A block of mass is attached to a spring of stiffness and slides on a frictionless…

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

A block of mass m=1.0 kgm = 1.0\,\mathrm{kg} is attached to a spring of stiffness k=4.0 N m−1k = 4.0\,\mathrm{N\,m^{-1}} and slides on a frictionless horizontal table, so L(q,q˙)=12mq˙2−12kq2L(q,\dot q) = \tfrac12 m\dot q^2 - \tfrac12 k q^2 with qq the displacement from the spring's natural length. At t1=0t_1 = 0 the block is at q=0q = 0 with velocity q˙(0)=v0=2.0 m s−1\dot q(0) = v_0 = 2.0\,\mathrm{m\,s^{-1}}. The final time t2>0t_2 > 0 is fixed but the value of q(t2)q(t_2) is left free. ((a)) Derive the natural boundary condition that stationarity of S=∫0t2L dtS = \int_0^{t_2} L\,\mathrm{d}t imposes at the free end, and use it to find the smallest t2>0t_2 > 0 for which the given initial data are consistent with that condition, together with q(t2)q(t_2) at that time. ((b)) Show that adding f(q)=αq2f(q) = \alpha q^2 (for a constant α\alpha) as L′=L+df/dtL' = L + \mathrm{d}f/\mathrm{d}t leaves the equation of motion unchanged. ((c)) Since LL has no explicit tt, write down the Beltrami (Jacobi) first integral for this LL and check numerically that it takes the same value at t=0t=0 and at the t2t_2 found in part (a).

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