Coupled oscillations §

Matrix form. With $\mathbf{x} = (x_1, x_2)^T$, the equations become $\mathbf{M}\ddot{\mathbf{x}} + \mathbf{K}\mathbf{x} = 0$, where $\mathbf{M}$ is the diagonal mass matrix and $\mathbf{K}$ the stiffness matrix. Trying a solution $\mathbf{x}(t) = \mathbf{e}\,e^{i\omega t}$ gives the eigenvalue problem $\mathbf{M}^{-1}\mathbf{K}\,\mathbf{e} = \omega^2\mathbf{e}$: the squares of the normal-mode frequencies are the eigenvalues of $\mathbf{M}^{-1}\mathbf{K}$ (equivalently, the frequencies are the square roots of those eigenvalues), and the eigenvectors give the mode patterns. For $N$ coupled oscillators, $\mathbf{M}^{-1}\mathbf{K}$ is $N \times N$ and there are exactly $N$ normal modes.

Symmetric case. For $m_1 = m_2 = m$ and $k_1 = k_2 = k$, the eigenvectors are $(1, 1)$ and $(1, -1)$, giving the two normal frequencies:

The in-phase mode ($x_1 = x_2$, eigenvector $(1,1)$): the coupling spring is never stretched, so it exerts no force — each mass oscillates as if the other were absent, at the bare frequency $\sqrt{k/m}$. The out-of-phase mode ($x_1 = -x_2$, eigenvector $(1,-1)$): the coupling spring is stretched twice as much as either outer spring per unit displacement, adding $2k_c$ to the effective stiffness, so the frequency rises to $\sqrt{(k+2k_c)/m}$.

Why eigenvectors? In a mode, the ratio $x_1/x_2$ stays constant in time — that is the defining property of an eigenvector. Only then can both masses share a single $\cos(\omega_n t)$ factor. Any other motion mixes the two modes, so the ratio changes in time.

Orthogonality. Modes with different frequencies are orthogonal with respect to the mass matrix: $\sum_i M_i\, e_i^{(n)} e_i^{(m)} = 0$ for $n \neq m$. This lets you project any initial condition onto the modes independently, exactly like a Fourier decomposition (see Wave propagation).

Physical picture. In the in-phase mode the coupling spring does no work, so the mode cannot exchange energy with anything — it oscillates in isolation. In the out-of-phase mode the coupling spring alternately stores and releases energy, adding stiffness and raising the frequency. Coupling always splits the frequencies apart: degenerate identical oscillators become $\omega_+$ and $\omega_- \gt \omega_+$.

Sum-to-product form. Applying the identities $\cos\alpha + \cos\beta = 2\cos\frac{\alpha-\beta}{2}\cos\frac{\alpha+\beta}{2}$ and $\cos\alpha - \cos\beta = -2\sin\frac{\alpha-\beta}{2}\sin\frac{\alpha+\beta}{2}$:

Each mass oscillates rapidly at the average frequency $(\omega_+ + \omega_-)/2$, under a slowly varying envelope of angular frequency $(\omega_- - \omega_+)/2$. The two envelopes are in quadrature: when $|\cos|$ is maximal for mass 1, $|\sin|$ is zero for mass 2.

Beat frequency. The envelope amplitude repeats at the beat frequency

— the same formula as the beats heard between two detuned tuning forks, and the same interference mathematics as in Wave propagation.

Energy transfer period. The energy on each mass is proportional to the envelope squared, $\cos^2\!\bigl(\frac{\omega_- - \omega_+}{2}t\bigr)$ and $\sin^2\!\bigl(\frac{\omega_- - \omega_+}{2}t\bigr)$. Squaring doubles the angular frequency, so the energy sloshes completely from mass 1 to mass 2 and back with period

At $t = \pi/(\omega_- - \omega_+)$ (half period), mass 1 is momentarily at rest with zero energy and mass 2 carries all of it; at $t = 2\pi/(\omega_- - \omega_+)$ the energy is back on mass 1. Total energy is conserved — the coupling spring is the pipe through which it flows.

The bridge from particles to fields. The ladder is:

• 2 coupled masses → 2 normal modes (two discrete frequencies).
• $N$ coupled masses → $N$ normal modes, whose frequencies fill the band $\omega(k)$ between $\omega_0$ (in-phase, $k = 0$) and $\omega_0\sqrt{1 + 4\epsilon}$ (alternating, $k = \pi/a$).
• $N \to \infty$ → a continuum of modes: a wave with dispersion relation $\omega(k)$. For small $ka$, $\cos(ka) \approx 1 - (ka)^2/2$, so $\omega \approx \omega_0\sqrt{1 + \epsilon (ka)^2} \approx \omega_0 (1 + \tfrac{\epsilon}{2}(ka)^2)$ — nearly linear near $k = 0$, where the continuum approximation is best.

Where the same band structure appears. Atoms in a crystal (phonon dispersion — the $k = 0$ acoustic mode is the “in-phase” mode, and the speed of sound is its slope), coupled LC circuits (filter passbands), a guitar string (harmonics are the normal modes), and a drum membrane (2D modes). In every case, the continuum limit of coupled oscillators is a field.

Problem. Two identical masses ($m = 1$ kg) are connected to fixed walls by springs of stiffness $k = 100$ N/m and to each other by a coupling spring $k_c = 50$ N/m. Mass 1 is displaced by 10 cm and released from rest, with mass 2 held at its equilibrium position and then released.

1. Find the two normal-mode frequencies.
2. Find the beat frequency and the time for the energy to transfer completely to mass 2.
3. Sketch (or describe) the displacement of each mass at that moment.