A quiet snowy tree cricket chirps at its own natural angular rate

PHYSICS · Multi-part · Fermi University Physics · Mechanics & Dynamical Systems · Thermal, Statistical & Complex Systems · Nonlinear dynamics and chaos

Many Parts, One Behaviour, Ch Synchronisation, cascades and the resilience of networks

Problem

A quiet snowy tree cricket chirps at its own natural angular rate ω\omega, but it also hears a large chorus of dominant crickets chirping steadily at angular rate ωref\omega_{\rm ref}; the chorus is loud enough that the quiet cricket's chirping does not affect it, so the coupling acts one way only. The quiet cricket's phase obeys θ˙=ω+Ksin⁡(θref−θ),\dot\theta = \omega + K\sin(\theta_{\rm ref} - \theta), where θref=ωreft\theta_{\rm ref} = \omega_{\rm ref}t and KK (in rad s−1\mathrm{rad\,s^{-1}}) measures how strongly the quiet cricket's chirp rate responds to the chorus. The natural-rate difference is Δω=ω−ωref=0.5 rad s−1\Delta\omega = \omega - \omega_{\rm ref} = 0.5\,\mathrm{rad\,s^{-1}}. ((a)) Writing ϕ=θ−θref\phi = \theta - \theta_{\rm ref} for the phase of the quiet cricket relative to the chorus (positive when the cricket is ahead), derive Adler's equation for ϕ\phi and state the condition on KK under which the quiet cricket can lock to the chorus. ((b)) For K=0.3 rad s−1K = 0.3\,\mathrm{rad\,s^{-1}}, show that locking fails and find the mean angular rate Ω\Omega at which ϕ\phi slips, so that one slip by 2π2\pi takes a time 2π/Ω2\pi/\Omega. ((c)) For K=0.6 rad s−1K = 0.6\,\mathrm{rad\,s^{-1}}, find the locked phase difference ϕ∗\phi^* and confirm, by linear stability, that it is the stable fixed point.

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