Past exam of the year 2 of the earth sciences (geology) course of the University of Oxford / 2025 / Paper 3 / 8 / g / Solution

At $t=0$, the displacement expansion is $f(x)={\sum }_n{A}_n\sin (n\pi x/L)$. Fourier orthogonality givesDifferentiating the modal solution at zero gives $g(x)={\sum }_n{\omega }_n{B}_n\sin (n\pi x/L)$, soThe factor $1/{\omega }_n$ is required because the derivative of $\sin ({\omega }_{n}t)$ at zero is ${\omega }_n$. The dimensions agree: $A_n$ and $B_n$ are displacements, while the sine coefficients of $g$ are velocities. These projections also establish uniqueness within the finite-energy solution class.