Past exam of the mathematics course of the University of Cambridge 2009 ib Paper 2 15B Solution Created 2026-09-24 Updated 2026-10-14
For small transverse displacement, the wave equation on a string iswhere is the tension and is mass per unit length. Using separation of variables, put , the ratios equal a constant. The fixed-end conditions exclude nontrivial solutions for nonnegative separation constants; for they give and , . The associated angular frequencies are . Thus the general superposition of separated normal modes isThe string is initially straight and at rest before the impulse, so . Assume the hammer interval lies inside . The imposed velocity has Fourier sine series coefficientTherefore , giving the amplitude of every excited normal mode.
The string energy is . At the impulse it is . Orthogonality of the sine modes gives initially; each mode subsequently exchanges kinetic and elastic energy while retaining that total. Thus the modal energy of a localized velocity impulse on a string has fractionsFor and , a mode is absent exactly when or . Hence the unexcited modes are precisely
Assume the struck interval lies within the string, so . The initially stationary straight string has . Write its initial velocity as a Fourier sine series . ThenThus . Immediately after the strike all the energy is kinetic energy:The orthogonality integral assigns energy to mode . Since the ideal modes evolve independently, their total kinetic-plus-elastic energies remain constant. Therefore the modal energy of a localized velocity impulse on a string givesThe centre factor selects modes having significant motion at the impact location; the width factor accounts for cancellation across the finite hammer footprint. Parseval's identity ensures . The velocity discontinuities are understood in the mean-square sense and do not alter the energy integrals.