Past exam of the mathematical methods II course of the University of Oxford / 2015 / 10 / a

Let $x_1,x_2$ be displacements from equilibrium, with positive directions chosen alike. Hooke's law and Newton's second law giveFor a normal mode $x_j=a_j\cos \Omega t$, the determinant condition isSolving the quadratic for ${\Omega }^2$ givesThe angular frequencies are the positive square roots. Positive masses and spring constants make the stiffness quadratic form $k_1{x}_1^2+k_2(x_2-x_1{)}^2+k_3{x}_2^2$ positive definite, ensuring both squared frequencies are positive. Distinct spring constants do not require equal masses or any symmetry approximation.

Put $k_1=k_3=k$ and $m_1=m_2=m$. The mass-normalized stiffness matrix isIts symmetric eigenvector $(1,1)$ has eigenvalue $k/m$, while its antisymmetric eigenvector $(1,-1)$ has eigenvalue $(k+2k_2)/m$. ThusThe symmetric frequency does not depend on the middle spring constant because that spring has no extension in an in-phase normal mode. The antisymmetric motion stretches the middle spring and has the higher frequency for $k_2>0$.

With the same positive displacement convention for both masses, the first eigenvector equation is $(k+k_2-m{\Omega }^2)a_1-k_2{a}_2=0$. At the symmetric frequency it reduces to $k_2(a_1-a_2)=0$, while at the antisymmetric frequency it becomes $-k_2(a_1+a_2)=0$. HenceIn both normal modes the displacement magnitudes are equal. For the first, the masses move in phase and their separation stays unchanged; for the second they move in opposite phase, alternately approaching and separating. A negative signed amplitude ratio records a phase difference of $\pi$, not a negative displacement magnitude. The left panel of the later figure plots these signed normal mode shapes with the fixed wall values included.