A ring of equal masses is joined by six identical springs

PHYSICS · Multi-part · Fermi University Physics · Oscillations, Waves & Acoustics · Simple, damped and driven oscillations · Coupled oscillators, normal modes and Fourier analysis

Modes, Packets and Media, Ch Normal modes of many coupled oscillators: eigenvalues, symmetry and response

Problem

A ring of N=6N=6 equal masses mm is joined by six identical springs, giving the linear ring-mode frequencies ωp=2ω0∣sin⁡(πp/N)∣\omega_p = 2\omega_0|\sin(\pi p/N)| with ω0=κ/m=40.0 rad s−1\omega_0 = \sqrt{\kappa/m} = 40.0\,\mathrm{rad\,s^{-1}}. Each spring is now slightly anharmonic, contributing a force per unit mass on its stretch xx of the Duffing form ω02x+εx3\omega_0^2x + \varepsilon x^3, with ε=8000 m−2 s−2\varepsilon = 8000\,\mathrm{m^{-2}\,s^{-2}}. Mode p=1p=1 is excited on its own, to a peak amplitude A=0.10 mA = 0.10\,\mathrm{m} in its normal coordinate. Treating this mode, in isolation, as a single Duffing oscillator with its own linear frequency and the same coefficient ε\varepsilon (an idealisation that neglects the coupling of this mode to the others, valid before any Fermi-Pasta-Ulam-type sharing of energy has had time to develop), use the amplitude-dependent frequency result ω(A)≈ω1(1+3εA28ω12)\omega(A) \approx \omega_1\left(1+\dfrac{3\varepsilon A^2}{8\omega_1^2}\right) to estimate its shifted frequency. ((a)) Find the linear ring-mode frequency ω1\omega_1 (p=1p=1, N=6N=6). ((b)) Find the amplitude-shifted frequency ω(A)\omega(A) for A=0.10 mA=0.10\,\mathrm{m}.

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