Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2018 / Paper 2 / 7

Choose generalized coordinates $q_i$ measured from a static equilibrium point, and assume a conservative system with no velocity-dependent forces. Keeping terms of degree two in displacements and velocities gives the small oscillation LagrangianHere $A$ is the positive-definite symmetric mass matrix evaluated at equilibrium, and $B$ is the symmetric Hessian matrix of the potential energy there. The linear potential term is zero by equilibrium, and an additive constant has been dropped. Coordinate-dependent corrections to the kinetic matrix contribute only at higher order.

The Euler-Lagrange equations are $A\ddot{\mathbf{q}}+B\mathbf{q}=0$. For a normal mode $\mathbf{q}=\mathbf{v}{e}^{i\omega t}$, they give the generalized eigenvalue problem $(B-{\omega }^{2}A)\mathbf{v}=0$. A nonzero mode shape exists exactly whenThis is a degree-$n$ polynomial in ${\omega }^2$. At a stable equilibrium its roots are nonnegative; zero roots correspond to zero modes.

Choose positive $\varphi$ so the bob is displaced to the right relative to the cart. Its Cartesian coordinates areDifferentiation gives ${\dot{X}}^2+{\dot{Y}}^2={\dot{x}}^2+2L\cos \varphi \dot{x}\dot{\varphi }+L^2{\dot{\varphi }}^2$. Thus the exact kinetic energy and potential energy areRetain quadratic terms, using $\cos \varphi =1+O({\varphi }^2)$ in the kinetic cross term and $1-\cos \varphi ={\varphi }^2/2+O({\varphi }^4)$ in the potential. The Lagrangian isIts Euler-Lagrange equations giveThe kinetic cross term is essential: acceleration of the support couples the pendulum and cart. These equations define the normal modes of a spring-mounted pendulum cart.

With the stated parameters, the mass and restoring matrices areThe generalized eigenvalue problem givesTherefore the two normal angular frequencies areThe second row of the modal equation is $-{\omega }^{2}x+(1-{\omega }^2)\varphi =0$. It gives the mode shapesEach normal mode has arbitrary small amplitude $a$ and phase $\delta$. The bob's horizontal displacement is $X=x+\varphi$ to first order. Thus in the lower mode, bob and cart move in phase, with bob amplitude $1+\sqrt{2}$ times the cart amplitude. In the higher mode they move oppositely, because $X=(1-\sqrt{2})x$; the bob amplitude is $\sqrt{2}-1$ times the cart amplitude. Its vertical displacement is quadratic in the angle and is absent at this order.