Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2021 / Paper 2 / 7 / b / iv / Solution

For distinct nonzero recurrence roots, the determinant formula givesThe ratio one is excluded: at that repeated root, part (iii) gives $(N+1)(\pm k{)}^N\ne 0$. Thus the naive extra root corresponding to equal recurrence roots is spurious.

To count the physical solutions without ambiguity from exchanging ${\beta }_1$ and ${\beta }_2$, write ${\beta }_{1,2}=k{e}^{\pm iq}$ and $d=2k\cos q$ with $0<q<\pi$. ThenIts $N$ distinct zeros occur at $q=j\pi /(N+1)$, $j=1,\dots ,N$. They give exactly $N$ distinct positive normal frequencies, or equivalently $N$ distinct roots in ${\omega }^2$:If signed $\omega$ is counted, the characteristic equation has $2N$ roots $\pm {\omega }_j$. The sine quotient is the Chebyshev polynomial of the second kind representation of the fixed-end spring-chain determinant and its repeated-root limit.