# Normal mode of a coupled pendulum: why the constant $\psi_1$, $\psi_2$

I need to solve a problem that tells me to find out the motion of both the pendulums that appear in the first 45 seconds of this video

I think this kind of motion is described by a system of differential equation of the form:

$$\ddot{x} + \omega^2x = \epsilon y$$ $$\ddot{y} + \omega^2 y = \epsilon x$$

where all constants like the mass of the pendulum and so on are missing and $x$ describes the motion of the first pendulum and $y$ the motion of the second one.

To solve the problem one needs to assume $\epsilon$ very small and $\epsilon \lt \omega^2$.

I've tried to solve this problem analytically, but it was a little too complicated so i tried the physical approach by following the example given here under the section coupled oscillator.

I think that i've understood almost everything except the fact that when we evaluate the normal modes we add the constant $\psi_1$ and $\psi_2$ and get the solutions

$$\vec{\nu}_1 = c_1 \begin{pmatrix} 1 \\ 1 \end{pmatrix}\cos(\omega_1 t + \psi_1)$$ $$\vec{\nu}_2 = c_2 \begin{pmatrix} 1 \\ -1 \end{pmatrix}\cos(\omega_2 t + \psi_2)$$

My question is: why do we have those two constant $\psi_1, \psi_2$? I understand the fact that we are interested only in the real valued part of the solution $Ae^{i\omega t}$, but I don't understand where do these constant come from, is there a physical or better mathematical explanation for this?

• It is the solution of two second-order differential equations, then why you shouldn't have 4 integrating constants? – Antonio Ragagnin May 8 '14 at 10:53
• @AntonioRagagnin You are right, I know there must be 4 values that must than be found with the initial condition, but why do we add these two constant $\psi_1, \psi_2$? I mean, I've done the calculus in the same way as in the example and they don't appear, they are just there. The only values that appear are $c_1, c_2$ for this reason I've thought there might have been a physical explanation to these two, for me, misterious consants. – Bman72 May 8 '14 at 10:56

The general solution of your second-order differential eqaution is not $Ae^{i\omega t}$. In fact, a general solution is:

$$f_t=Ae^{i\omega t}+Be^{-i\omega t}.$$

If you start with a solution of this kind for your system, you'll end up with 4 integrating constants.

## Physical point of view:

From a physical point of view, you expect four counstant to describe the initial conditions:

• $\psi_1$ and $\psi_2$ are the angle the pendulum-s start from.

This is why you need $\psi_1$ and $\psi_2$ as initial conditions: you expect different a evolution if you change the initial angle.