 Standing wave phase and energy

A stationary mode shape does not imply absence of energy. Write $A=2a$ and $y=A\sin (kx)\cos (\omega t)$. For the ideal wave equation on a string, the kinetic energy and elastic potential energy per unit reference length add toHere $P$ is signed power crossing a position on the string, positive to the right; $e$ is an energy density per length, not per volume. Differentiation and $\mu y_{tt}=\tau y_{xx}$ give the local conservation of energy identity ${\partial }_{t}e+{\partial }_{x}P=0$. Substitution into $P$ givesIt can be nonzero away from a node of a standing wave or an antinode, but its time average over an oscillation period is zero. Energy flows locally back and forth even though this pure standing wave transports no net energy across a section over a complete cycle. At a node of a standing wave, $y_t=0$, so kinetic energy density is zero, but $y_x$ need not vanish: elastic potential energy density can be positive there.

For a fixed-end normal mode, integration over $0\le x\le L$ givesThe last equality uses $\tau k^2=\mu {\omega }^2$. At a straight-string snapshot, $U=0$ while $K=E$; at an extreme snapshot, $K=0$ while $U=E$. This is normal-mode energy exchange. A driven real string with damping requires continuing input power to replace lost energy, so a visually stationary pattern does not establish zero average supplied power.