ADVANCED DYNAMICS — PHY 4241/5227 HOME AND CLASS WORK – SET 7

Solution for assignment 10b: Double Pendulum (continuation of 10).

For small oscillation we have derived the Euler-Lagrange equations, which read in matrix notation

( \begin{matrix}2 & 1 \\ 1 & 1 \end{matrix} ) ( ϕ̈ ; ψ̈ ) + ( \begin{matrix}2g/l & 0 \\ 0 & g/l \end{matrix} ) ( \begin{matrix}ϕ \\ ψ \end{matrix} ) = 0 .

This is solved by the exponential ansatz (physical is the real part of the solution):

Φ(t) = ( \begin{matrix}ϕ \\ ψ \end{matrix} ) = e^{i ω t} ( \begin{matrix}ϕ_{0} \\ ψ_{0} \end{matrix} ) ⇒ ( ϕ̈ ; ψ̈ ) = −ω^{2} e^{i ω t} ( \begin{matrix}ϕ_{0} \\ ψ_{0} \end{matrix} )

( \begin{matrix}−2ω^{2} + 2g/l & −ω^{2} \\ −ω^{2} & −ω^{2} + g/l \end{matrix} ) ( \begin{matrix}ϕ \\ ψ \end{matrix} ) = 0 .

0 = det | \begin{matrix}−2ω^{2} + 2g/l & −ω^{2} \\ −ω^{2} & −ω^{2} + g/l \end{matrix} | = ω^{4} − \frac{4g}{l} ω^{2} + \frac{2g^{2}}{l^{2}}

with eigenmodes

ω_{±} = \sqrt{\frac{g}{l}} \sqrt{2 ± \sqrt{2}} .