Past exam of the mathematical methods II course of the University of Oxford / 2018 / 10

Let $\delta =x_2-x_1$. The inextensible-string constraint gives the upward displacement $y_2=l-\sqrt{l^2-{\delta }^2}={\delta }^2/(2l)+O({\delta }^4/l^3)$. Thus the quadratic potential energy is $U=k{x}_1^2/2+mg(x_2-x_1{)}^2/(2l)$. The vertical velocity of the second mass is of second order in the small motion, so the quadratic kinetic energy is $m({\dot{x}}_1^2+{\dot{x}}_2^2)/2$.

The Euler-Lagrange equations for this quadratic Lagrangian giveSmall displacements are used both in the quadratic approximation to the gravitational potential energy and in neglecting the higher-order vertical kinetic energy. These are the spring-supported pendulum normal modes at linear order. The corresponding force picture sets the string tension to $mg$ at leading order and its horizontal angle to $\delta /l$.

Put $\kappa =k/m$, $q=g/l$. The positive stiffness-per-mass matrix isSubstitution of a normal mode $x=v\cos \omega t$ requires $Kv={\omega }^{2}v$. Its characteristic equation is ${\lambda }^2-(\kappa +2q)\lambda +\kappa q=0$. ThereforeFor $k,m,g,l>0$, their sum and product are positive, so both squared angular frequencies are positive and ${\omega }_{+}>{\omega }_{-}>0$.

Define $D=\sqrt{{\kappa }^2/4+q^2}$. Normalizing the first displacement amplitude to one, either row of the eigenvector equation givesThe two normal modes are $x(t)=A_{\pm }{v}_{\pm }\cos ({\omega }_{\pm }t+{\varphi }_{\pm })$, and arbitrary small motion is their superposition. The lower-frequency mode has $r_{-}>0$, so the masses move in phase; the upper-frequency mode has $r_{+}<0$, so they move in opposite phase. The identity $r_{+}{r}_{-}=-1$ verifies orthogonality of these eigenvectors for equal masses.

The product-to-sum identity gives $\sin \Omega t\sin \omega t=[\cos (\Omega -\omega )t-\cos (\Omega +\omega )t]/2$. Thus the first mass must have equal and opposite cosine-mode amplitudes and no sine-mode amplitudes. Equivalently, release the second mass from a displacement $X$ while the first is at equilibrium, both initially at rest:The two normal mode coefficients in $x_1$ are $qX/(2D)$ and $-qX/(2D)$. Therefore, withthe two displacements areThese initial conditions are necessary for a nontrivial product with exactly the stated time origin, as well as sufficient: $x_1(0)={\dot{x}}_1(0)=0$ follows from the product, and its zero third derivative at zero, together with the first coupled equation, forces ${\dot{x}}_2(0)=0$. Since both normal frequencies are positive, $\Omega >\omega >0$. This is the product form used to describe beating, although a well-separated slow envelope requires the two normal frequencies to be close. No such extra near-degeneracy is assumed in the exact formula.