Past exam of the mathematical methods II course of the University of Oxford / 2016 / 11 / e / Solution

For small transverse displacements, the string's kinetic energy and extra elastic potential energy areThe orthogonality of trigonometric functions removes cross terms between the two normal modes. For a mode $B_n\sin (k_{n}x)\sin ({\omega }_{n}t)$, use $T{k}_n^2=\rho {\omega }_n^2$ and ${\int }_0^L{\sin }^2(k_{n}x)dx={\int }_0^L{\cos }^2(k_{n}x)dx=L/2$ to obtainConsequently the modal energy distribution isThe second mode contributes more total energy at every time, despite its smaller displacement amplitude. The total energy is $13\rho L{V}_0^2/64$, 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.