ADVANCED DYNAMICS — PHY 4241/5227

HOME AND CLASS WORK – SET 7

Solution for assignment 10c: Double pendulum solution and plot.

Let us take minors with respect to the first row of the determinant. For the ω_{+} frequency the ratio of the two minors is

\frac{△_{1+}}{△_{2+}} = \frac{−1 − \sqrt{2}}{2 + \sqrt{2}} = \frac{(−1 − \sqrt{2})}{(2 + \sqrt{2})} \frac{(1 − \sqrt{2})}{(1 − \sqrt{2})} = \frac{1}{−\sqrt{2}}

and for ω_{−} it is

\frac{△_{1+}}{△_{2+}} = \frac{−1 + \sqrt{2}}{2 − \sqrt{2}} = \frac{(−1 + \sqrt{2})}{(2 − \sqrt{2})} \frac{(1 + \sqrt{2})}{(1 + \sqrt{2})} = \frac{1}{\sqrt{2}} .

Therefore, the solutions (real part) can be written

ϕ_{+}(t) = A_{+} cos(ω_{+}t) + B_{+} sin(ω_{+}t) ,
ϕ_{−}(t) = A_{−} cos(ω_{+}t) + B_{−} sin(ω_{+}t) ,
ϕ(t) = ϕ_{+}(t) + ϕ_{−}(t) ,
ψ(t) = −\sqrt{2} ϕ_{+}(t) + \sqrt{2} ϕ_{−}(t) .

The four constants are determined by the four initial value, e.g., ϕ_{0}, ϕ̇_{0}, ψ_{0}, ψ̇_{0} at time t = 0:

ϕ_{0} = A_{+} + A_{−} ,  ψ_{0} = \sqrt{2} (−A_{+} + A_{−}) ,
ϕ̇_{0} = ω_{+}B_{+} + ω_{−}B_{−} , ψ̇_{0} = \sqrt{2} (−ω_{+}B_{+} + ω_{−}B_{−}) ,

which gives

A_{+} = \frac{ϕ_{0}}{2} − \frac{ψ_{0}}{2\sqrt{2}} ,  A_{−} = \frac{ϕ_{0}}{2} + \frac{ψ_{0}}{2\sqrt{2}} ,
B_{+} = \frac{ϕ̇_{0}}{2ω_{+}} − \frac{ψ̇_{0}}{2\sqrt{2} ω_{+}} , B_{−} = \frac{ϕ̇_{0}}{2ω_{−}} + \frac{ψ̇_{0}}{2\sqrt{2} ω_{−}} .

For the initial conditions

ϕ_{0} = 0 , ϕ̇_{0} = 1 , ψ_{0} = 0 , ψ̇_{0} = −1

at time t = 0 the figure gives a plot up to t = 50 \sqrt{l/g} (next page).