 Solution

In the small-slope string model, the additional elastic potential energy is $V=(T/2){\int }_0^L{y}_x^{2}dx$. For the fundamental mode, $y_x=A_1{k}_1\cos (k_{1}x)\cos ({\omega }_{1}t)$. Since ${\int }_0^L{\cos }^2(\pi x/L)dx=L/2$,Using $T{k}_1^2=\mu {\omega }_1^2$ from the dispersion relation givesThe 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.