Past exam of the theoretical physics I course of the University of Cambridge / 2018 / 2

Use the downward $y$ direction of the PDF diagram. The bob's Cartesian coordinates areThusThe hinge and rod are massless, so only the bob contributes kinetic energy:Because gravity points along positive $y$, its gravitational potential energy is $-mgy=-mg\ell \cos \theta$. The zero-rest-length spring has elastic potential energy $k{s}^2/2$. Put $s=\ell \eta$ and divide $T-V$ by the positive constant $m{\ell }^2$. The equivalent Lagrangian isMultiplying the action by a nonzero constant leaves its stationary paths unchanged. This derivation is a direct use of kinetic energy from Cartesian position vectors; no independent kinetic term should be assigned to the massless support.

For the horizontal generalized coordinate $\eta$,The corresponding Euler-Lagrange equation isFor $\theta$, the terms proportional to $\dot{\eta }\dot{\theta }\sin \theta$ cancel between $d(\partial L/\partial \dot{\theta })/dt$ and $\partial L/\partial \theta$. Hence the other exact equation isKeeping only terms linear in small coordinates and their derivatives givesThe equilibrium is $\theta =\eta =0$. These characteristic frequencies label gravitational and spring restoring scales; they are not two independent normal mode frequencies of this massless-support system.

Subtracting the two linearized equations eliminates their acceleration terms:For $g,k>0$, the assumed nontrivial simple harmonic motion therefore hasThe support displacement and pendulum angle oscillate in phase. Inserting this relation into either Euler-Lagrange equation yieldsso the one positive finite angular frequency isThe negative frequency describes the same real oscillation after a phase change. For a general phase, replace the sine in both coordinates by $A\cos (\omega t)+B\sin (\omega t)$.

There is no missing second finite branch. The quadratic Lagrangian iswhose kinetic matrix is $\left(\begin{array}{cc}1& 1\\ 1& 1\end{array}\right)$ and has rank one. Its characteristic determinant is ${\omega }_0^2{\omega }_1^2-({\omega }_0^2+{\omega }_1^2){\omega }^2$, not a quadratic polynomial in ${\omega }^2$. This singular kinetic matrix in small oscillations results from taking the support mass to be exactly zero. Adding a small support mass restores a second mode that runs to infinite frequency as that mass vanishes.

For a stiff spring,The hinge becomes fixed, recovering the simple pendulum frequency, as expected for the pendulum with a massless spring-mounted support. At finite $k$, horizontal support motion makes the frequency smaller.