Physics Library
 An open source physics library
Encyclopedia | Forums | Docs | Random  

Small Oscillations and the Normal-Mode Eigenvalue Problem

(Topic)

Small Oscillations and the Normal-Mode Eigenvalue Problem

Many mechanical systems are nonlinear when written exactly.

Near a stable equilibrium, however, their motion often simplifies dramatically.

To second order in small displacements, a broad class of conservative systems takes the form

|---------------|
M  ¨η + K η = 0. |
-----------------
(1)

Here

  • η is the displacement from equilibrium in generalized coordinates,
  • M is the mass matrix evaluated at equilibrium,
  • K is the stiffness matrix, equal to the Hessian of the potential at equilibrium.

The associated normal modes satisfy

|--------------|
-Ka--=-ω2Ma.---|
(2)

This is the generalized normal-mode eigenvalue problem.

Its eigenvalues

ω2r
(3)

determine the small-oscillation frequencies, while its eigenvectors determine the collective mode shapes.

M06-04 introduced this structure using coupled springs.

The present article derives it from the local mechanics of a general conservative system and develops the mathematical properties that make modal analysis work.

PIC

Figure 1. Near a stable equilibrium, a smooth potential surface is approximated by a quadratic form. Its local curvature determines the stiffness matrix and therefore the small-oscillation frequencies.

1 Equilibrium and displacement coordinates

Let the system have generalized coordinates

q1,...,qN .
(4)

Suppose

q0
(5)

is an equilibrium configuration.

Define the small displacement vector

-------------
|η = q − q0. |
-------------|
(6)

At equilibrium,

   |
∂U-||
∂q |  =  0
  i q0
(7)

for every independent coordinate.

This removes the linear term from the local expansion of the potential energy.

2 Quadratic expansion of the potential

Expand

U(q0 + η )
(8)

in a Taylor series:

U = U0 + ∑ i∂U  ||
----||
∂qi0ηi (9)
+ 1
2- ∑ i,j      |
 ∂2U  |
∂q-∂q-||
  i  j0ηiηj + ⋅⋅⋅. (10)

The first derivatives vanish at equilibrium.

Therefore, to quadratic order,

|--------------------|
|            1       |
U (2) = U0 + -ηT K η,|
-------------2--------
(11)

where

|-----------------|
|       ∂2U  ||    |
Kij =  ------||  . |
-------∂qi∂qj-q0---
(12)

The matrix

K
(13)

is the Hessian of the potential at equilibrium.

For a smooth potential,

|---------|
| T       |
K---=-K.---
(14)

3 Stable equilibrium and the stiffness matrix

For a strict stable equilibrium, the quadratic change in potential must be positive for every nonzero sufficiently small displacement:

  T
η  K η > 0.
(15)

Thus, in the nondegenerate stable case,

|----------------------|
|K  is positive de finite.|
-----------------------
(16)

If

K
(17)

is only positive semidefinite, zero-curvature directions can occur.

These directions lead to zero modes.

If the quadratic form has a negative direction, the equilibrium is linearly unstable.

4 Quadratic kinetic energy

For a broad class of mechanical systems,

       ∑
T =  1-   M   (q)˙q ˙q.
     2       ij    i j
       i,j
(18)

Near equilibrium,

q = q0 + η.
(19)

The velocities are already first order in the small motion.

Therefore replacing

Mij (q)
(20)

by its equilibrium value introduces only higher-order corrections.

To quadratic order,

|----------------|
|T(2) = 1η˙T M η˙, |
--------2--------|
(21)

where

|------------|
M  =  M (q0).|
--------------
(22)

For independent physical degrees of freedom,

|--T-------|
-M---=--M--|
(23)

and normally

|----------------------|
-M--is positive-definite.|
(24)

5 Why coordinate-dependent inertia can be frozen at equilibrium

Suppose

M (q0 + η ) = M0  + M1 (η) + ⋅⋅⋅ .
(25)

The correction

M1
(26)

is first order in

η.
(27)

But

˙ηT M1 ˙η
(28)

contains one factor of displacement and two factors of velocity.

It is third order in the small-motion bookkeeping.

Therefore it is omitted in the quadratic small-oscillation theory.

