z=\frac Tv=\rho v=\sqrt{T\rho},\qquad
\langle P\rangle=\frac12Z\omega^2|A_0|^2}.
$$
This real <mechanical impedance> is <force> divided by transverse velocity for a <travelling wave>, with units $\mathrm{kg\,s^{-1}}$.
\b[Matching and <power> ratios.] At an ordinary massless junction without an external transverse <force>, displacement is continuous and transverse <tension> is balanced: $\psi_1=\psi_2$, $T_1\psi_{1x}=T_2\psi_{2x}$. For the same <tension>, the slope is therefore continuous even if density changes. With real positive impedances,
$$
r_\psi=\frac{Z_1-Z_2}{Z_1+Z_2},\quad t_\psi=\frac{2Z_1}{Z_1+Z_2},
\qquad
\boxed{P_r=\frac{(Z_1-Z_2)^2}{(Z_1+Z_2)^2}P_i,\quad
P_t=\frac{4Z_1Z_2}{(Z_1+Z_2)^2}P_i}.
$$
Their sum equals $P_i$; the transmitted <power> includes the impedance ratio and is not simply $t_\psi^2P_i$.
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