Two identical pendulums

PHYSICS · Multi-part · Fermi University Physics · Mathematical Methods for Physics · Linear algebra and vector spaces · Numerical and computational mathematics

Error, Chance and the Machine, Ch Rounding, backward error and the sensitivity of a computation

Problem

Two identical pendulums, weakly coupled, have squared normal-mode frequencies ω12\omega_1^2 and ω22\omega_2^2 that are the two roots of the characteristic quadratic p(λ)=λ2+a1λ+a0,p(\lambda) = \lambda^2 + a_1\lambda + a_0, with a1=−(ω12+ω22)a_1 = -(\omega_1^2+\omega_2^2) and a0=ω12ω22a_0 = \omega_1^2\omega_2^2. For the pair under study ω12=4.00 s−2\omega_1^2 = 4.00\ \mathrm{s^{-2}} and ω22=4.02 s−2\omega_2^2 = 4.02\ \mathrm{s^{-2}}, so a1=−8.02 s−2a_1 = -8.02\ \mathrm{s^{-2}} and a0=16.08 s−4a_0 = 16.08\ \mathrm{s^{-4}}. ((a)) Starting from the identity p(r)=0p(r)=0 and treating a simple root rr as a function of one coefficient aja_j at a time (the others held fixed), derive the relative condition number κj=∣ajrj−1/p′(r)∣\kappa_j = |a_j r^{j-1}/p'(r)| of rr with respect to aja_j. ((b)) Show that for this monic quadratic, with roots r1r_1 and r2r_2, p′(r1)=r1−r2p'(r_1) = r_1 - r_2, and explain why κj\kappa_j for the root ω12\omega_1^2 must diverge as ω22→ω12\omega_2^2 \to \omega_1^2. ((c)) Evaluate κ0\kappa_0 and κ1\kappa_1, the condition numbers of the root ω12\omega_1^2 with respect to a0a_0 and a1a_1, for the numbers above. ((d)) Let a1a_1 alone change by a relative amount uu, an absolute change of 8.02u s−28.02u\ \mathrm{s^{-2}}. Use κ1\kappa_1 to find how far ω12\omega_1^2 can move, to first order, and compare this with the largest shift Weyl's theorem allows for an eigenvalue of a symmetric matrix (entries in s−2\mathrm{s^{-2}}) perturbed by a symmetric E\mathsf{E} with ∥E∥2=8.02u s−2\|\mathsf{E}\|_2=8.02u\ \mathrm{s^{-2}}.

Browse the problem library