= Solution
For either <normal mode>, let $k_n=n\pi/L$ and write its contribution as $B_n\sin(k_nx)\sin(\omega_nt)$. The product-to-sum identity gives
$$
\boxed{B_n\sin(k_nx)\sin(\omega_nt)=\frac{B_n}{2}\left[\cos(k_nx-\omega_nt)-\cos(k_nx+\omega_nt)\right].}
$$
This <standing wave> is a <superposition> of two <travelling waves> of equal <amplitude> $|B_n|/2$, moving oppositely with <phase velocity> $\omega_n/k_n=c$. Their cancellation at $x=0,L$ enforces the fixed-end <boundary conditions> at every time.
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