Coupled Oscillators and Normal Modes
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You should be comfortable with solving systems of linear differential equations and understand the concept of a generalized eigenvalue problem \(\mathbf{A}\mathbf{x} = \lambda \mathbf{B}\mathbf{x}\) where \(\mathbf{B}\) is positive-definite.
Why this matters
Normal modes are the fundamental building blocks of all small oscillations in physics.
Whether you are analyzing the vibrations of a guitar string, the normal modes of a molecule, or the collective excitations in a solid, the underlying mathematics is identical. By decomposing a complex coupled system into independent, uncoupled oscillators, we transform a difficult problem of coupled motion into a set of simple harmonic oscillators. This lecture bridges the gap between the abstract spectral theory of self-adjoint operators and the concrete mechanics of coupled masses and springs.
The Intuition of Decoupling
Imagine two masses connected by springs. If you pull one mass and release it, the motion is complicated: the first mass moves, pulls the second, which pulls back, and so on. The equations of motion are coupled; the acceleration of mass one depends on the position of mass two, and vice versa. However, there are special ways to move the system where the coupling effectively disappears. If you move both masses in the same direction by the same amount, the spring between them does not stretch or compress. In this specific pattern of motion, each mass behaves as if it were attached to a single spring, oscillating independently. These special patterns of motion are called normal modes. The key insight is that any arbitrary motion of the system can be viewed as a superposition of these independent normal modes. Just as a complex chord on a piano is a sum of pure notes, the complex motion of coupled oscillators is a sum of pure normal modes.
From Equations of Motion to Eigenvalues
To formalize this, we consider a system with \(n\) degrees of freedom. For small oscillations around a stable equilibrium, the kinetic energy \(T\) and potential energy \(V\) can be approximated by quadratic forms. Let \(\mathbf{q}\) be the vector of generalized coordinates. The kinetic energy is \(T=\tfrac12\,\dot{\mathbf{q}}^{\mathsf T}\mathbf{M}\dot{\mathbf{q}}\) and the potential energy is \(V=\tfrac12\,\mathbf{q}^{\mathsf T}\mathbf{K}\mathbf{q}\). Here, \(\mathbf{M}\) is the mass matrix (symmetric and positive-definite) and \(\mathbf{K}\) is the stiffness matrix (symmetric). Applying Lagrange's equations leads to the linearized equations of motion:
We seek harmonic solutions of the form \(\mathbf{q}(t)=\mathbf{a}\,e^{i\omega t}\). Substituting this into the equation of motion, noting that \(\ddot{\mathbf{q}} = -\omega^2 \mathbf{a}\,e^{i\omega t}\), we obtain:
This is a generalized eigenvalue problem. The eigenvalues are the squared frequencies \(\omega_k^2\), and the eigenvectors \(\mathbf{a}_k\) are the normal mode shapes. Because \(\mathbf{M}\) is positive-definite and \(\mathbf{K}\) is symmetric, the eigenvalues \(\omega_k^2\) are real and positive (assuming stability), and the eigenvectors can be chosen to be orthogonal with respect to the mass matrix \(\mathbf{M}\).
M-Orthogonality and Normalization
In standard linear algebra, eigenvectors of a symmetric matrix are orthogonal with respect to the standard dot product. However, here we have a generalized problem. The orthogonality condition is weighted by the mass matrix. Two distinct normal modes \(\mathbf{a}_j\) and \(\mathbf{a}_k\) satisfy:
This is called \(\mathbf{M}\)-orthogonality. Physically, the modes are also \(\mathbf{K}\)-orthogonal (\(\mathbf{a}_j^{\mathsf T}\mathbf{K}\mathbf{a}_k = 0\) for \(j \neq k\)), so both the kinetic and the potential energy split into a sum of separate terms, one per mode, with no cross terms at any instant. This orthogonality is crucial because it allows us to decouple the equations of motion. We can define new coordinates, the normal coordinates \(Q_k\), such that \(\mathbf{q} = \sum_k Q_k \mathbf{a}_k\). If we normalize the eigenvectors such that \(\mathbf{a}_j^{\mathsf T}\mathbf{M}\mathbf{a}_k = \delta_{jk}\), the kinetic energy becomes \(T = \tfrac12 \sum_k \dot{Q}_k^2\) and the potential energy becomes \(V = \tfrac12 \sum_k \omega_k^2 Q_k^2\). The equations of motion for each \(Q_k\) are now completely independent simple harmonic oscillators.
Physical Interpretation of Modes
The eigenvectors \(\mathbf{a}_k\) tell us how the different parts of the system move relative to each other in a given mode. The lowest frequency mode (the "ground state" of the oscillator system) typically involves all masses moving in the same direction, resulting in minimal stretching of the internal springs. Higher frequency modes involve more rapid changes in direction between adjacent masses, leading to greater internal stress and higher energy for the same amplitude. The number of normal modes equals the number of degrees of freedom \(n\). This decomposition is not just a mathematical trick; it reflects the physical reality that energy can be transferred between modes only through non-linear interactions or external driving. In the linear approximation, energy in one normal mode stays in that mode forever.
