A system displaced from an unstable equilibrium has a displacement obeying y - k^2y = s(t)

PHYSICS · Subjective · Fermi University Physics · Mathematical Methods for Physics · Ordinary differential equations · Ordinary differential equations

Mathematical Methods for Physics, Ch Higher-order linear equations, the Wronskian and variation of parameters

Problem

A system displaced from an unstable equilibrium has a displacement y(t)y(t) obeying y¨−k2y=s(t),t≥0 s,\ddot y - k^2y = s(t), \qquad t \ge 0\,\mathrm{s}, where k=2.0 s−1k = 2.0\,\mathrm{s^{-1}} is a positive constant set by the system's geometry, yy is in radians, and s(t)s(t) is an applied generalised force per unit generalised inertia (units rad s−2\mathrm{rad\,s^{-2}}), switched on at t=0 st=0\,\mathrm{s}. The system starts from rest at equilibrium: y(0)=0y(0)=0, y˙(0)=0\dot y(0)=0. The hyperbolic sine and cosine are sinh⁡u=12(eu−e−u)\sinh u = \tfrac12(e^{u}-e^{-u}) and cosh⁡u=12(eu+e−u)\cosh u = \tfrac12(e^{u}+e^{-u}). (a) Using y1(t)=ekty_1(t)=e^{kt} and y2(t)=e−kty_2(t)=e^{-kt} as the two independent solutions of the undriven equation, use variation of parameters to show that the response from rest to any drive s(t)s(t) can be written as the single integral y(t)=1k∫0tsinh⁡[k(t−t′)] s(t′) dt′.y(t) = \frac{1}{k}\int_0^{t}\sinh[k(t-t')]\,s(t')\,\mathrm{d}t'. (b) Evaluate this integral for a constant drive s(t)=s0=0.40 rad s−2s(t) = s_0 = 0.40\,\mathrm{rad\,s^{-2}} switched on at t=0t=0, and give y(1.0 s)y(1.0\,\mathrm{s}) to two significant figures. (c) Check the result of part (b) against the general solution built directly from the complementary function plus a constant particular integral, applying the same initial conditions.

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