Two-coupled oscillator: Doubt in finding normal modes and natural frequency

Question

I want to find the natural frequency of a two coupled oscillator system like this-

My book does it this way but I don't really get it.

The equations of motion for the pendula are-

$$I\frac{d^2\theta_1}{dt^2}=−M_\text{eff}\ gL\sin \theta_1− \kappa l^2(\sin \theta_1−\sin \theta_2)$$

$$I\frac{d^2\theta_2}{dt^2}=−M_\text{eff}\ gL\sin \theta_2+ \kappa l^2(\sin \theta_1−\sin\theta_2)$$

To find the natural frequencies of the system, we take the sum and subtraction of the equations and we obtain (Using small angle approximation):

$$I\left(\frac{d^2\theta_1}{dt^2}+\frac{d^2\theta_2}{dt^2}\right)=−M_\text{eff}\ gL(\theta_1+\theta_2)$$

and

$$I\left(\frac{d^2\theta_1}{dt^2}-\frac{d^2\theta_2}{dt^2}\right)=−M_\text{eff}\ gL(\theta_1−\theta_2)−2\kappa l^2(\theta_1−\theta_2)$$

The two equations above are uncoupled and represent the two normal modes of the coupled system. The $\theta_1+\theta_2$ mode or ‘+’ mode represent the in-phase motion of the pendula where both the pendula are moving with same phase (same direction). The $\theta_1−\theta_2$ mode or ‘−’ mode represent the out-of-phase motion of the pendula where the pendula are moving with opposite phase (opposite direction).

I have marked the portions I don't understand in bold above.

Doubts:

1. What is meant by uncoupled?
2. Why does the two equations represent the normal modes?
3. Why does $\theta_1+\theta_2$ represent in-phase and $\theta_1-\theta_2$ represent out of phase motion?

Answer 1

1. Suppose $\alpha = \theta_{1} +\theta_{2}$ and $\beta = \theta_{1}-\theta_{2}$ then the first equation only depends on single variable $\alpha$ and the second on $\beta$, thus the equations are uncoupled.
2. Stating that the equations are uncoupled, or that the matrix of the system is diagonal, or that the equations define the eigenvectors and eigenvalues, or that the equations define the normal modes and natural frequencies, it all means the same, as far as I know.
3. $\theta_{1}=(\alpha + \beta)/2$, $\theta_{2}=(\alpha-\beta)/2$, if the system oscilates in its first normal mode only then $\beta=0$, meaning $\theta_{1}=\theta_{2}$, the two masses have an in-phase motion. With second mode only $\alpha=0$, $\theta_{1}=-\theta_{2}$, the two masses have an out-of-phase motion.

Answer 2

Consider $\theta_{\rm sum} = \theta_1 + \theta_2$ as one variable, and $\theta_{\rm diff} = \theta_1 - \theta_2$ as a second variable.

The two equations become

$$ \begin{aligned} I \frac{{\rm d}^2}{{\rm d}t^2} \theta_{\rm sum} & = − \left(M_\text{eff}\ g L\right) \theta_{\rm sum} \\ I \frac{{\rm d}^2}{{\rm d}t^2} \theta_{\rm diff} & = − \left( M_\text{eff}\ gL−2\kappa l^2\right) \theta_{\rm diff} \end{aligned} $$

Now it is clear they are two decoupled equations. Each differential equation is only in terms of one unknown.

• Now if they two pendulums where out of phase by the same amount $\theta_{\rm sum} = \theta + (-\theta) =0 $ and $\theta_{\rm diff} = \theta - (-\theta) = 2 \theta$. So the second equation describes the out of phase vibration.
• Conversely, if they are in phase $\theta_{\rm sum} = 2 \theta$ and $\theta_{\rm diff} = 0$, which means the first equation describes the in-phase motion.