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PU-103 / L10 · lecture

Two Coupled Oscillators and Normal Modes

~6 min read · Tier toggle changes depth (T1 plain · T2 standard · T3 rigour)

BEFORE YOU START

You should be comfortable with simple harmonic motion, the concept of resonance, and basic linear algebra (eigenvalues and eigenvectors of symmetric matrices).

Why this matters

Understanding how coupled systems decompose into independent modes is the foundation for analyzing everything from molecular vibrations to bridges.

In the previous lecture, we examined a single driven oscillator and its response to external forces. Now we move to a system of two masses connected by springs. This is the simplest case where the motion of one part directly influences the other. By solving this two-degree-of-freedom problem, we discover that the complex motion can be broken down into two simple, independent patterns called normal modes. This concept generalizes to any number of coupled oscillators and is crucial for understanding wave propagation in continuous media.

The Idea: Independent Patterns of Motion

Imagine two identical masses connected by three springs: one spring between the left wall and mass 1, one between mass 1 and mass 2, and one between mass 2 and the right wall. If you pull mass 1 and release it, mass 2 will eventually start moving too. The motion looks complicated because the masses are constantly exchanging energy. However, there are two special ways the system can move where this complexity disappears. In the first special case, both masses move in the same direction with the same amplitude. In the second special case, the masses move in opposite directions with the same amplitude. In both cases, the distance between the masses changes in a simple, predictable way, and the system behaves exactly like a single simple harmonic oscillator. These special patterns are the normal modes. Any arbitrary motion of the two masses can be described as a combination (superposition) of these two modes.

Equations of Motion and Matrix Form

Let the two masses be \( m \) and the spring constants be \( k_1 \) (left), \( k_2 \) (middle), and \( k_3 \) (right). Let \( x_1 \) and \( x_2 \) be the displacements from equilibrium. Applying Newton's second law to each mass gives:

For mass 1: \( m\ddot{x}_1 = -k_1 x_1 + k_2(x_2 - x_1) \)

For mass 2: \( m\ddot{x}_2 = -k_3 x_2 + k_2(x_1 - x_2) \)

Rearranging these into a standard linear form, we can write the system in matrix notation. The mass matrix \( \mathbf{M} \) is diagonal because the masses are independent, and the stiffness matrix \( \mathbf{K} \) is symmetric because the force on one mass due to the other is equal and opposite. The equation of motion is:

\[ \mathbf{M}\ddot{\mathbf{x}} + \mathbf{K}\mathbf{x} = \mathbf{0} \]

where \( \mathbf{x} = \begin{pmatrix} x_1 \\ x_2 \end{pmatrix} \), \( \mathbf{M} = m \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \), and \( \mathbf{K} = \begin{pmatrix} k_1 + k_2 & -k_2 \\ -k_2 & k_2 + k_3 \end{pmatrix} \). This matrix form is powerful because it allows us to use linear algebra to find the solutions.

Normal Modes and Eigenvalues

We seek solutions where both masses oscillate at the same frequency \( \omega \). We assume a solution of the form \( \mathbf{x}(t) = \mathbf{a} e^{i\omega t} \), where \( \mathbf{a} \) is a constant vector describing the relative amplitudes and phases. Substituting this into the matrix equation yields the generalized eigenvalue problem:

\[ \mathbf{K}\mathbf{a} = \omega^2 \mathbf{M}\mathbf{a} \]

The values of \( \omega^2 \) that satisfy this equation are the eigenvalues, and the corresponding vectors \( \mathbf{a} \) are the eigenvectors. The eigenvectors define the normal modes. For the symmetric case where \( k_1 = k_3 = k \), the two normal modes are: 1) The symmetric mode where \( x_1 = x_2 \), with frequency \( \omega_1 = \sqrt{k/m} \). 2) The antisymmetric mode where \( x_1 = -x_2 \), with frequency \( \omega_2 = \sqrt{(k+2k_2)/m} \) (which is \( \sqrt{3k/m} \) when all three springs are equal). The antisymmetric mode has a higher frequency because the middle spring is compressed or stretched, adding to the restoring force.

Superposition and General Motion

Since the equations of motion are linear, any linear combination of solutions is also a solution. Therefore, the general motion of the system is the sum of the two normal modes:

\[ \mathbf{x}(t) = \mathbf{a}^{(1)} (A_1 \cos \omega_1 t + B_1 \sin \omega_1 t) + \mathbf{a}^{(2)} (A_2 \cos \omega_2 t + B_2 \sin \omega_2 t) \]

The coefficients \( A_i \) and \( B_i \) are determined by the initial conditions (initial positions and velocities). Physically, this means that if you start the system with an arbitrary configuration, it will oscillate as a mixture of the two fundamental patterns. The energy sloshes back and forth between the two masses when both modes are excited and their frequencies are close (weak coupling), a phenomenon known as beats, which we will explore in the next lecture.