6 Quadratic Lagrangian

Ignoring the irrelevant constant

U0,
(29)

the quadratic Lagrangian is

|-------1---------1--------|
|L(2) = -˙ηT M ˙η − --ηT K η.|
--------2---------2---------
(30)

The Euler-Lagrange equations give

|---------------|
M  ¨η + K η = 0. |
-----------------
(31)

This is the universal linear system for conservative small oscillations.

7 Normal-mode ansatz

Seek a collective motion

|--------------------|
η-(t) =-a-cos(ωt-+-ϕ).
(32)

Then

       2
¨η = − ω a cos(ωt + ϕ).
(33)

Substitution yields

|-------2----------|
-(K-−--ω-M-)a-=--0.|
(34)

For a nonzero mode vector,

|--------------------|
|det(K −  ω2M  ) = 0.|
---------------------
(35)

PIC

Figure 2. The quadratic kinetic and potential energies define M and K. The small-oscillation equations then reduce to a generalized eigenvalue problem whose eigenvalues are ω2 and whose eigenvectors are mode shapes.

8 Why the eigenvalues are frequencies squared

The generalized eigenvalue equation is

Ka  = λMa.
(36)

Comparison with the mode equation identifies

|--------|
-λ-=-ω2.-|
(37)

Thus a positive eigenvalue gives an oscillatory angular frequency

     √ --
ω  =   λ.
(38)

A zero eigenvalue gives a zero mode.

A negative eigenvalue corresponds to an imaginary

ω
(39)

and therefore exponential behavior rather than stable oscillation.

9 Rayleigh quotient

Multiply

Ka  =  ω2Ma
(40)

on the left by

aT .
(41)

Then

aTKa  = ω2aT Ma.
(42)

Therefore any exact mode satisfies

|-------------|
|     aT Ka   |
ω2 =  -T----. |
------a-Ma----
(43)

This ratio motivates the Rayleigh quotient

|----------T-----|
|R (y ) = y--Ky--.|
---------yT-My---|
(44)

The normal modes are stationary directions of this quotient.

For a positive-definite conservative system, the lowest mode minimizes the Rayleigh quotient.

10 Reality and positivity of stable frequencies

If

M
(45)

is positive definite and

K
(46)

is positive definite, then for every nonzero

a,
(47)

aT Ma  > 0,
(48)

and

aT Ka  > 0.
(49)

Hence

|--------|
|  2     |
-ω-->-0.-
(50)

The mass-weighting transformation below also shows that the eigenvalues are real.

11 Mass-weighted coordinates

Because

M
(51)

is positive definite, it has a positive square root

M1 ∕2.
(52)

Define

|-------1∕2---|
-y-=-M----η.-|
(53)

Then

       −1∕2
η =  M     y.
(54)

The equation of motion becomes

|------------|
y¨+  Dy  = 0,|
--------------
(55)

where

|--------------------|
|D =  M −1∕2KM  −1∕2.|
----------------------
(56)

Since

M
(57)

and

K
(58)

are symmetric,

|---------|
DT--=-D.---
(59)

Thus the generalized eigenvalue problem becomes an ordinary symmetric eigenvalue problem.

PIC

Figure 3. Mass weighting converts the generalized problem Ka = ω2Ma into an ordinary symmetric eigenvalue problem for D = M−1∕2KM−1∕2.

12 Orthogonality of distinct modes

Let

ar
(60)

and

as
(61)

satisfy

         2
Kar  = ω rMar,
(62)

and

Kas  = ω2sMas.
(63)

Multiply the first equation by

aT
  s
(64)

and the second by

aTr .
(65)

Using symmetry,

aT Kar =  aTKas.
 s         r
(66)

Therefore

(ω2 − ω2)aT Ma   = 0.
  r     s  r    s
(67)

For distinct eigenfrequencies,

|------------|
| T          |
-ar Mas-=--0.-
(68)

Distinct modes are orthogonal in the mass inner product.

13 Stiffness orthogonality

For distinct mass-orthogonal modes,

aTr Kas = ω2saTr Mas.
(69)

Therefore

|-----------|
aT Kas =  0 |
-r-----------
(70)

when

r ⁄= s
(71)

and the frequencies are distinct.

