A ring of equal masses is joined by six identical springs of stiffness

PHYSICS · Multi-part · Fermi University Physics · Mechanics & Dynamical Systems · Oscillations, Waves & Acoustics · 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 of stiffness κ\kappa, giving ring-mode frequencies ωp=2ω0∣sin⁡(πp/N)∣\omega_p = 2\omega_0|\sin(\pi p/N)| with ω0=κ/m=50.0 rad s−1\omega_0=\sqrt{\kappa/m}=50.0\,\mathrm{rad\,s^{-1}}, p=0,1,…,5p=0,1,\dots,5. Every mode has the same small damping rate β=1.5 rad s−1\beta = 1.5\,\mathrm{rad\,s^{-1}}. The ring's springs are then modulated so that κ(t)=κ[1+hcos⁡(Ωt)]\kappa(t) = \kappa\left[1+h\cos(\Omega t)\right]. Recall from the parametric-oscillator analysis that a mode of frequency ω\omega and damping β\beta becomes unstable (grows without bound) when the pump is tuned to Ω=2ω\Omega = 2\omega and the modulation depth exceeds the threshold hc=2/Q=4β/ωh_{\rm c} = 2/Q = 4\beta/\omega. ((a)) Find the frequency ω2\omega_2 of ring mode p=2p=2. ((b)) Find the pump frequency Ω\Omega that puts mode p=2p=2 at parametric resonance. ((c)) Find the minimum modulation depth hch_{\rm c} needed to destabilise mode p=2p=2.

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