= Solution
The instantaneous <energy> flux towards positive $x$ is the rate at which the string on the left does work on the string on the right. To first order its transverse <force> is $-T y_x$, so
$$
P(x,t)=-T y_x y_t.
$$
For a right-travelling sinusoidal wave $y=A\cos(kx-\omega t)$, $y_x=-Ak\sin(kx-\omega t)$ and $y_t=A\omega\sin(kx-\omega t)$. Thus
$$
P(x,t)=T\omega kA^2\sin^2(kx-\omega t),\qquad\boxed{\langle P\rangle=\frac12T\omega kA^2}.
$$
The printed constant formula is therefore the \b[time-average <power>], not the instantaneous <power>. The latter oscillates between zero and twice its average. A left-travelling wave carries negative signed flux through the same positively oriented section; its transported <power> magnitude has the same average formula with positive $k$.
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