= Standing wave phase and energy
{title2=$\langle P\rangle_t=0$}
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 to
$$
e=\frac{\mu}{2}y_t^2+\frac{\tau}{2}y_x^2,
\qquad P=-\tau y_x y_t.
$$
Here $P$ 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 $P$ gives
$$
P(x,t)=\tau A^2k\omega\sin(kx)\cos(kx)\sin(\omega t)\cos(\omega t).
$$
It 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 $0\leq x\leq L$ gives
$$
K(t)=\frac{\mu A^2\omega^2L}{4}\sin^2(\omega t),\qquad U(t)=\frac{\tau A^2k^2L}{4}\cos^2(\omega t),\qquad E=K+U=\frac{\mu A^2\omega^2L}{4}.
$$
The 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>.
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