= Normal-mode energy exchange
{title2=$\mathrm{KE}=E\sin^2(\omega t),\quad\mathrm{PE}=E\cos^2(\omega t)$}
For the undamped first <normal mode> of a fixed string, start at maximum <displacement> with $u(x,t)=A\sin(\pi x/L)\cos(\omega_1t)$. In the small-slope <wave equation on a string>, its <kinetic energy> and excess <potential energy> are
$$
\mathrm{KE}(t)=\frac{\mu}{2}\int_0^L u_t^2\,dx,\qquad \mathrm{PE}(t)=\frac{\tau}{2}\int_0^L u_x^2\,dx.
$$
The <potential energy> comes from the extra length under <tension>: $\sqrt{1+u_x^2}-1\simeq u_x^2/2$. It depends on slope, so a <node of a standing wave> need not have zero local <potential energy> density. These are physical <energies>, with the mass and <tension> factors restored to the normalized <wave energy>. https://farside.ph.utexas.edu/teaching/315/Waveshtml/node43.html[Richard Fitzpatrick, Oscillations and Waves, Energy Conservation, equations 6.24-6.29]{ref}
Integrating $\sin^2(\pi x/L)$ and $\cos^2(\pi x/L)$ over the string gives $L/2$ in both cases. Using $\mu\omega_1^2=\tau\pi^2/L^2$ yields
$$
E=\frac{\tau A^2\pi^2}{4L},\qquad \mathrm{KE}(t)=E\sin^2(\omega_1t),\qquad \mathrm{PE}(t)=E\cos^2(\omega_1t),\qquad \mathrm{KE}+\mathrm{PE}=E.
$$
For the model $\mu=1\,\mathrm{kg\,m^{-1}}$, $\tau=1\,\mathrm N$, $L=1\,\mathrm m$, and $A=0.010\,\mathrm m$, the <angular frequency> is $\omega_1=\pi\,\mathrm{rad\,s^{-1}}$, the <frequency> is $0.5\,\mathrm{Hz}$, and $E\simeq2.4674\times10^{-4}\,\mathrm J$. The maximum slope $A\pi/L\simeq0.0314$ is small. At $t=0$ all the <energy> is <potential energy>; at $t=T_1/4=0.5\,\mathrm s$ the string is straight and all the <energy> is <kinetic energy>. At $t=T_1/2=1\,\mathrm s$ the <displacement> is reversed and all the <energy> is again <potential energy>. The energy curves repeat twice during one <displacement> cycle, while <conservation of energy> keeps their sum constant.
\Image[/normal-mode-energy-exchange.png]
{title=First-mode displacement in metres and kinetic and potential energies divided by the conserved total energy}
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