Past exam of the theoretical physics I course of the University of Cambridge / 2013 / Paper 1 / 1

Measure both angles from the downward vertical, as in the PDF. With the support as origin and downward coordinate $z$,Thus $|AC{|}^2=2{\ell }^2[1+\cos (\theta -\varphi )]$. The requested expression implicitly uses a zero-rest-length spring: its energy is $\kappa |AC{|}^2/2$. The exact Lagrangian for equal masses isExpand the cosines and count angles and velocities as first order. The cosine correction multiplying $\dot{\theta }\dot{\varphi }$ is fourth order and is dropped. After removing constants,The positive spring term in $L_2$ means the spring lowers its energy when the straight assembly bends. If its natural length were $s_0$, this term would instead be $\kappa \ell (2\ell -s_0)(\theta -\varphi {)}^2/4$. An unspecified ordinary natural length cannot simply be ignored; the displayed target fixes the zero-length idealization.

Put $G=g/\ell$ and $s=\kappa /m$. Applying the Euler-Lagrange equations to the quadratic Lagrangian givesA nonzero motion satisfying the specified equal-and-opposite initial displacements can have one frequency only if $(1,-1)$ is a normal-mode vector. Its acceleration terms are proportional to $(1,0)$, while the restoring terms are proportional to $(2G-2s,2s-G)$. HenceandThe rods swing oppositely. The hinge mass moves sideways, while the final mass has no first-order displacement: the two horizontal contributions cancel and its vertical change is quadratic in angle. In the corresponding geometric ansatz it moves vertically at twice the small-oscillation frequency to leading nonzero order, but the displayed single-frequency solution is a linearized result. The trivial case $\xi =0$ imposes no parameter condition.

Removing the hinge mass gives the exact LagrangianThe stated equations again concern its small-angle quadratic approximation. In the sum and difference coordinates this becomesConsequentlyFor unequal squared frequencies, $\nu =0$: the rods remain aligned to first order and the end mass swings like a pendulum of length $2\ell$. This degenerate Lagrangian gives the difference coordinate no quadratic kinetic term, so its equation is an algebraic constraint, not a second oscillator. If ${\omega }_0={\omega }_1$, the quadratic action loses all dependence on $\nu$. Its motion is then undetermined at this order; nonlinear terms or a small hinge mass must resolve the degeneracy. For larger spring stiffness the straight constrained branch is not a minimum of the bending potential, despite the absence of a finite-inertia growing mode in this strictly massless quadratic description.

Write $u=t-t^{\prime }$ and $\Omega =\sqrt{{\omega }_0^2-{\gamma }^2/4}$. With the printed Fourier transform convention, a time derivative transforms to $i\omega$. The retarded Green function isIts poles are $\omega =\pm \Omega +i\gamma /2$, both in the upper half-plane. For $u>0$ close the contour above; for $u<0$ close below, where there are no poles. Residues giveIt is continuous at $u=0$ and has derivative jump $G^{\prime }(0^{+})-G^{\prime }(0^{-})=1$, reproducing the Dirac delta function source. Thus the solution initially at rest isand vanishes for $t<0$. Set $D={\alpha }^2-\gamma \alpha +{\omega }_0^2=(\alpha -\gamma /2{)}^2+{\Omega }^2>0$. Evaluating the integral givesThe homogeneous solution constants are $C_1=-A/D$ and $C_2=A(\alpha -\gamma /2)/(D\Omega )$. Substitution at zero gives $\eta (0)=\dot{\eta }(0)=0$. Causality selects this response; an arbitrarily excited prehistory would add a homogeneous solution.