= Solution
For a rightgoing harmonic <travelling wave> in the lossless string, its <force> and transverse-velocity <complex amplitudes> satisfy $F=Zu$ with real $Z=\sqrt{T\rho}$. Thus the supplied mean-power expression gives
$$
\boxed{\langle P\rangle=\frac12Z|u|^2=\frac{|F|^2}{2Z}}
=\boxed{\frac12\sqrt{T\rho}\,\omega^2|A|^2},
$$
where $A$ is the displacement <complex amplitude> and $u=-i\omega A$ for convention $e^{i(kx-\omega t)}$. The same result follows directly from instantaneous string <power> $P=-T\Psi_x\Psi_t$ averaged over a cycle. The amplitudes here are peak amplitudes; with <root mean square> <force> or velocity the factor $1/2$ is absent. If reflections return to the driver, its net mean input is still $\operatorname{Re}(Fu^*)/2$, but its input <mechanical impedance> need not equal the travelling-wave value $Z$.
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