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Coupled Oscillators and Normal Modes

(Topic)

Coupled Oscillators and Normal Modes

A single ideal oscillator has one displacement coordinate and one natural frequency.

A system with several interacting degrees of freedom behaves differently.

Its coordinates are coupled, and the system generally possesses several special collective patterns of motion called normal modes.

In a normal mode:

  • every coordinate oscillates at the same angular frequency,
  • the ratios of the coordinate amplitudes remain fixed,
  • the entire many-coordinate system behaves like one effective harmonic oscillator.

The central linear equation is

|---------------|
M--¨q-+-Kq--=-0,-|
(1)

where

The normal-mode ansatz

q(t) = acos(ωt + ϕ )
(2)

gives

|-------2----------|
-(K-−--ω-M-)a-=--0.|
(3)

A nonzero mode shape

a
(4)

exists only when

|--------------------|
|          2         |
-det(K-−--ω-M--) =-0.
(5)

This is a generalized eigenvalue problem.

The eigenvalues are

ω2n,
(6)

and the eigenvectors are the corresponding mode shapes.

PIC

Figure 1. Two equal masses attached to outer springs and coupled by a middle spring form the standard two-degree-of-freedom oscillator. The coupling spring makes the motion of one mass influence the other.

1 Two identical masses with three springs

Consider two equal masses

m
(7)

with displacements

x1,     x2.
(8)

Each mass is attached to a wall by a spring of constant

k,
(9)

and the masses are joined by a coupling spring of constant

kc.
(10)

When mass 1 moves by

x1
(11)

and mass 2 by

x ,
 2
(12)

the extension of the coupling spring is

x −  x .
 2    1
(13)

The coupling force on mass 1 is

kc(x2 − x1).
(14)

The coupling force on mass 2 is

− kc(x2 − x1).
(15)

2 Equations of motion

For mass 1,

m ¨x1 = − kx1 + kc(x2 − x1).
(16)

Thus

|----------------------------|
m ¨x1 + (k + kc)x1 − kcx2 = 0.|
------------------------------
(17)

For mass 2,

m ¨x2 = − kx2 − kc(x2 − x1).
(18)

Thus

|----------------------------|
m ¨x2 + (k + kc)x2 − kcx1 = 0.|
------------------------------
(19)

Each equation contains both coordinates.

That is the defining feature of a coupled system.

3 Matrix form

Define

    [  ]
     x1
x =  x2  .
(20)

Then

|--------[----]--|
|M  = m   1  0  ,|
|         0  1   |
-----------------
(21)

and

|-----[--------------]-|
|      k + kc   − kc   |
K  =    − k    k + k  .|
-----------c--------c---
(22)

The equations become

|---------------|
M--¨x-+-Kx--=-0.-|
(23)

4 Normal-mode ansatz

Seek a motion in which both coordinates oscillate at one common frequency:

|--------------------|
x-(t) =-a-cos(ωt-+-ϕ),
(24)

where

    [   ]
a =   a1
      a2
(25)

is a constant vector.

Then

¨x = − ω2a cos(ωt + ϕ).
(26)

Substitution gives

(K − ω2M  ) a = 0.
(27)

For this system,

[                            ][   ]
 k + kc − mω2       − kc        a1
     − kc       k + kc − mω2    a2  = 0.
(28)

5 Characteristic equation

A nonzero vector

a
(29)

requires

   (      2   )
det K  − ω M   =  0.
(30)

Thus

(            2)2    2
 k + kc − mω    −  kc = 0.
(31)

Factor:

0 = [             2    ]
 k + kc − m ω  − kc (32)
×[                  ]
 k + kc − m ω2 + kc. (33)

Therefore

|--------|
| 2   k  |
ω 1 = --,|
------m---
(34)

and

|--------------|
|ω2 = k-+-2kc-.|
--2------m-----|
(35)

The coupled two-mass system has two natural frequencies.

