Finding the constants of the general approximate solution to a double pendulum

[giovanna]

Homework Statement

Given the (approximate) general solution to the double pendulum, find the constant $A_1$, $A_2$, $\delta_1$ and $\delta_2$ by stating some initial conditions for the double pendulum.

Relevant Equations

$\phi(t) = A_1 \begin{pmatrix} 1 \\ \sqrt{2} \end{pmatrix} \cos(\omega_1 t - \delta_1) + A_2 \begin{pmatrix} 1 \\ -\sqrt{2} \end{pmatrix} \cos(\omega_2 t - \delta_2)
$

EDIT: My Latex is not showing... Sorry. I attached a file with my "solution".

I though this would be quite easy, but I can't seem to solve this system of equations. Should I solve for each mode, or both of them together? I tried to solve them together, here's how far I get:

$\text{General solution:} \\

\phi(t)=A_1\binom{1}{\sqrt{2}}\cos\left(\omega_1t-\delta_1\right)+A_2\binom{1}{-\sqrt{2}}\cos\left(\omega_2t-\delta_2\right)

\text{Expanding the components, we get} \\

\phi_1(t)=A_1\cos\left(\omega_1t-\delta_1\right)+A_2\cos\left(\omega_2t-\delta_2\right) \\
\phi_2(t)=\sqrt{2}A_1\cos\left(\omega_1t-\delta_1\right)-\sqrt{2}A_2\cos\left(\omega_2t-\delta_2\right)

\text{Neglecting the amplitudes, this can be written as} \\

\phi_1(t)=A_1\cos\left(\omega_1t-\delta_1\right)+A_2\cos\left(\omega_2t-\delta_2\right) \\
\phi_2(t)=A_1\cos\left(\omega_1t-\delta_1\right)-A_2\cos\left(\omega_2t-\delta_2\right)

\text{Differentiating, we obtain} \\

\dot{\phi}_1(t)=-\omega_1A_1\sin\left(\omega_1t-\delta_1\right)-\omega_2A_2\sin\left(\omega_2t-\delta_2\right) \\
\dot{\phi}_2(t)=-\omega_1A_1\sin\left(\omega_1t-\delta_1\right)+\omega_2A_2\sin\left(\omega_2t-\delta_2\right)

\text{To find the particular solution, we need to determine } A_1, A_2, \delta_1, \text{ and } \delta_2. \text{ Choosing the initial conditions} \\

\phi_1(0)=\phi_2(0)=\frac{\pi}{6} \\
\dot{\phi}_1(0)=\dot{\phi}_2(0)=0

\text{We obtain a system of equations with four equations and four unknowns:} \\

A_1\cos(\delta_1)+A_2\cos(\delta_2)=\frac{\pi}{6} \\
A_1\cos(\delta_1)-A_2\cos(\delta_2)=\frac{\pi}{6} \\
-\omega_1A_1\sin(\delta_1)-\omega_2A_2\sin(\delta_2)=0 \\
-\omega_1A_1\sin(\delta_1)+\omega_2A_2\sin(\delta_2)=0

\text{Since } \cos(-x)=\cos(x) \text{ and } \sin(-x)=-\sin(x), \text{ we can rewrite this as} \\

A_1\cos(\delta_1)+A_2\cos(\delta_2)=\frac{\pi}{6}\ (1) \\
A_1\cos(\delta_1)-A_2\cos(\delta_2)=\frac{\pi}{6}\ (2) \\
\omega_1A_1\sin(\delta_1)+\omega_2A_2\sin(\delta_2)=0\ (3) \\
\omega_1A_1\sin(\delta_1)-\omega_2A_2\sin(\delta_2)=0\ (4)$

 

[renormalize]

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