The derivation

◆ DERIVATION EMBED · D-033 · not re-derived here

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 \( \mathbf{q}(t)=\sum_r \mathbf{a}^{(r)}\,(A_r\cos\omega_r t + B_r\sin\omega_r t) \).

\[ \mathbf{K}\,\mathbf{a}^{(r)}=\omega_r^2\,\mathbf{M}\,\mathbf{a}^{(r)},\qquad \mathbf{q}(t)=\sum_{r=1}^{n}\mathbf{a}^{(r)}\big(A_r\cos\omega_r t+B_r\sin\omega_r t\big),\qquad \big(\mathbf{a}^{(r)}\big)^{\mathsf T}\mathbf{M}\,\mathbf{a}^{(s)}=\delta_{rs} \]
Open full derivation →
Tier 3 ยท watch out

A common misconception is that the normal modes are the only possible motions. In reality, the normal modes are just a convenient basis for describing the motion. If you excite only one mode, the system stays in that mode forever (in the absence of damping). However, if you excite a general motion, the relative phases of the two modes determine how the energy is distributed. The orthogonality of the modes with respect to the mass matrix, \( (\mathbf{a}^{(r)})^{\mathsf T} \mathbf{M} \mathbf{a}^{(s)} = \delta_{rs} \), ensures that the modes are independent and do not exchange energy directly. C

Worked example

Problem. Two masses of \( m = 1.0\ \mathrm{kg} \) are connected by springs with constants \( k_1 = k_3 = 100\ \mathrm{N/m} \) and \( k_2 = 200\ \mathrm{N/m} \). Find the two normal frequencies and the ratio of amplitudes for each mode.

  1. Identify the stiffness matrix \( \mathbf{K} \) for the symmetric case: \( \mathbf{K} = \begin{pmatrix} 300 & -200 \\ -200 & 300 \end{pmatrix} \ \mathrm{N/m} \).
  2. Set up the eigenvalue equation \( \det(\mathbf{K} - \omega^2 \mathbf{M}) = 0 \). This gives \( (300 - \omega^2)^2 - (-200)^2 = 0 \).
  3. Solve for \( \omega^2 \): \( 300 - \omega^2 = \pm 200 \). Thus, \( \omega_1^2 = 100 \) and \( \omega_2^2 = 500 \).
  4. Calculate the frequencies: \( \omega_1 = \sqrt{100} = 10\ \mathrm{rad/s} \) and \( \omega_2 = \sqrt{500} \approx 22.4\ \mathrm{rad/s} \).
  5. Determine the eigenvectors. For \( \omega_1^2 = 100 \), \( (300-100)a_1 - 200a_2 = 0 \implies 200a_1 = 200a_2 \implies a_1 = a_2 \). For \( \omega_2^2 = 500 \), \( (300-500)a_1 - 200a_2 = 0 \implies -200a_1 = 200a_2 \implies a_1 = -a_2 \).
\[ \omega_1 = 10\ \mathrm{rad/s},\ \omega_2 \approx 22.4\ \mathrm{rad/s} \]

Units check. \( \sqrt{\mathrm{N/m} / \mathrm{kg}} = \sqrt{(\mathrm{kg \cdot m/s^2}) / (\mathrm{m \cdot kg})} = \sqrt{1/s^2} = 1/s = \mathrm{rad/s} \). The units are consistent.

CHECK YOURSELF

Three quick tests

  1. What is the physical significance of the eigenvectors in the context of coupled oscillators?
    AnswerThey represent the normal modes, which are the independent patterns of motion where all parts of the system oscillate at the same frequency.
  2. If the two masses are identical and the springs are symmetric, what is the ratio of the amplitudes in the higher frequency mode?
    AnswerThe ratio is -1, meaning the masses move in opposite directions with equal amplitude.
  3. Why can the general motion be expressed as a superposition of normal modes?
    AnswerBecause the equations of motion are linear, the principle of superposition applies, allowing any solution to be written as a sum of independent solutions.
PROBLEMS

Problem set for PU-103 →

Each derivation in this unit ends with five graded problems with full solutions.

NOTES

Lecture notes drafted with a local model and checked line by line by the editor; every number in the worked example was recomputed.

Derivations cited: D-033 · Threads: force, energy, waves