6 First normal mode: in-phase motion

For

  2   k
ω 1 = --,
      m
(36)

the eigenvalue equation becomes

k a −  k a =  0.
 c 1    c 2
(37)

Therefore

|--------|
-a1 =-a2.-
(38)

A convenient normalized mode shape is

---------------
|         [ ]  |
|e1 = √1-- 1  .|
--------2--1---|
(39)

Both masses move together.

The coupling spring is not stretched because

x2 − x1 = 0.
(40)

Therefore the coupling stiffness does not contribute to this mode’s restoring force.

PIC

Figure 2. In the symmetric or in-phase normal mode, both masses move together. The coupling spring retains its equilibrium length, giving ω1 = ∘k-∕m--.

7 Second normal mode: out-of-phase motion

For

 2   k-+-2kc-
ω2 =    m    ,
(41)

the eigenvalue equation gives

a1 + a2 = 0.
(42)

Thus

|----------|
-a2-=-−-a1.|
(43)

A normalized mode shape is

|---------[---]--|
|     -1--  1    |
|e2 = √2-- − 1  .|
-----------------
(44)

The masses move oppositely.

The coupling spring changes length strongly, increasing the restoring stiffness.

Therefore

|--------|
|ω2 > ω1 |
---------
(45)

for

kc > 0.
(46)

PIC

Figure 3. In the antisymmetric or out-of-phase normal mode, the masses move oppositely. The coupling spring stretches and compresses strongly, increasing the mode frequency to ω2 = ∘ ------------
  (k + 2kc)∕m.

8 Normal coordinates

The original coordinates

x1,    x2
(47)

are physically intuitive but dynamically coupled.

Define instead

|--------------|
|Q  =  x1 +√-x2,|
| 1        2   |
----------------
(48)

and

|--------------|
|      x1 − x2 |
|Q2 =  --√----.|
-----------2----
(49)

These are the normal coordinates.

The inverse transformation is

|--------------|
|     Q1-+-Q2--|
x1 =    √ --  ,|
----------2-----
(50)

and

|--------------|
x  =  Q1√−-Q2-.|
| 2       2    |
----------------
(51)

The coordinate

Q1
(52)

measures the in-phase motion.

The coordinate

Q2
(53)

measures the out-of-phase motion.

PIC

Figure 4. The transformation from physical coordinates (x1,x2) to normal coordinates (Q1,Q2) rotates the coordinate description into the two mode directions.

9 Decoupling the equations

Add the two physical-coordinate equations:

m (¨x  + ¨x ) + k(x  + x ) = 0.
     1   2       1    2
(54)

Therefore

|----------------|
| ¨     2        |
-Q1-+-ω-1Q1-=-0,-
(55)

where

  2   k-
ω 1 = m .
(56)

Subtract the second physical equation from the first:

m (x¨1 − x¨2 ) + (k + 2kc)(x1 − x2) = 0.
(57)

Therefore

-----------------
| ¨     2        |
-Q2-+-ω-2Q2-=-0,-|
(58)

where

ω2 = k-+-2kc-.
 2      m
(59)

The coupled system has become two independent harmonic oscillators.

This is the central purpose of normal coordinates.

10 General solution in normal coordinates

Each normal coordinate has its own SHO solution:

|----------------------|
|Q  = A  cos(ω t + ϕ ),|
--1-----1-----1-----1---
(60)

and

|----------------------|
|Q2 = A2 cos(ω2t + ϕ2).|
------------------------
(61)

Transforming back,

|----------------------------------------|
|     A1-cos(ω1t +-ϕ1-) +-A2-cos(ω2t-+-ϕ2)
x1 =                 √ --               ,|
-----------------------2------------------
(62)

and

|-----------------------------------------|
|     A1 cos(ω1t + ϕ1 ) − A2 cos(ω2t + ϕ2)|
x2 =  ---------------√------------------. |
-----------------------2------------------
(63)

A general motion is therefore a superposition of normal modes.

PIC

Figure 5. A general coupled motion is a superposition of the system’s normal modes. Each normal coordinate evolves independently at its own Normal frequency.

