= Solution
For small transverse <displacements>, the string's <kinetic energy> and extra elastic <potential energy> are
$$
K=\frac\rho2\int_0^Ly_t^2\,dx,\qquad U=\frac T2\int_0^Ly_x^2\,dx.
$$
The <orthogonality of trigonometric functions> removes cross terms between the two <normal modes>. For a mode $B_n\sin(k_nx)\sin(\omega_nt)$, use $Tk_n^2=\rho\omega_n^2$ and $\int_0^L\sin^2(k_nx)\,dx=\int_0^L\cos^2(k_nx)\,dx=L/2$ to obtain
$$
K_n=\frac{\rho L B_n^2\omega_n^2}{4}\cos^2(\omega_nt),\quad U_n=\frac{\rho L B_n^2\omega_n^2}{4}\sin^2(\omega_nt),\quad E_n=\frac{\rho L B_n^2\omega_n^2}{4}.
$$
Consequently the <modal energy distribution> is
$$
\boxed{E_1=\frac{\rho LV_0^2}{16},\qquad E_2=\frac{9\rho LV_0^2}{64},\qquad \frac{E_2}{E_1}=\frac94.}
$$
\b[The second mode contributes more total energy at every time], despite its smaller displacement amplitude. The total <energy> is $13\rho LV_0^2/64$, with fractions $4/13$ and $9/13$ in the first and second modes. Each mode exchanges <kinetic energy> and <potential energy>, while its sum is conserved; the comparison is of these sums, not of instantaneous kinetic energies alone.
\Image[/past-exam-of-the-mathematical-methods-ii-course-of-the-university-of-oxford/2016-string-modal-energy.png]
{title=String modal coordinates and separate kinetic and potential energies, showing the larger energy of the second mode}
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