Thus normal modes diagonalize both quadratic forms.

14 Mass normalization

The scale of an eigenvector is arbitrary.

A particularly useful convention is

|------------|
|aTMa   =  1.|
--r----r------
(72)

Then

|--------------|
|aTMa    = δ  ,|
--r----s----rs-
(73)

and

|----------------|
|aT Ka  =  ω2δ  .|
--r----s----r-rs-
(74)

15 Modal matrix and simultaneous diagonalization

Collect the mass-normalized mode vectors into

|----⌊------------⌋---|
|      |         |    |
Φ  = ⌈ a1  ⋅⋅⋅  aN⌉ . |
|      |         |    |
-----------------------
(75)

Then

|-T----------|
Φ---M-Φ-=--1,-
(76)

and

|--T---------2-|
-Φ--K-Φ--=-Ω--,|
(77)

where

  2          2      2
Ω   = diag(ω1,...,ω N).
(78)

Set

|---------|
η-=--ΦQ.---
(79)

Then

|--------------|
-¨Q-+-Ω2Q---=-0.-
(80)

The coupled small-oscillation system becomes a collection of independent harmonic oscillators.

16 Quadratic energy in normal coordinates

The quadratic energy is

E (2) = 1˙ηT M η˙+  1ηT K η.
        2          2
(81)

Use

η =  ΦQ.
(82)

For mass-normalized modes,

|------1-∑---(-----------)--|
E (2) = --     Q˙2r + ω2rQ2r . |
|      2  r                 |
-----------------------------
(83)

Each normal coordinate carries the energy of an independent SHO.

17 Projecting arbitrary initial conditions

Suppose

η (0) = η0,
(84)

and

˙η(0) = v0.
(85)

Since

η0 =  ΦQ0,
(86)

multiply by

 T
Φ  M.
(87)

Using mass normalization,

|--------------|
Q0  = ΦT M  η0.|
----------------
(88)

Similarly,

|--------------|
Q˙0  = ΦT Mv0.  |
----------------
(89)

For every positive-frequency mode,

|--------------------------------|
|                     Q˙         |
Qr (t) = Qr0cos ωrt + --r0-sin ωrt.|
----------------------ωr----------
(90)

PIC

Figure 8. Initial displacement and velocity are projected onto the mass-normalized mode basis. Each modal initial state then evolves independently as a one-dimensional harmonic oscillator.

18 Degenerate normal frequencies

Two or more linearly independent modes can share the same frequency.

This is a degeneracy.

If

ω =  ω ,
 r    s
(91)

the earlier orthogonality proof does not force

aTMa   =  0.
 r    s
(92)

However, within the degenerate eigenspace, one can choose a mass-orthonormal basis.

Thus degeneracy means the mode basis is not unique.

Any suitable linear combination within the degenerate eigenspace is also a mode with the same frequency.

19 Zero modes

If

ωr = 0,
(93)

then

Kar  = 0.
(94)

The potential has no quadratic restoring curvature in that direction.

The modal equation becomes

|--------|
|¨Qr =  0.|
---------
(95)

Therefore

|--------------------|
|Qr (t) = Qr0 + Q˙r0t. |
---------------------
(96)

A zero mode is not an oscillator.

It represents a neutral direction, often associated with a continuous symmetry.

20 Unstable modes

Suppose an eigenvalue is negative:

ω2 = − σ2,     σ > 0.
(97)

Then the modal equation is

¨Q −  σ2Q =  0.
(98)

Its solution is

|----------------------|
|Q (t) = Ce σt + De −σt.
-----------------------
(99)

The growing exponential signals linear instability.

Thus the spectrum of the quadratic problem classifies local behavior:

  • ω2 > 0: stable oscillatory mode,
  • ω2 = 0: neutral zero mode,
  • ω2 < 0: unstable mode.

PIC

Figure 7. The sign of the generalized eigenvalue classifies the local motion: positive eigenvalues give oscillations, zero eigenvalues give neutral directions, and negative eigenvalues give exponentially growing and decaying solutions.

