For a simple case of frictionless coupled oscillators shown in the figure below:

Coupled pendula

(Image: two pendula of equal length and equal masses suspended from a level ceiling and connected by a spring) (and concerning small oscillations only)

The two normal modes are $$\xi_1(t) = \frac{y_1(t) + y_2(t)}{2}$$ which is the average of the two displacements i.e. the co-ordinate of the centre of mass of the system, if you forget about the distance between the equilibrium points, and $$\xi_2(t) = \frac{y_1(t) - y_2(t)}{2}$$ which concerns the relative displacement of the pendula, and describes a motion where the centre of mass of the system does not move and the pendula swing in concert towards or away from each other.

I don't understand how you can get from this notion of relative displacement to the idea that the centre of mass of the system is not moving. I can only imagine this relative displacement varying in time and can't see how this quantity directly links to the centre of mass remaining stationary. Any pointers would be appreciated.


1 Answer 1


If $\xi_1(t) = 0$, then the center of mass is not moving. If $\xi_2(t)=0$, then the relative displacement of the two pendula is constant, so they're moving in phase. Generically the motion will be some combination of the two, but the "amount" of each can be specified independently.

If you think about two pendula next to each other which are completely uncoupled, you would describe them with the variables $y_1$ and $y_2$. Generically the motion will be some combination of the two, but neither one influences the other. This should be fairly obvious. The point is that when you add a spring between them, then $y_1$ and $y_2$ cannot be specified independently of one another, but $\xi_1$ and $\xi_2$ can.

  • $\begingroup$ I understand that our vectors $\xi_1$ and $\xi_2$ are a way to "decouple" the equations of motion by a certain transformation of co-ordinates, but what if we end up with $\xi_i(t) = at, a \in \mathbb{R}$ or some arbitrary function involving $t$ on either/both of our normal modes? Would this physical situation ever come about, for one, and if it does then how would we interpret this changing quantity, now that it is no longer constant? Many thanks. $\endgroup$
    – Poo2uhaha
    Commented Jan 25, 2021 at 18:07
  • $\begingroup$ @Poo2uhaha How is that possible? $\xi_1$ obeys the equation of motion $\ddot \xi_1 = -\omega^2 \xi_1$, and $\xi_2$ obeys the same equation with a different frequency. They're not randomly chosen out of the air, right? $\endgroup$
    – J. Murray
    Commented Jan 25, 2021 at 18:15
  • $\begingroup$ Sorry, it was an injudicious choice by me. So we have $\xi$'s behaving as decoupled harmonic oscillators, but their values will not be constant in time (depending on initial conditions). If, for example, we have $\xi_1(t) = 0$ and $\xi_2(t) = Acos(\omega t + \phi), A \in \mathbb{R}$ and some phase angle $\phi$, we can no longer say the relative displacement is constant, so we cannot say that the centre of mass of the system is stationary? Is that correct? $\endgroup$
    – Poo2uhaha
    Commented Jan 25, 2021 at 18:19
  • 1
    $\begingroup$ @Poo2uhaha No. $\xi_1(t)$ is the position of the center of mass, so obviously $\xi_1(t)=0$ means the center of mass position is constant. $\endgroup$
    – J. Murray
    Commented Jan 25, 2021 at 18:32

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