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

The four spring extensions are $x_1$, $x_2-x_1$, $x_3-x_2$ and $-x_3$. Hooke's law therefore gives potential energy $k[x_1^2+(x_2-x_1{)}^2+(x_3-x_2{)}^2+x_3^2]/2$, while the kinetic energy is $m{\sum }_i{\dot{x}}_i^2/2$. Thuswith the matrices in the requested quadratic formThe equilibrium offsets disappear because the expansion is in displacements from static equilibrium.

Write $d=2k-m{\omega }^2$. The characteristic determinant of the tridiagonal matrix isThe three values of $d$ are $\sqrt{2}k$, $0$ and $-\sqrt{2}k$. Therefore the normal mode frequencies and convenient Euclidean-normalized generalized eigenvectors areEach pair satisfies $V\mathbf{u}={\omega }^{2}T\mathbf{u}$. The first mode moves all masses in phase, the middle mode leaves the central mass fixed, and the last has the central mass moving opposite to the outer masses. All squared frequencies are positive, consistent with the positive spring potential and fixed endpoints.

Using the three real eigenvectors and positive frequencies in part (ii), the general motion isThere are six independent real constants, $A_1,A_2,A_3,B_1,B_2,B_3$, corresponding to the three initial displacements and three initial velocities. With the displayed orthonormal eigenvectors, $A_j={\mathbf{u}}_j^T\mathbf{x}(0)$ and $B_j={\mathbf{u}}_j^T\dot{\mathbf{x}}(0)/{\omega }_j$. This spans all solutions because the normal-mode eigenvectors form a basis.

Introduce the fixed endpoint values $x_0=x_{N+1}=0$. The Lagrangian isThus $T=m{I}_N$, $V_{ij}=k(2{\delta }_{ij}-{\delta }_{i,j+1}-{\delta }_{i,j-1})$, andThis is a symmetric tridiagonal matrix, reflecting that only nearest-neighbour masses are coupled.

Expand the tridiagonal determinant along its first row. The diagonal entry contributes $d{D}_{N-1}$. In the off-diagonal cofactor, expansion along its first column supplies the other factor $-k$, and the cofactor sign makes the net contribution $-k^2{D}_{N-2}$. ThereforeThe initial values are $D_1=d$ and $D_2=d^2-k^2$. Writing the same formula at $N=2$ forces $\boxed{D_0=1}$, the empty determinant convention. This is the tridiagonal determinant recurrence.

Substitute $D_N={\beta }^N$ into the recurrence. Since $k>0$, its roots are nonzero and satisfyFor distinct roots, the general recurrence solution is $D_N=c_1{\beta }_1^N+c_2{\beta }_2^N$. The initial values require $c_1+c_2=1$ and $c_1{\beta }_1+c_2{\beta }_2=d={\beta }_1+{\beta }_2$, givingAt a repeated root, $d=\pm 2k$ and ${\beta }_1={\beta }_2=\beta =\pm k$, this quotient must be replaced by its limit:It also follows by solving the repeated-root recurrence as $(c_1+c_{2}N){\beta }^N$. This exceptional case is necessary to avoid introducing a false root in the next part; it is included in the fixed-end spring-chain determinant and its repeated-root limit.

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.