Normal modes of the coupled oscillator

What happens when two or more oscillators are linked together? The simplest system is that of two identical masses $m$ connected by three springs: two springs to the walls (constant $K$) and one central spring (constant $\kappa$) coupling them.

Two identical masses connected to the walls by springs $K$ and to each other by the coupling spring $\kappa$ (in gold).

The equations of motion for the two masses are coupled: each one contains both $x_1$ and $x_2$.

$\begin{cases} m\,\ddot x_1 = -K\,x_1 - \kappa\,(x_1 - x_2) \\ m\,\ddot x_2 = -K\,x_2 - \kappa\,(x_2 - x_1) \end{cases}$

They are no longer two independent oscillators. The trick for solving them is a change of variables: $X = (x_1 + x_2)/2$ (motion of the centre of mass) and $\xi = x_1 - x_2$ (relative difference).

Principle — Normal modes of the coupled oscillator

Adding and subtracting the two equations gives two decoupled equations: $m\,\ddot X = -K\,X \qquad m\,\ddot\xi = -(K + 2\kappa)\,\xi$ The system therefore has two independent normal modes:

• symmetric mode ($x_1 = x_2$, the masses oscillate in phase): $\omega_+ = \sqrt{K/m}$; the central spring does not deform.
• antisymmetric mode ($x_1 = -x_2$, in phase opposition): $\omega_- = \sqrt{(K + 2\kappa)/m}$.

Key formula — The two normal modes

$\omega_+ = \sqrt{\frac{K}{m}} \ \text{(in phase)}, \qquad \omega_- = \sqrt{\frac{K + 2\kappa}{m}} \ \text{(out of phase)}$ Every motion of the system is a superposition of the two modes. The antisymmetric mode is faster because the central spring, deforming, adds a restoring effect.

▶ The Surprising Secret of Synchronization — Veritasium

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Topics: Oscillations and harmonic motion Concepts: Normal modes Skills: Symbolic set-up Objects: Spring

Related exercises: Problem — Two coupled swings · Worked exercise — Two pendulums coupled by a string · Worked exercise — Finding A and φ from initial conditions