The derivation
Normal Modes from the Generalized Eigenvalue Problem
For a conservative mechanical system with \(n\) degrees of freedom executing small oscillations about a stable equilibrium, with kinetic energy \(T=\tfrac12\,\dot{\mathbf{q}}^{\mathsf T}\mathbf{M}\dot{\mathbf{q}}\) and potential energy \(V=\tfrac12\,\mathbf{q}^{\mathsf T}\mathbf{K}\mathbf{q}\) (both quadratic forms, \(\mathbf{M}\) symmetric positive-definite, \(\mathbf{K}\) symmetric), the linearized equations of motion \(\mathbf{M}\ddot{\mathbf{q}}+\mathbf{K}\mathbf{q}=\mathbf{0}\) admit harmonic solutions \(\mathbf{q}(t)=\mathbf{a}\,e^{i\omega t}\) precisely when the amplitude vector \(\mathbf{a}\) and frequency \(\omega\) solve the generalized eigenvalue problem \(\mathbf{K}\mathbf{a}=\omega^2\mathbf{M}\mathbf{a}\). The \(n\) eigenvalues \(\omega_k^2\) are real, and the eigenvectors can be chosen \(\mathbf{M}\)-orthonormal, \(\mathbf{a}_j^{\mathsf T}\mathbf{M}\mathbf{a}_k=\delta_{jk}\) and \(\mathbf{a}_j^{\mathsf T}\mathbf{K}\mathbf{a}_k=\omega_k^2\,\delta_{jk}\), so the normal modes simultaneously diagonalize both the inertia and the stiffness.
A common misconception is that the eigenvectors are orthogonal in the standard Euclidean sense (\(\mathbf{a}_j^{\mathsf T}\mathbf{a}_k = 0\)). This is generally false for coupled oscillators unless the mass matrix \(\mathbf{M}\) is proportional to the identity matrix. You must always use the \(\mathbf{M}\)-inner product for orthogonality checks and normalization. C
Worked example
Problem. Consider two identical masses \(m = 1.0\ \mathrm{kg}\) connected by three identical 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 and the mode shapes.
- Define the coordinates \(x_1\) and \(x_2\) as the displacements of the left and right masses from equilibrium. The mass matrix is \(\mathbf{M} = \begin{pmatrix} m & 0 \\ 0 & m \end{pmatrix}\).
- Determine the stiffness matrix \(\mathbf{K}\). For mass 1, the force is \(-k x_1 + k(x_2 - x_1) = -2k x_1 + k x_2\). For mass 2, the force is \(-k x_2 + k(x_1 - x_2) = k x_1 - 2k x_2\). Thus, \(\mathbf{K} = \begin{pmatrix} 2k & -k \\ -k & 2k \end{pmatrix}\).
- Solve the generalized eigenvalue problem \(\mathbf{K}\mathbf{a} = \omega^2 \mathbf{M}\mathbf{a}\). Since \(\mathbf{M} = m\mathbf{I}\), this simplifies to \(\mathbf{K}\mathbf{a} = (m\omega^2)\mathbf{a}\). Let \(\lambda = m\omega^2\).
- Find the eigenvalues of \(\mathbf{K}\). The characteristic equation is \(\det(\mathbf{K} - \lambda \mathbf{I}) = (2k-\lambda)^2 - k^2 = 0\). This gives \(\lambda_1 = k\) and \(\lambda_2 = 3k\).
- Calculate the frequencies. \(\omega_1 = \sqrt{k/m} = \sqrt{100/1} = 10\ \mathrm{rad/s}\) and \(\omega_2 = \sqrt{3k/m} = \sqrt{300/1} \approx 17.32\ \mathrm{rad/s}\).
- Find the eigenvectors. For \(\lambda_1 = k\), \((2k-k)a_1 - k a_2 = 0 \implies a_1 = a_2\). Mode 1 is \(\mathbf{a}_1 = \begin{pmatrix} 1 \\ 1 \end{pmatrix}\). For \(\lambda_2 = 3k\), \((2k-3k)a_1 - k a_2 = 0 \implies -a_1 = a_2\). Mode 2 is \(\mathbf{a}_2 = \begin{pmatrix} 1 \\ -1 \end{pmatrix}\).
Units check. \(\sqrt{\mathrm{N/m} / \mathrm{kg}} = \sqrt{(\mathrm{kg \cdot m/s^2}) / (\mathrm{m \cdot kg})} = \sqrt{1/\mathrm{s^2}} = 1/\mathrm{s} = \mathrm{rad/s}\). The units are consistent.
Three quick tests
- What is the physical significance of the eigenvector \(\mathbf{a}_1 = \begin{pmatrix} 1 \\ 1 \end{pmatrix}\) in the worked example?
Answer
It represents the mode where both masses move in the same direction with the same amplitude. The internal spring does not stretch, so this mode has the lower frequency. - Why are the eigenvalues \(\omega_k^2\) guaranteed to be real?
Answer
Because the mass matrix \(\mathbf{M}\) is symmetric positive-definite and the stiffness matrix \(\mathbf{K}\) is symmetric, the generalized eigenvalue problem has real eigenvalues. - If the system is released with initial conditions that match exactly the shape of the second normal mode, what happens?
Answer
The system will oscillate purely in the second normal mode at frequency \(\omega_2\), with no energy transfer to the first mode, assuming the system remains linear.
Problem set for PU-104 →
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.