11 Initial conditions determine modal amplitudes

Suppose

x1(0) = x10,    x2(0) = x20,
(64)

and

v1(0) = v10,    v2(0) = v20.
(65)

Then

|--------x---+-x---|
Q1 (0) = -10√---20,|
--------------2-----
(66)

|------------------|
|        x10 − x20 |
Q2 (0) = ---√-----,|
--------------2-----
(67)

and similarly

|------------------|
|˙Q (0) = v10√+-v20,|
| 1           2    |
--------------------
(68)

|------------------|
|˙       v10-−-v20 |
|Q2(0) =    √2--  .|
--------------------
(69)

Each modal initial state then determines its own amplitude and phase using the SHO results from M06-01.

12 Example 1: one mass displaced and released

Suppose

x1(0) = A,     x2(0) = 0,
(70)

and

v1(0) = v2(0) = 0.
(71)

Then

Q  (0) = √A-,
  1        2
(72)

and

Q  (0) = √A-.
  2        2
(73)

Because both modal velocities initially vanish,

         A
Q1 (t) = √---cosω1t,
          2
(74)

and

Q  (t) = √A--cosω  t.
  2       2      2
(75)

Therefore

|-----------------------------|
|       A                     |
x1(t) = -- (cos ω1t + cosω2t), |
--------2----------------------
(76)

and

|-----------------------------|
|       A-                    |
x2(t) = 2 (cos ω1t − cosω2t). |
-------------------------------
(77)

13 Beats and energy exchange

Use

                   (      )     (      )
                     a-+-b        a −-b
cosa + cosb = 2 cos    2    cos     2    .
(78)

Then

|-------------(---------)-----(---------)--|
|               ω1 + ω2        ω2 −  ω1    |
|x1(t) = A cos   -------t  cos  --------t  .|
-------------------2---------------2--------
(79)

Similarly,

|-------------(---------)----(----------)--|
|               ω1-+-ω2-       ω2-−-ω1-    |
|x2(t) = A sin      2   t  sin     2    t  ,|
-------------------------------------------
(80)

up to an overall sign convention.

When the coupling is weak,

ω2 − ω1
(81)

is small.

The fast oscillation is modulated by a slow envelope.

Motion appears to transfer back and forth between the two masses.

This is the coupled-oscillator origin of beats and energy exchange.

PIC

Figure 6. When the two normal frequencies are close, a displacement initially concentrated on one mass evolves into a slow exchange of oscillation amplitude between the two masses.

14 Weak-coupling frequency splitting

If

kc ≪ k,
(82)

then

     ∘ -k-
ω1 =   -- =  ω0,
       m
(83)

and

       ∘  --------
              2kc-
ω2 = ω0   1 +  k .
(84)

Using

√ -----       𝜖
  1 + 𝜖 ≈ 1 + -,
              2
(85)

|-------(-------)---|
|             kc    |
ω2 ≈  ω0  1 + k   . |
--------------------
(86)

Thus

|----------------|
|             kc |
|ω2 − ω1 ≈ ω0 k .|
------------------
(87)

The normal-frequency splitting is therefore a direct measure of weak coupling strength.

15 Beat time scale

The slow envelope contains

    ( Δ ω )
cos   ---t  ,
       2
(88)

where

Δ ω =  ω2 − ω1.
(89)

The envelope magnitude repeats after

|--------------|
|        -2π-- |
|Tbeat = |Δ ω |.|
---------------
(90)

Complete transfer from one mass to the other occurs on the half-beat time scale

|----------------|
|          --π-- |
|Ttransfer = |Δω |.|
-----------------
(91)

16 Energy of the two-mass system

The kinetic energy is

     1-  2   1-   2
T =  2m ˙x1 + 2 m ˙x2.
(92)

The potential energy is

    1       1       1
U = --kx21 + --kx22 + -kc(x2 − x1)2.
    2       2       2
(93)

Using the normal-coordinate transformation,

x  = Q1-+√--Q2-,    x  = Q1-√−-Q2-,
 1        2         2        2
(94)

one obtains

|--------------------|
|    1-  ˙2   1-  ˙2 |
T  = 2 m Q1 + 2 m Q2,|
----------------------
(95)

and

|---------------------------|
U  = 1-kQ21 + 1(k + 2kc)Q22. |
-----2--------2--------------
(96)

Therefore

|--------------|
|E =  E1 + E2, |
---------------
(97)

where

E  =  1m Q˙2 +  1m ω2Q2 ,
 1    2    1   2   1  1
(98)

and

      1-  ˙2   1-  2  2
E2 =  2m Q 2 + 2m ω2Q 2.
(99)

The normal coordinates diagonalize the energy as well as the equations of motion.

