Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge 2009 Paper 1 2 Solution Created 2026-10-18 Updated 2026-10-20
Use separation of variables . The fixed-end Dirichlet boundary conditions select , , and each mode satisfies . In the physical resistance range , all normal modes are underdamped, with . The modal factor normalized by , isConsequently the linearly damped string has solutionThe coefficients are those of the initial Fourier sine series. Differentiating each normalized modal factor at zero confirms that the string starts from rest. Suitable initial data give convergence in the string's energy space; the piecewise linear profile here need not give a twice differentiable solution along every characteristic from its corner.
For the specified tent profile, integrating its two linear pieces by parts givesFor example the first piece gives , and the second gives the opposite cosine term and the same sine term. HenceThe zero even coefficients also follow from reflection symmetry about the midpoint. This is the Fourier sine series of a symmetric tent. If were read without the physical assumption , the same modal solution extends by replacing imaginary-frequency sine and cosine by their hyperbolic counterparts, and taking when .
