= Solution
The instantaneous rightward <energy flux> along a string is the transverse <force> times the transverse <velocity>,
$$
P=-Ty_xy_t.
$$
For a rightgoing <time-harmonic wave> of peak displacement <wave amplitude> $|A|$, averaging over a <wave period> gives
$$
\langle P\rangle=\frac12Tk\omega|A|^2=\frac12Z\omega^2|A|^2.
$$
A leftgoing <time-harmonic wave> has the opposite signed <energy flux>. The positive ratios of reflected and transmitted <power> to incident <power> are therefore
$$
\mathcal R=|r|^2=\frac{(Z_L-Z_R)^2}{(Z_L+Z_R)^2},\qquad
\boxed{\mathcal T=\frac{Z_R}{Z_L}|t|^2=\frac{4Z_LZ_R}{(Z_L+Z_R)^2}.}
$$
All explicit <angular frequency> factors cancel. In the ideal <wave equation on a string>, $T$ and the two <linear mass densities> are constant, so the <mechanical impedances> have no <frequency> or <wavelength> dependence. Consequently $\mathcal T$ is independent of <wavelength> even though the two string <wavelengths> generally differ.
Finally,
$$
\boxed{\mathcal R+\mathcal T=
\frac{(Z_L-Z_R)^2+4Z_LZ_R}{(Z_L+Z_R)^2}=1.}
$$
This is <conservation of energy> at a lossless, massless junction. The transmitted displacement <wave amplitude> can exceed the incident one when $Z_R<Z_L$; the factor $Z_R/Z_L$ is essential when comparing <power>.
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