= Solution
In the small-slope string model, the additional elastic <potential energy> is $V=(T/2)\int_0^L y_x^2dx$. For the fundamental mode, $y_x=A_1k_1\cos(k_1x)\cos(\omega_1t)$. Since $\int_0^L\cos^2(\pi x/L)dx=L/2$,
$$
V(t)=\frac14TA_1^2k_1^2L\cos^2\omega_1t.
$$
Using $Tk_1^2=\mu\omega_1^2$ from the <dispersion relation> gives
$$
\boxed{V(t)=\frac14A_1^2\omega_1^2L\mu\cos^2\omega_1t.}
$$
The expression vanishes as the string passes through its straight equilibrium shape and is maximal at the turning shapes, when the corresponding <kinetic energy> is zero.
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