Normal Modes: The Generalized Eigenproblem
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You should be comfortable with the quadratic expansion of the Lagrangian around a stable equilibrium and the definition of the kinetic and potential energy matrices \( M \) and \( K \).
Why this matters
Normal modes transform a complex coupled system into a set of independent, simple oscillators.
In the previous lecture, we established that for small oscillations around a stable equilibrium, the equations of motion take the form \( M \ddot{\eta} + K \eta = 0 \). This is a system of coupled linear differential equations. Solving this directly is difficult because the coordinates \( \eta_i \) are mixed together. The concept of normal modes provides a systematic way to "untangle" these coordinates. By identifying the specific patterns of motion where all parts of the system oscillate at the same frequency, we can decouple the equations entirely. This is not just a mathematical trick; it is the fundamental language used to describe vibrations in molecules, structures, and even quantum fields.
The Physical Idea
Imagine a system of masses connected by springs. If you pull one mass and release it, the motion is complicated because energy sloshes back and forth between the masses. However, there are special ways to displace the system. If you displace the masses in a specific ratio and release them from rest, every single mass will oscillate up and down (or back and forth) with the exact same frequency. They will all reach their maximum displacement at the same time, and they will all pass through the equilibrium position at the same time. These special patterns are called normal modes. In a normal mode, the system behaves as if it were a single, simple harmonic oscillator, even though it is made of many parts. The key insight is that any arbitrary motion of the system can be built by adding together these simple normal modes. This is the principle of superposition, and it allows us to break a hard problem into several easy ones.
The Generalized Eigenvalue Problem
To find these special modes mathematically, we assume a solution where all coordinates oscillate with the same frequency \( \omega \). We propose a trial solution of the form \( \eta(t) = a \cos(\omega t + \phi) \), where \( a \) is a constant vector representing the relative amplitudes of the coordinates. Substituting this into the equation of motion \( M \ddot{\eta} + K \eta = 0 \) yields the algebraic equation:
This is a homogeneous linear equation. For a non-trivial solution (where \( a \) is not the zero vector), the determinant of the coefficient matrix must vanish. This leads to the characteristic equation \( \det(K - \omega^2 M) = 0 \). This is a polynomial equation in \( \omega^2 \). The roots of this equation, \( \omega_r^2 \), are the squared normal frequencies. Because the system is stable, these roots are real and positive. The corresponding vectors \( a^{(r)} \) are the eigenvectors, which define the shape of the \( r \)-th normal mode. This is known as a generalized eigenvalue problem because the matrix \( M \) is not the identity matrix.
Orthogonality and Mass-Weighted Inner Product
A crucial property of the normal modes is their orthogonality. 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 involving two matrices, \( M \) and \( K \). The orthogonality condition is modified. Two distinct normal modes \( a^{(r)} \) and \( a^{(s)} \) satisfy:
This is called the mass-weighted inner product. Physically, this means that if the system is set into motion in one normal mode, it does no work on the other modes. The kinetic energy of mode \( r \) is independent of the kinetic energy of mode \( s \). This orthogonality is what guarantees that the modes are truly independent and do not exchange energy with each other. It is the mathematical foundation that allows us to decouple the system.
General Solution via Superposition
Once we have found the \( N \) normal frequencies \( \omega_r \) and their corresponding eigenvectors \( a^{(r)} \), we can construct the general solution for any initial condition. The total displacement vector \( \eta(t) \) is a linear combination of all the normal modes:
Here, \( C_r \) and \( \phi_r \) are constants determined by the initial position and velocity of the system. Each term in the sum represents one normal mode oscillating independently. The coefficient \( C_r \) tells us how much of that specific mode is present in the motion. This expression is the most important result of this lecture. It shows that the complex motion of a coupled system is simply the sum of simple, independent oscillations. The "coupling" is an illusion created by looking at the original coordinates; in the space of normal modes, the system is completely free.
The derivation
Normal Modes as a Generalized Eigenproblem
Derives that oscillations about stable equilibrium are governed by the generalized eigenvalue problem (K - omega^2 M)a = 0, giving normal modes and frequencies.
Do not confuse the eigenvectors \( a^{(r)} \) with the normal coordinates \( q_r \). The eigenvectors define the shape of the mode in the original coordinate space, while the normal coordinates are the time-dependent amplitudes \( C_r \cos(\omega_r t + \phi_r) \). A common error is to assume that the components of the eigenvector \( a^{(r)} \) are the amplitudes; they are actually the relative ratios of the displacements. The actual amplitude is determined by the initial conditions and the normalization of the eigenvector. C
Worked example
Problem. Consider a system of two identical masses \( m = 1.0\ \mathrm{kg} \) connected by three identical springs with spring constant \( k = 100\ \mathrm{N/m} \). The masses are on a frictionless surface, with the outer springs attached to fixed walls. Find the normal frequencies and the mode shapes.
- Write the equations of motion. Let \( x_1 \) and \( x_2 \) be the displacements. The potential energy is \( V = \frac{1}{2}k x_1^2 + \frac{1}{2}k (x_2 - x_1)^2 + \frac{1}{2}k x_2^2 \). Expanding this gives \( V = k x_1^2 - k x_1 x_2 + k x_2^2 \). The kinetic energy is \( T = \frac{1}{2}m \dot{x}_1^2 + \frac{1}{2}m \dot{x}_2^2 \).
- Identify the matrices. The mass matrix is \( M = m \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \). The stiffness matrix is \( K = k \begin{pmatrix} 2 & -1 \\ -1 & 2 \end{pmatrix} \).
- Solve the eigenvalue problem \( \det(K - \omega^2 M) = 0 \). This gives \( \det \begin{pmatrix} 2k - m\omega^2 & -k \\ -k & 2k - m\omega^2 \end{pmatrix} = 0 \). So, \( (2k - m\omega^2)^2 - k^2 = 0 \).
- Find the frequencies. \( 2k - m\omega^2 = \pm k \). Case 1: \( 2k - m\omega^2 = k \implies m\omega_1^2 = k \implies \omega_1 = \sqrt{k/m} \). Case 2: \( 2k - m\omega^2 = -k \implies m\omega_2^2 = 3k \implies \omega_2 = \sqrt{3k/m} \).
- Find the eigenvectors. For \( \omega_1^2 = k/m \), the equation is \( (2k - k)a_1 - k a_2 = 0 \implies a_1 = a_2 \). Mode 1: \( a^{(1)} = \begin{pmatrix} 1 \\ 1 \end{pmatrix} \). For \( \omega_2^2 = 3k/m \), the equation is \( (2k - 3k)a_1 - k a_2 = 0 \implies -a_1 = a_2 \). Mode 2: \( a^{(2)} = \begin{pmatrix} 1 \\ -1 \end{pmatrix} \).
Units check. \( \sqrt{\mathrm{N/m} / \mathrm{kg}} = \sqrt{(\mathrm{kg \cdot m/s^2}) / (\mathrm{m \cdot kg})} = \sqrt{1/s^2} = 1/s = \mathrm{rad/s} \). Correct.
Three quick tests
- What condition must the determinant satisfy for non-trivial normal modes to exist?
Answer
The determinant of \( (K - \omega^2 M) \) must be zero. - Why are the normal modes orthogonal?
Answer
Because the matrices \( M \) and \( K \) are symmetric, leading to orthogonality with respect to the mass-weighted inner product. - Can two different normal modes have the same frequency?
Answer
Yes, if the system has degeneracy, but they can still be chosen to be orthogonal.
Problem set for PU-201 →
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.