Why does my general solution for a double pendulum have three unknowns?

[zenterix]

Homework Statement

This question is about some calculations I saw in a video lecture about coupled oscillators.

Relevant Equations

In particular, there is the derivation of the solutions to the problem of the double pendulum.

It seems to me the derivation ends up with three unknown parameters to be determined by initial conditions. Shouldn't it be four?

For the record, here is the lecture I am speaking of.

Consider a double pendulum.

I'm going to skip the setup of the problem and jump to the system of differential equations

$$m\ddot{x}_1=-\frac{3g}{l}x_1+\frac{g}{l}x_2\tag{1}$$

$$m\ddot{x}_2=\frac{g}{l}x_1-\frac{g}{l}x_2\tag{2}$$

where $\omega_0^2=\frac{g}{l}$.

The lecturer then says that any normal mode solution will have to have the form

$$x_1=A_1\cos{(\omega t+\phi_1)}\tag{3}$$

$$x_2=A_2\cos{(\omega t+\phi_2)}\tag{4}$$

We plug these into the system and we reach

$$A_1(\omega^2-3\omega_0^2)+A_2\omega_0^2=0\tag{5}$$

$$A_1\omega_0^2+A_2(\omega^2-\omega_0^2)=0\tag{6}$$

The only way this has a non-zero solution for $A_1$ and $A_2$ is if the system matrix has a zero determinant.

Given $\omega_0$, this happens for two positive values of $\omega$. Let's call them $\omega_A$ and $\omega_B$ (I won't post the actual values because it isn't important for my question I think).

For each of these these angular frequencies, the system represented by (5) and (6) is singular: there is only one equation. We can only determine the ratio $A_1/A_2=r$.

The normal mode solutions are then

$$x_1=r_1A_2\cos{(\omega_A t+\phi_1)}\tag{s1}$$

$$x_2=A_2\cos{(\omega_A t+\phi_2)}\tag{s1}$$

and

$$x_1=r_2A_2\cos{(\omega_B t+\phi_1)}\tag{s2}$$

$$x_2=A_2\cos{(\omega_B t+\phi_2)}\tag{s2}$$

The general solution seems to be the sum of these solutions.

My question is about the number of unknown parameters to be determined by initial conditions.

As far as I can tell we have $A_2, \phi_1,$ and $\phi_2$.

Shouldn't there be four?

 

[zenterix]

Okay, so I think the answer to my question is that actually $A_2$ is distinct for each of the normal modes. I noticed this just after posting.

The normal mode solutions are then

$$x_1=r_1A_{2,1}\cos{(\omega_A t+\phi_1)}\tag{s1}$$

$$x_2=A_{2,1}\cos{(\omega_A t+\phi_2)}\tag{s1}$$

and

$$x_1=r_2A_{2,2}\cos{(\omega_B t+\phi_1)}\tag{s2}$$

$$x_2=A_{2,2}\cos{(\omega_B t+\phi_2)}\tag{s2}$$

So the unknown parameters are actually $A_{2,1}, A_{2,2}, \phi_1,$ and $\phi_2$.

 

[TSny]

The normal mode solutions are then$$x_1=r_1A_{2,1}\cos{(\omega_A t+\phi_1)}\tag{s1}$$ $$x_2=A_{2,1}\cos{(\omega_A t+\phi_2)}\tag{s1}$$

For a normal mode, the phase constant $\phi$ is the same for $x_1$ and $x_2$. See your video link at time 37:35. So, for the first normal mode (mode $A$), let $\phi_A$ be the common phase constant of $x_1$ and $x_2$. For the second mode (mode $B$), let $\phi_B$ be the phase constant. Thus, for mode $A$

$$x_1=r_1A_2\cos{(\omega_A t+\phi_A)}\tag{s1}$$ $$x_2=A_2\cos{(\omega_A t+\phi_A)}\tag{s1}.$$ For mode $B$, $$x_1=r_1B_2\cos{(\omega_B t+\phi_B)}\tag{s1}$$ $$x_2=B_2\cos{(\omega_B t+\phi_B)}\tag{s1}.$$
What are the four parameters that are determined by the initial conditions?