Past exam of the mathematics course of the University of Cambridge 2012 iii Paper 74 2 b Solution Created 2026-10-03 Updated 2026-10-14
A positive-eigenvalue normal mode satisfies . Solving the constant-coefficient equation gives the sine, cosine, hyperbolic sine and hyperbolic cosine family. To avoid confusing modal coefficients with the filament bending modulus, write those coefficients as .
The free conditions at zero require and . With , the remaining conditions areIts determinant is . Therefore the free-free bending spectrum isA convenient corresponding eigenfunction iswhere gives unit norm. The stable characteristic equation for free-free bending modes is . Intersections or numerical bracketing giveFor the entire positive sequence, let . Expanding near its zero and using givesThere are also two zero-energy filament modes, not captured by substituting into the finite-coefficient trigonometric expression. Solving with free endpoints gives the affine kernel. An orthonormal basis of that kernel isThe positive normal modes are orthogonal to both. Including these two rigid-motion modes is essential for completeness; the regular self-adjoint operator on a finite interval has a complete discrete eigenbasis.