21 Example 1: equal-mass double pendulum

Consider a planar double pendulum with:

  • two equal point masses m,
  • two massless rods of equal length ℓ,
  • small angular displacements 𝜃1 and 𝜃2 from the downward vertical.

The exact kinetic energy is

T = 1
--
2mℓ2[2𝜃12 + 𝜃 22 (100)
+ 2𝜃1𝜃2 cos(𝜃1 − 𝜃2)]. (101)

For small oscillations,

cos(𝜃1 − 𝜃2) ≈ 1.
(102)

Thus

|--------------------------------|
|  (2)   1   2[   2           2]  |
|T   =  -m ℓ   2˙𝜃1 + 2 ˙𝜃1 ˙𝜃2 + 𝜃˙2 .
--------2------------------------
(103)

The gravitational potential is

U  = − mg ℓ(2 cos𝜃1 + cos𝜃2).
(104)

Using

           1
cos𝜃 ≈ 1 − --𝜃2,
           2
(105)

the quadratic part is

|------------------------|
|U(2) = 1mg ℓ (2𝜃2+  𝜃2),|
--------2--------1----2---
(106)

apart from an additive constant.

PIC

Figure 4. Small angles 𝜃1 and 𝜃2 describe the double pendulum near its downward equilibrium. The kinetic energy couples the angular velocities even though the quadratic gravitational stiffness matrix is diagonal in these coordinates.

22 Double-pendulum mass and stiffness matrices

Define

    [  ]
     𝜃1
𝜃 =  𝜃   .
      2
(107)

Then

|------------------|
|          [    ]  |
|M  = m ℓ2  2  1  ,|
------------1--1---
(108)

and

|------------------|
|          [2  0]  |
|K  = mg ℓ       . |
------------0--1---
(109)

This example is important because

M
(110)

is not diagonal.

Coupling can arise through kinetic energy as well as potential energy.

23 Double-pendulum eigenvalue equation

The mode condition is

          2
det(K −  ω M  ) = 0.
(111)

Define the dimensionless quantity

     ω2ℓ
μ =  ---.
      g
(112)

Then

    [[     ]    [     ]]
det   2  0  −  μ 2  1    = 0.
      0  1       1  1
(113)

Thus

(2 − 2μ)(1 − μ) − μ2 = 0.
(114)

Therefore

μ2 − 4μ + 2 = 0.
(115)

The roots are

|----------√---|
|μ  =  2 ±   2.|
--±------------
(116)

Hence

|------------------|
| 2   g-(    √ -)  |
ω ± = ℓ   2 ±  2  .|
--------------------
(117)

24 Double-pendulum mode shapes

For the lower frequency,

          √ --
μ− =  2 −   2.
(118)

The eigenvector equation gives

|-----[---]--|
|      √1--  |
|a− ∝    2  .|
--------------
(119)

The two angles have the same sign.

For the higher frequency,

          √--
μ+ =  2 +  2,
(120)

and

|-----[-----]--|
|        1     |
|a+ ∝   − √2- .|
----------------
(121)

The two angles have opposite signs.

PIC

Figure 5. The equal-mass double pendulum has a lower-frequency same-sign angular mode and a higher-frequency opposite-sign mode. The relative amplitude magnitude is √ --
  2 in these angular coordinates.

25 Example 2: numerical double-pendulum frequencies

Let

ℓ = 1.00m,     g =  9.81 m ∕s2.
(122)

Then

ω− = ∘ ---------√---
  9.81(2 −   2) (123)
≈ 2.40 rad∕s, (124)

and

ω+ = ∘ -------------
           √ --
  9.81(2 +   2) (125)
≈ 5.79 rad∕s. (126)

The corresponding periods are approximately

|------------|
|T− ≈  2.62 s,|
-------------
(127)

and

|------------|
|T+ ≈  1.09 s.|
-------------
(128)

26 Example 3: Rayleigh estimate of the lowest double-pendulum frequency

Choose the simple trial direction

     [1]
y =     .
      1
(129)

The Rayleigh quotient is

          T
R (y ) = y--Ky--.
        yT My
(130)

For the double pendulum,

yT Ky  = 3mg ℓ,
(131)

while

yT My  =  5m ℓ2.
(132)

Therefore

|------------|
|        3g  |
|R (y) = 5ℓ-.|
-------------
(133)

The trial estimate is

|--------------|
|        ∘ --- |
|ω    =    3g-.|
--trial-----5ℓ--
(134)

The exact lowest squared frequency is

ω2  = g-(2 − √2-) ≈ 0.586 g.
  −   ℓ                  ℓ
(135)

The trial value is

     g
0.600--,
     ℓ
(136)

which is already close.

