A double pendulum has an upper bob of mass and a lower bob of mass

PHYSICS · Subjective · Fermi University Physics · Mathematical Methods for Physics · 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 double pendulum has an upper bob of mass mm and a lower bob of mass 2m2m, joined by two light rigid rods each of length ll, swinging under gravity gg. For small angles θ1\theta_1 (upper rod from vertical) and θ2\theta_2 (lower rod from vertical), the linearised equations of motion are the standard small-oscillation result ml(3222)(θ¨1θ¨2)+mg(3002)(θ1θ2)=0.ml\begin{pmatrix}3&2\\2&2\end{pmatrix}\begin{pmatrix}\ddot\theta_1\\\ddot\theta_2\end{pmatrix} + mg\begin{pmatrix}3&0\\0&2\end{pmatrix}\begin{pmatrix}\theta_1\\\theta_2\end{pmatrix}=0. Here the mass matrix is not diagonal, but it is symmetric and positive definite, so det⁡(K−ω2M)=0\det(\mathbf{K}-\omega^2\mathbf{M})=0 still gives real, non-negative ω2\omega^2. Take l=0.80 ml = 0.80\,\mathrm{m} and g=9.8 m s−2g=9.8\,\mathrm{m\,s^{-2}}. Find the two normal-mode angular frequencies ω1<ω2\omega_1 < \omega_2.

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