advanced30 min studyLesson 128 in Physics

Coupled small-oscillation eigenmodes

Linearize a mechanical Lagrangian around equilibrium and solve a mass-weighted generalized eigenvalue problem.

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What you’ll understand

  • Expand kinetic and potential energy to quadratic order near rest equilibrium.
  • Solve K a=ω² M a with correct mass weighting.
  • Interpret positive, negative, and zero squared frequencies cautiously.

A coupled mechanical system rarely oscillates one coordinate at a time. Its simple motions are collective patterns in which all coordinates share a frequency and keep fixed amplitude ratios. Finding these patterns turns differential equations into a matrix problem. The mass matrix is part of that problem: diagonalizing the potential-energy matrix alone gives the wrong frequencies when masses or coordinate scales differ. Small oscillations make this structure visible without claiming that a nonlinear system stays linear at every amplitude.

Expand about a genuine equilibrium

Consider a stationary natural Lagrangian with kinetic energy quadratic in rates and a configuration-only potential. Choose an equilibrium q₀ where the potential gradient vanishes and the system can remain at rest. Write q=q₀+η. To second order in small displacement and velocity, evaluate the mass matrix at q₀ and retain the Hessian K of the potential. Terms from the coordinate variation of M multiply two velocities and are third order in a uniform small-amplitude expansion, so they do not enter this leading quadratic model.

The approximation assumes a regular positive-definite mass matrix and a smooth potential in a neighborhood of the equilibrium. A moving reference solution, a rotating frame with gyroscopic terms, or a velocity-dependent magnetic coupling requires a more general linearization. Do not delete such terms by copying the rest-equilibrium formula. Check dimensions as well: for length coordinates, M has mass units and K has spring-constant units. For angular coordinates the corresponding coefficients carry different units but produce the same dimension of inverse time squared.

L2=12η˙TMη˙−12ηTKη,Mη¨+Kη=0,Kab=∂2V∂qa∂qb∣q0L_2=\frac12\dot\eta^{\mathsf T}M\dot\eta-\frac12\eta^{\mathsf T}K\eta,\qquad M\ddot\eta+K\eta=0,\qquad K_{ab}=\left.\frac{\partial^2V}{\partial q_a\partial q_b}\right|_{q_0}
This quadratic model is for small motion around a rest equilibrium of a stationary natural mechanical system. M and K are evaluated at that equilibrium.↔ Scroll wide equations horizontally.

Solve a generalized rather than ordinary eigenproblem

Insert a trial collective motion with a fixed vector of relative amplitudes and a sinusoidal time dependence. The equation becomes K a=ω² M a. Nonzero amplitude vectors exist when the determinant of K−ω²M vanishes. If M is positive definite and K is symmetric, the problem can be converted to a symmetric ordinary eigenproblem by mass whitening. This explains why the squared frequencies are real even though M inverse times K need not look symmetric in ordinary coordinate components.

For distinct squared frequencies, the eigenvectors are orthogonal in the M-weighted inner product, not generally in the unweighted Cartesian dot product. Normalize them so aᵀ M a equals one when convenient. The normalization changes modal-coordinate units and amplitudes, but not the physical displacement pattern. Degenerate frequencies permit a choice of mass-orthogonal basis within their eigenspace. Such freedom does not create additional independent frequencies or make one particular drawing of a degenerate mode physically unique.

det⁡(K−ω2M)=0,Kas=ωs2Mas,arTMas=δrs\det(K-\omega^2M)=0,\qquad K a_s=\omega_s^2 M a_s,\qquad a_r^{\mathsf T}M a_s=\delta_{rs}
The final normalization can be chosen for a complete real mode basis when M is positive definite and K is symmetric, including an orthogonal basis inside degeneracies.↔ Scroll wide equations horizontally.

Interpret the spectrum within the approximation

A positive squared frequency gives an oscillatory linear mode. A negative one gives exponential growth and decay directions, indicating a linear instability of this rest equilibrium. A zero value needs additional analysis: it can represent an exact symmetry direction, free drift, or an equilibrium whose leading restoring term appears beyond quadratic order. A nonnegative spectrum alone is not a complete nonlinear stability proof. The smallness of the displacement and the neglected terms must remain compatible over the time interval being interpreted.

The oscillator lab represents one already-decoupled positive-frequency mode. Set damping to zero and predict how increasing natural frequency from one to two changes the period. It has no mass-matrix controls and does not compute eigenvectors or coupled frequencies. Its normalized initial displacement is one, so interpreting it as a physical small mode requires a separate small amplitude scale. Use the worked matrix example to derive frequencies first, then use the lab only to inspect the time dependence of one allowed positive-frequency mode.

Two unequal masses, two collective frequencies

Use length coordinates with M=diag(2,1) kg and K=[[6,−2],[−2,2]] N/m. Find both squared frequencies and unnormalized mode vectors.

  1. Put λ=ω². The determinant is (6−2λ)(2−λ)−4=2λ²−10λ+8=2(λ−1)(λ−4). Thus λ=1 or 4 in s⁻².

  2. For λ=1, the first row gives 4a₁−2a₂=0, so take a=(1,2). For λ=4, it gives −2a₁−2a₂=0, so take b=(1,−1).

  3. The frequencies are 1 and 2 rad/s. Check mass orthogonality: aᵀMb=(1,2)·(2,−1)=0, although the ordinary dot product a·b=−1 is nonzero.

The result

The lower mode moves both masses in the same direction; the higher mode moves them oppositely. Their correct orthogonality is mass weighted.

Explore the model

Interactive lab

Interactive lab · oscillator

Inspect one decoupled mode

Set Damping ratio to zero. Compare Natural frequency one and two after predicting the period ratio. This analytic one-mode oscillator does not solve the two-mass eigenproblem or display a physical mode shape.

Inspect one decoupled modeUnit mass, x(0) = 1 and v(0) = 0. The analytic solution includes underdamped, critically damped and overdamped cases; time is measured in seconds. Current readings: Displacement x(t): 1; Velocity v(t): 0; Regime: underdamped. All parameters are adjustable with the labeled controls below.02.557.510−1.15−0.5700.571.15Time · sDisplacementunderdamped
Displacement x(t)Current time
Displacement x(t)
1
Velocity v(t)
0
Regime
underdamped
t = 0 s

Unit mass, x(0) = 1 and v(0) = 0. The analytic solution includes underdamped, critically damped and overdamped cases; time is measured in seconds. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forCompute collective frequencies with both stiffness and mass before interpreting an individual oscillation.

Check your understanding

For symmetric K and positive-definite M, distinct mode vectors are orthogonal in which inner product?

Try it yourself

Normalize the two mode vectors a=(1,2) and b=(1,−1) from the example using M=diag(2,1). Verify each mass norm and their cross term.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

Compute aᵀMa and bᵀMb before rescaling; ordinary Euclidean normalization is not the goal.

Compare with the solution

The mass norms are 6 and 3, so choose e₁=a/√6 and e₂=b/√3. Then e₁ᵀMe₁=e₂ᵀMe₂=1 and e₁ᵀMe₂=0. The dimensionful interpretation of these normalizations belongs to the chosen coordinate convention; physical displacements are obtained by multiplying modes by their modal amplitudes.

Take it with you

  • Linearize around a stated rest equilibrium and retain both M and K.
  • The eigenvalues are squared frequencies, not frequencies.
  • Mass orthogonality replaces ordinary dot-product orthogonality.

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