This illustrates why the Rayleigh quotient is useful for estimating low modes of systems whose exact eigenproblem is difficult.

PIC

Figure 6. The Rayleigh quotient assigns an effective squared frequency to any trial direction. Its stationary directions are the normal modes, and its minimum gives the lowest mode for a positive-definite conservative system.

27 Coordinate changes and physical invariance

Let

η =  Sz
(137)

for an invertible constant matrix

S.
(138)

Then

M
(139)

and

K
(140)

transform by congruence:

|--------------|
|M ′ = ST MS,  |
---------------
(141)

and

|-′----T-----|
K---=-S--KS.--
(142)

The component form changes, but the physical Normal frequencies do not.

Thus normal frequencies are properties of the local mechanical system, not artifacts of a particular generalized-coordinate basis.

28 Degeneracy and symmetry

Degenerate frequencies frequently reflect symmetry.

For example, an isotropic two-dimensional oscillator has equal stiffness in two independent directions.

Its two mode frequencies are equal.

Within that two-dimensional eigenspace, many different orthogonal pairs can be chosen as mode shapes.

Symmetry therefore often explains both degeneracy and freedom in the choice of mode basis.

29 Beyond the quadratic approximation

The small-oscillation theory neglects cubic and higher terms such as

η η η .
  ij k
(143)

At larger amplitude these terms can produce:

  • amplitude-dependent frequencies,
  • coupling between normal modes,
  • internal Resonances,
  • nonlinear energy transfer,
  • chaotic behavior in sufficiently nonlinear systems.

Normal modes are exact for the quadratic linearized system and approximate for the original nonlinear system near equilibrium.

30 Why the theory is so general

The same generalized eigenvalue structure appears in:

  • molecular vibrations,
  • structural vibration analysis,
  • lattice dynamics,
  • acoustic cavities,
  • elastic bodies,
  • finite-element models,
  • electrical small-signal systems,
  • wave equations after spatial discretization.

The normal-mode eigenproblem is therefore one of the central mathematical structures linking classical mechanics to much broader physics.

31 Common mistakes

  1. Expanding about a configuration that is not an equilibrium.
  2. Keeping the linear term in the potential even though the expansion point is an equilibrium.
  3. Treating the mass matrix as automatically diagonal in generalized coordinates.
  4. Forgetting that coordinate dependence of M contributes only beyond quadratic order in the standard small-oscillation expansion.
  5. Solving det K = 0 instead of det(K − ω2M) = 0.
  6. Interpreting the generalized eigenvalue itself as ω rather than ω2.
  7. Using ordinary Euclidean orthogonality when the mass matrix is not proportional to the identity.
  8. Forgetting that eigenvectors have arbitrary overall scale.
  9. Forgetting to mass normalize before using Q = ΦT Mη.
  10. Applying the positive-frequency SHO formula to a zero mode.
  11. Calling a negative eigenvalue an oscillation frequency instead of recognizing an instability.
  12. Assuming degenerate mode vectors are unique.
  13. Confusing a diagonal stiffness matrix with a fully decoupled problem when the mass matrix is not diagonal.
  14. Assuming normal modes exchange energy in the exact conservative quadratic theory.
  15. Extending the quadratic approximation to amplitudes where higher-order terms are no longer negligible.

