A uniform heavy rope of length hangs from a hook and swings in small transverse…

PHYSICS · Multi-part · Fermi University Physics · Mathematical Methods for Physics · Variational and asymptotic methods · Function spaces and operator theory

Conditions of Validity, Ch Eigenfunction expansions, convergence and the Rayleigh quotient

Problem

A uniform heavy rope of length L=2.00 mL=2.00\ \mathrm{m} hangs from a hook and swings in small transverse oscillations. With xx measured up from the free lower end, its mode shapes satisfy −(xY′)′=λY,0<x≤L,λ=ω2g,-\big(xY'\big)'=\lambda Y,\qquad 0<x\le L,\qquad \lambda=\frac{\omega^2}{g}, with Y(L)=0Y(L)=0 at the hook and YY bounded at the free end, where p=xp=x vanishes. Take g=9.81 m s−2g=9.81\ \mathrm{m\,s^{-2}}. Take as given that R[Y]≥λ1R[Y]\ge\lambda_1 for every bounded trial function with Y(L)=0Y(L)=0 and xY′→0xY'\to0 as x→0x\to0, where R[Y]=∫0LxY′2 dx∫0LY2 dx.R[Y]=\frac{\int_0^LxY'^2\,\mathrm dx}{\int_0^LY^2\,\mathrm dx}. ((a)) Evaluate RR for the straight-line trial Y=1−x/LY=1-x/L. ((b)) For the one-parameter family Ys=1−(x/L)sY_s=1-(x/L)^s with s>0s>0, show that R[Ys]=(s+1)(2s+1)4sLR[Y_s]=\dfrac{(s+1)(2s+1)}{4sL}. Find the ss that gives the best bound, and the bound itself. ((c)) Turn the best bound into an upper bound on the lowest angular frequency ω1\omega_1. The exact eigenvalue, from Bessel functions, is λ1=1.4458/L\lambda_1=1.4458/L. By what percentage does your bound on ω1\omega_1 exceed the exact value, and why can it not fall below it?

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