Coupled Oscillators and Normal Modes
~5 min read · Tier toggle changes depth (T1 plain · T2 standard · T3 rigour)
You should be comfortable with solving linear second-order differential equations and understanding the concept of a harmonic oscillator from previous lectures.
Why this matters
Complex mechanical systems can be decomposed into simple, independent vibrations known as normal modes.
When multiple masses are connected by springs, their motions are coupled, making the equations of motion difficult to solve directly. However, by identifying specific patterns of motion where all parts of the system oscillate at the same frequency, we can decouple the system into independent simple harmonic oscillators. This concept of normal modes is fundamental not only to mechanics but also to molecular vibrations, structural engineering, and even quantum field theory.
The Idea of Normal Modes
Imagine two masses connected by springs. If you push one mass, it moves, pulling the other. The motion is complex because the forces on each mass depend on the position of the other. However, there are special ways to move the system where this complexity disappears. In these special modes, every part of the system moves up and down (or back and forth) with the same frequency and the same phase relationship. For example, in a two-mass system, one mode might involve both masses moving in the same direction, while another involves them moving in opposite directions. In each of these modes, the system behaves exactly like a single simple harmonic oscillator. The general motion of the system is just a combination (superposition) of these independent modes.
Linearized Equations of Motion
Consider a system with \( n \) degrees of freedom described by generalized coordinates \( \mathbf{q} = (q_1, q_2, \dots, q_n)^{\mathsf T} \). Near a stable equilibrium point, the potential energy \( V \) can be approximated by a quadratic form, and the kinetic energy \( T \) is also quadratic in the velocities. The Lagrangian \( L = T - V \) leads to the linearized equations of motion:
Here, \( \mathbf{M} \) is the mass matrix, which is symmetric and positive-definite, and \( \mathbf{K} \) is the stiffness matrix, which is symmetric and positive-semidefinite. These matrices encapsulate the inertial and restoring force properties of the system, respectively.
The Generalized Eigenvalue Problem
To find the normal modes, we assume a solution of the form \( \mathbf{q}(t) = \mathbf{a} e^{i\omega t} \), where \( \mathbf{a} \) is a constant vector representing the mode shape and \( \omega \) is the angular frequency. Substituting this into the equations of motion yields:
Rearranging gives the generalized eigenvalue problem:
The eigenvalues \( \omega_r^2 \) are the squared normal frequencies, and the corresponding eigenvectors \( \mathbf{a}^{(r)} \) are the normal modes. Because \( \mathbf{M} \) and \( \mathbf{K} \) are symmetric, the eigenvectors can be chosen to be orthogonal with respect to the mass matrix, satisfying the orthogonality condition:
Superposition of Modes
Since the equations of motion are linear, any linear combination of solutions is also a solution. Therefore, the general motion of the system is a superposition of all normal modes:
Here, \( A_r \) and \( B_r \) are constants determined by the initial conditions. Each term in the sum represents an independent harmonic oscillator with frequency \( \omega_r \) and amplitude determined by the projection of the initial state onto the mode \( \mathbf{a}^{(r)} \). This decomposition simplifies the analysis significantly, as each mode can be treated independently.
The derivation
Coupled Oscillators and Normal Modes
For a mechanical system near a stable equilibrium, the linearized equations of motion \( \mathbf{M}\ddot{\mathbf{q}} = -\mathbf{K}\mathbf{q} \), with symmetric positive-definite mass matrix \( \mathbf{M} \) and symmetric positive-semidefinite stiffness matrix \( \mathbf{K} \), are decoupled by the generalized eigenvalue problem \( \mathbf{K}\mathbf{a}^{(r)} = \omega_r^2\,\mathbf{M}\mathbf{a}^{(r)} \): the eigenvectors \( \mathbf{a}^{(r)} \) are the normal modes (rigid spatial patterns that oscillate in unison) and the eigenvalues \( \omega_r^2 \) are the squared normal frequencies, so every free motion is a superposition \( \mathbf{q}(t)=\sum_r \mathbf{a}^{(r)}\,(A_r\cos\omega_r t + B_r\sin\omega_r t) \).
A common misconception is that normal modes are unique. In fact, if two modes have the same frequency (degeneracy), any linear combination of their eigenvectors is also a valid normal mode. The choice of basis within the degenerate subspace is arbitrary and depends on the symmetry of the system. C
Worked example
Problem. Two identical masses \( m = 2\ \mathrm{kg} \) are connected by three springs, each with spring constant \( k = 100\ \mathrm{N/m} \). The outer springs are attached to fixed walls, and the inner spring connects the two masses. Find the normal frequencies of the system.
- Equations of motion: \(m\ddot{x}_1 = -kx_1 - k(x_1 - x_2)\) and \(m\ddot{x}_2 = -kx_2 - k(x_2 - x_1)\).
- Matrix form: \(\mathbf{M} = m\mathbf{I}\), \(\mathbf{K} = \begin{pmatrix} 2k & -k \\ -k & 2k \end{pmatrix}\).
- Eigenvalues of \(\mathbf{K}\mathbf{a} = \omega^2 m\mathbf{a}\): \(\omega^2 = k/m\) with \(\mathbf{a} \propto (1, 1)\) (masses move together, the middle spring is not stretched) and \(\omega^2 = 3k/m\) with \(\mathbf{a} \propto (1, -1)\) (masses move oppositely).
- Numbers: \(k/m = 100/2 = 50\ \mathrm{s^{-2}}\), so \(\omega_1 = \sqrt{50} \approx 7.07\ \mathrm{rad/s}\) and \(\omega_2 = \sqrt{150} \approx 12.2\ \mathrm{rad/s}\).
Units check. The spring constant \( k \) has units \( \mathrm{N/m} = \mathrm{kg/s^2} \). The ratio \( k/m \) has units \( (\mathrm{kg/s^2}) / \mathrm{kg} = 1/\mathrm{s^2} \). Taking the square root gives \( 1/\mathrm{s} = \mathrm{rad/s} \), which is correct for angular frequency.
Three quick tests
- What is the physical significance of the eigenvectors in the normal mode analysis?
Answer
They represent the relative displacements of the masses in each mode, showing how the system moves as a whole. - Why must the mass matrix \( \mathbf{M} \) be positive-definite?
Answer
Positive-definiteness ensures that the kinetic energy is always positive for any non-zero velocity, which is a physical requirement for a stable system. - How does the orthogonality condition \( \big(\mathbf{a}^{(r)}\big)^{\mathsf T}\mathbf{M}\,\mathbf{a}^{(s)} = \delta_{rs} \) simplify the calculation of initial conditions?
Answer
It allows the initial conditions to be projected directly onto each mode independently, without cross-terms, making it easy to determine the amplitudes \( A_r \) and \( B_r \).
Problem set for PU-101 →
Each derivation in this unit ends with five graded problems with full solutions.
Lecture notes drafted with a local model and checked line by line by the editor; every number in the worked example was recomputed.