17 Why normal modes do not exchange energy

In the ideal linear conservative system,

Q1
(100)

and

Q2
(101)

satisfy independent SHO equations.

Thus

E1
(102)

and

E2
(103)

are separately conserved.

The apparent energy exchange between the physical masses is not energy transfer between normal modes.

It is interference between fixed-energy modal contributions when transformed back into the physical coordinates.

This distinction is important.

18 Orthogonality of the equal-mass mode shapes

The two mode shapes are

         [ ]
     -1-- 1
e1 = √ -- 1  ,
       2
(104)

and

         [   ]
     -1--  1
e2 = √ 2  − 1  .
(105)

Their ordinary dot product is

-----------
| T        |
-e1e2-=--0.|
(106)

For equal masses,

M  = m1,
(107)

so ordinary orthogonality and mass orthogonality are equivalent.

19 General mass orthogonality

For a general conservative system,

Kar  = ω2 Mar,
         r
(108)

and

Kas  = ω2sMas.
(109)

Assume

  T
M   =  M
(110)

and

  T
K   = K.
(111)

Multiply the first equation on the left by

aTs ,
(112)

and the second on the left by

aTr .
(113)

Symmetry gives

ω2raTs Mar =  ω2saTr Mas.
(114)

Since

aTMar  =  aTMas,
 s         r
(115)

(ω2r − ω2s)aTr Mas = 0.
(116)

For distinct frequencies,

|------------|
|aTMas  =  0.|
--r-----------
(117)

Distinct normal modes are orthogonal with respect to the mass matrix.

20 Mass normalization

A mode shape can be multiplied by any nonzero constant without changing the physical mode.

A convenient convention is

|------------|
|aTr Mar =  1.|
--------------
(118)

Then the mode shapes are mass normalized.

For mass-normalized modes,

|--------------|
|aTMa    = δ  .|
--r----s----rs-
(119)

This normalization is especially useful for many-degree-of-freedom systems.

21 General modal transformation

Collect the normal-mode vectors into the modal matrix

|----⌊------------⌋---|
|      |         |    |
Φ  = ⌈ a   ⋅⋅⋅  a ⌉ . |
|       1        N    |
-------|---------|-----
(120)

Write

|---------|
q =  ΦQ.  |
-----------
(121)

For mass-normalized modes,

|------------|
ΦT--M-Φ-=--1,-
(122)

and

|----------⌊------------⌋--|
|           ω21       0    |
|ΦT K Φ =  |     ..     | .|
|          ⌈      .    2⌉  |
|            0       ωN    |
----------------------------
(123)

Therefore

|--------------|
|¨Q + Ω2Q   = 0.|
----------------
(124)

The coupled physical-coordinate equations become independent modal equations.

22 Generalized eigenvalue problem

For

N
(125)

degrees of freedom,

M  ¨q + Kq  = 0.
(126)

Use

q = aeiωt.
(127)

Then

|--------------|
-Ka--=-ω2Ma.---|
(128)

The normal frequencies satisfy

|--------------------|
|det(K  − ω2M  ) = 0.|
----------------------
(129)

This is the standard small-oscillation eigenvalue problem.

23 Mass-weighted coordinates

When

M
(130)

is positive definite, define

y = M1  ∕2q.
(131)

Then

q =  M −1∕2y.
(132)

The equations become

|------------|
y¨+  Dy  = 0,|
--------------
(133)

where

|--------------------|
-D-=--M-−1∕2KM--−1∕2.-
(134)

If

K
(135)

is symmetric, then

D
(136)

is symmetric.