32 Practice exercises

  1. Starting from a Taylor expansion of U, derive the quadratic stiffness matrix.
  2. Explain why the first derivatives of the potential vanish at equilibrium.
  3. Show why coordinate dependence of the mass matrix can be neglected to quadratic order.
  4. Derive the small-oscillation equations from the quadratic Lagrangian.
  5. Derive the generalized normal-mode eigenvalue problem.
  6. Prove mass orthogonality of distinct modes.
  7. Show that distinct modes are also stiffness orthogonal.
  8. Derive the mass-weighted symmetric matrix D.
  9. Derive the modal projection formulas for initial displacement and velocity.
  10. Derive the modal energy as a sum of independent oscillator energies.
  11. Explain the difference among positive, zero, and negative eigenvalues.
  12. Derive the double-pendulum mass and stiffness matrices.
  13. Solve the double-pendulum characteristic equation and mode shapes.
  14. Reproduce the numerical double-pendulum frequencies for ℓ = 1 m.
  15. Use the Rayleigh quotient with a trial vector of your choice and compare the estimate with an exact normal frequency.

33 Summary

Near a stable equilibrium, a conservative mechanical system has the quadratic Lagrangian

|--------------------------|
| (2)   1 T       1  T     |
|L   =  2 ˙η M ˙η − 2-η  K η.|
----------------------------
(144)

The equations of motion are

|---------------|
M  ¨η + K η = 0. |
-----------------
(145)

Normal modes satisfy

|--------------|
-Ka--=-ω2Ma.---|
(146)

Distinct modes obey

|------------|
|aTr Mas =  0.|
--------------
(147)

With mass-normalized modes,

|------------|
ΦT  M Φ =  1,|
--------------
(148)

and

|--------------|
-ΦT-K-Φ--=-Ω2.-|
(149)

The Rayleigh quotient is

|----------------|
|        yT Ky   |
|R (y ) = -T-----.|
---------y--My---
(150)

Positive eigenvalues give stable oscillations, zero eigenvalues give neutral modes, and negative eigenvalues indicate linear instability.

The next step is to apply this finite-dimensional mode theory to extended discrete systems, where a chain of oscillators develops a dispersion relation and begins to behave like a wave-bearing medium.

References

References

[1]   PhysicsLibrary, M06-04, Coupled Oscillators and Normal Modes.

[2]   PhysicsLibrary, M06-03, Driven Harmonic Motion and Resonance.

[3]   PhysicsLibrary, M06-01, Simple Harmonic Motion.

[4]   A. P. French, Vibrations and Waves, W. W. Norton, 1971.

[5]   J. R. Taylor, Classical Mechanics, University Science Books, 2005.

[6]   H. Goldstein, C. Poole, and J. Safko, Classical Mechanics, 3rd ed., Addison-Wesley, 2002.


"Small Oscillations and the Normal-Mode Eigenvalue Problem" is owned by bloftin.
(view preamble)
View style:
Other names:  M06-05
Also defines:  small oscillation, quadratic Lagrangian, mass matrix, stiffness matrix, generalized normal-mode eigenvalue problem, mass-normalized mode, Rayleigh quotient, degenerate mode, unstable mode
Keywords:  small oscillations, normal modes, generalized coordinates, Lagrangian mechanics, mass matrix, stiffness matrix, Hessian, generalized eigenvalue problem, modal coordinates, mass orthogonality, Rayleigh quotient, zero modes, stability

Attachments:
GRE Physics Companion: Small Oscillations and the Normal-Mode Eigenvalue Problem (Example) by bloftin

Cross-references: dispersion relation, symmetric matrix, mass orthogonality, formula, identity, wave equations, vibrations, Resonances, two-dimensional, Normal, periods, magnitude, amplitude, angular velocities, kinetic energy, angular displacements, spectrum, force, normal coordinate, energy, vectors, inner product, mass, square, oscillation, angular frequency, generalized eigenvalue problem, degrees of freedom, velocities, zero modes, matrix, Taylor series, potential energy, vector, work, mechanics, M06-04, mode shapes, frequencies, normal modes, generalized coordinates, displacements, equilibrium, systems
There are 16 references to this object.

This is version 1 of Small Oscillations and the Normal-Mode Eigenvalue Problem, born on 2026-10-18.
Object id is 1496, canonical name is SmallOscillationsAndTheNormalModeEigenvalueProblem.
Accessed 18 times total.

Classification:
Physics Classification: 45.50.-j (Dynamics and kinematics of a particle and a system of particles)
 45.05.+x (General theory of classical mechanics of discrete systems)

Pending Errata and Addenda

None.

Discussion

No messages.

Interact