Past exam of the mathematical methods II course of the University of Oxford / 2015 / 11 / b / 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 $-T{y}_x$, soFor a right-travelling sinusoidal wave $y=A\cos (kx-\omega t)$, $y_x=-Ak\sin (kx-\omega t)$ and $y_t=A\omega \sin (kx-\omega t)$. ThusThe printed constant formula is therefore the 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 $k$.