The generalized eigenvalue problem has therefore been converted into an ordinary symmetric eigenvalue problem.

24 Zero modes

Not every normal frequency must be positive.

Consider two equal masses connected only to each other by a spring, with no springs to fixed walls.

The equations are

mẍ1 + kc(x1 − x2) = 0, (137)
mẍ2 + kc(x2 − x1) = 0. (138)

The in-phase mode

x1 = x2
(139)

does not stretch the spring.

Thus there is no restoring force.

Its frequency is

|ω--=-0.-|
--1------|
(140)

This is a zero mode.

It represents uniform translation of the entire system.

The out-of-phase mode has

|-----∘------|
|        2k  |
|ω2 =    --c.|
---------m----
(141)

PIC

Figure 7. Two masses connected only to each other possess a zero-frequency in-phase mode corresponding to rigid translation and a finite-frequency internal stretching mode.

25 Zero modes and continuous symmetry

The zero mode exists because translating both masses by the same amount does not change the spring extension or potential energy.

This reflects translational symmetry.

More generally, zero modes often arise when a continuous change of configuration costs no potential energy.

Examples include:

  • uniform translation of an isolated system,
  • rigid rotation of a rotationally invariant configuration,
  • symmetry motions in more complicated mechanical systems.

26 Unequal masses

Suppose

m1 ⁄=  m2.
(142)

Then

      [        ]
       m1    0
M  =    0   m2  .
(143)

The normal-mode equation remains

|-------2----------|
-(K-−--ω-M-)a-=--0.|
(144)

The ordinary Euclidean eigenvectors need not be orthogonal.

Instead, distinct modes satisfy

--------------
| T          |
-ar Mas-=--0.-
(145)

Thus the mass matrix defines the physically natural inner product.

27 Example 2: numerical equal-mass system

Let

m  = 2.0 kg,    k =  18N ∕m,     kc = 8.0 N∕m.
(146)

Then

      ∘ ---
        18-   |----------|
ω1 =     2 =  3.00-rad-∕s.-
(147)

The second frequency is

ω2 = ∘  --------
   18 + 16
   --------
      2 (148)
= √ ---
  17 (149)
≈ 4.12 rad∕s. (150)

Thus

Δ ω ≈  1.12 rad∕s.
(151)

The corresponding beat time is

Tbeat = 2π
----
Δ ω (152)
≈ 5.59 s. (153)

28 Small oscillations of a general mechanical system

Let generalized coordinates be

q1,...,qN
(154)

and let

q = 0
(155)

represent stable equilibrium.

Near equilibrium, the potential energy is approximately quadratic:

|------------------|
|          1-T     |
|U ≈ U0 +  2q  Kq. |
--------------------
(156)

The matrix

K
(157)

is the Hessian of the potential evaluated at equilibrium:

|------------|---|
|       ∂2U  |   |
Kij =  ------||  .|
-------∂qi∂qj-eq--
(158)

For many systems the kinetic energy near equilibrium is

|--------------|
|     1- T     |
|T =  2q˙ M q˙. |
---------------
(159)

The linearized equations are then

|---------------|
M--¨q-+-Kq--=-0.-|
(160)

Thus normal modes are the universal linear description of small oscillations about stable equilibrium.

29 From finite oscillators to waves

Consider a chain of many identical masses coupled by identical springs.

With

N
(161)

masses, there are

N
(162)

normal modes.

As

N
(163)

becomes large:

  • the allowed mode frequencies become dense,
  • the mode shapes resemble sampled sine or exponential patterns,
  • the system approaches a continuous elastic medium.

The normal modes of a continuous string are Standing Waves.

Thus the mathematics of coupled oscillators is the direct bridge from particle mechanics to wave physics.

PIC

Figure 8. Increasing the number of coupled oscillators produces a family of collective modes. In the continuum limit, these discrete mode shapes become the normal modes of wave systems.

