An undamped oscillator has mass and angular frequency (period )

PHYSICS · Multi-part · Fermi University Physics · Mathematical Methods for Physics · Mechanics & Dynamical Systems · Lagrangian mechanics

Action and Symmetry, Ch The action, the first variation and the boundary term

Problem

An undamped oscillator has mass m=1.0 kgm = 1.0\,\mathrm{kg} and angular frequency ω=π rad s−1\omega = \pi\,\mathrm{rad\,s^{-1}} (period T=2.0 sT = 2.0\,\mathrm{s}), released from x=0x=0 at t=0t=0. Someone argues: "Hamilton's principle says the physical path makes the action least. The stationary path from x=0x=0 at t=0t=0 to wherever it is at t2=1.5 st_2 = 1.5\,\mathrm{s} (three-quarters of a period) is the one the oscillator actually follows, so by Hamilton's principle it must be a minimum of the action, and indeed a least-action path, over that whole interval." ((a)) Using the trial variation δx=ϵsin⁡(πt/t2)\delta x = \epsilon\sin(\pi t/t_2) (which vanishes at both ends), find the second-order change ΔS\Delta S in the action for this oscillator over 0≤t≤t2=1.5 s0 \le t \le t_2 = 1.5\,\mathrm{s}, given that for this trial variation ΔS=14mϵ2t2[(π/t2)2−ω2]\Delta S = \tfrac14 m\epsilon^2 t_2\left[(\pi/t_2)^2 - \omega^2\right]. ((b)) State whether the claim is correct, and explain your reasoning using the conjugate point of this oscillator.

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