Past exam of the mathematics course of the University of Cambridge / 2012 / iii / Paper 74 / 2 / b

A positive-eigenvalue normal mode satisfies $K{W}_n=A{k}_n^4{W}_n$. 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 $c_1,c_2,c_3,c_4$.

The free conditions at zero require $c_3=c_1$ and $c_4=c_2$. With $q=kL$, the remaining conditions areIts determinant is $2(1-\cos q\cosh q)$. Therefore the free-free bending spectrum isA convenient corresponding eigenfunction iswhere $N_n$ gives unit $L^2$ norm. The stable characteristic equation for free-free bending modes is $\cos q=\mathrm{sech}q$. Intersections or numerical bracketing giveFor the entire positive sequence, let $q_n^0=(n+\frac{1}{2})\pi$. Expanding $\cos q_n$ near its zero and using $\mathrm{sech}{q}_n\sim 2e^{-q_n}$ givesThere are also two zero-energy filament modes, not captured by substituting $k=0$ into the finite-coefficient trigonometric expression. Solving $W^{(4)}=0$ 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.