Modal energy of a string ($E_n=\mu L{\omega }_n^2(A_n^2+B_n^2)/4$)

For a uniform fixed-end mechanical string of length $L$, linear mass density $\mu$ and wave speed $c$, let its $n$th normal mode be $y_n=[A_n\cos {\omega }_{n}t+B_n\sin {\omega }_{n}t]\sin (n\pi x/L)$, where ${\omega }_n=cn\pi /L$. The conserved energy is the sum of kinetic energy and elastic potential energy,The sine and cosine squared integrals are $L/2$, so substitution proves the title formula. Orthogonality makes energies of distinct normal modes add. For zero initial displacement and initial modal velocity $V_n$, $A_n=0$ and $B_n=V_n/{\omega }_n$, giving $E_n=\mu L{V}_n^2/4$.

For a uniform fixed-end mechanical string of length $l$ and tension $T$, a single sine normal mode of amplitude $A_n$ has conserved energy $E_n=n^2{\pi }^{2}T{A}_n^2/(4l)$. Hence amplitudes proportional to $1/n$ give equal energies, even though their displacement magnitudes differ. In particular $A_2=A_1/2$ makes the first two modal energies equal. Orthogonality removes cross terms in the integrated kinetic energy and potential energy, so the modal energies add.