30 Driven coupled systems

If an external force vector is applied,

M q¨+  Kq  = f(t).
(164)

In modal coordinates,

¨Q  + Ω2Q  = ΦT  f(t)
(165)

for mass-normalized modes.

Thus each normal coordinate behaves like a driven oscillator.

A drive couples strongly to a mode when:

  • its frequency is near that mode’s natural frequency,
  • its spatial force pattern has a nonzero projection onto that mode shape.

This introduces the idea of modal participation.

31 Damping in coupled systems

With damping,

M  ¨q + C ˙q + Kq  = 0.
(166)

The undamped normal modes still completely decouple the system only for special damping structures, such as proportional damping.

General damping can couple modal coordinates again.

Thus the clean normal-mode theory developed here is fundamentally the conservative linear starting point.

32 Normal modes versus arbitrary motion

A normal mode is not just any oscillatory motion.

A normal mode has:

  • one definite angular frequency,
  • a fixed coordinate-amplitude ratio,
  • a fixed relative phase pattern.

An arbitrary motion generally contains several normal modes simultaneously.

Its physical-coordinate motion can therefore look complicated even though every normal coordinate follows a simple harmonic equation.

33 Common mistakes

  1. Assuming coupled coordinates can be solved independently before finding normal coordinates.
  2. Confusing physical coordinates with normal coordinates.
  3. Treating the determinant condition alone as sufficient without finding the corresponding eigenvectors.
  4. Forgetting that a mode shape is defined only up to an overall scale.
  5. Assuming the higher-frequency mode must always be in phase.
  6. Forgetting that the coupling spring is unstretched in the symmetric equal-mass mode.
  7. Forgetting that the coupling spring contributes twice to the effective stiffness of the antisymmetric equal-mass mode.
  8. Confusing energy exchange between physical masses with energy exchange between ideal normal modes.
  9. Using ordinary dot-product orthogonality for unequal masses instead of mass orthogonality.
  10. Assuming every mode frequency must be nonzero.
  11. Missing zero modes associated with continuous symmetries.
  12. Forgetting that a general motion is a superposition of all excited normal modes.
  13. Using the normal-mode eigenvectors without transforming initial velocities as well as initial displacements.
  14. Assuming damping always preserves exact modal decoupling.
  15. Missing the connection between a long oscillator chain and wave normal modes.

34 Practice exercises

  1. Derive the coupled equations for the equal-mass three-spring system.
  2. Construct the mass and stiffness matrices.
  3. Derive the characteristic equation and both normal frequencies.
  4. Find and normalize the two mode shapes.
  5. Explain physically why the symmetric mode frequency is independent of kc.
  6. Explain physically why the antisymmetric mode has the larger frequency.
  7. Derive the normal-coordinate transformation and its inverse.
  8. Show directly that the equations decouple in Q1 and Q2.
  9. Show that kinetic and potential energy also diagonalize in the normal coordinates.
  10. For one mass initially displaced and the other at rest, derive x1(t) and x2(t).
  11. Derive the beat time and complete-transfer time from ω2 − ω1.
  12. Derive the weak-coupling frequency splitting.
  13. Prove the mass-orthogonality relation for two distinct modes.
  14. Derive the zero and nonzero frequencies of two free masses connected only by one spring.
  15. Show how mass-weighted coordinates convert the generalized eigenvalue problem into an ordinary symmetric eigenvalue problem.

35 Summary

Coupled linear oscillators satisfy

|---------------|
M--¨q-+-Kq--=-0.-|
(167)

A normal mode has the form

|--------------------|
q-(t) =-a-cos(ωt-+-ϕ).
(168)

Therefore

|--------2-----|
-Ka--=-ω--Ma.--|
(169)

The allowed frequencies satisfy

|--------------------|
-det(K-−--ω2M--) =-0.|
(170)

For two equal masses attached to outer springs

k
(171)

and coupled by

kc,
(172)

|-----∘----|
|       k  |
ω1 =    --,|
--------m---
(173)

with mode shape

[--]-|
| 1  |
| 1 ,|
------
(174)

and

|-----∘----------|
|        k +-2kc-|
|ω2 =      m    ,|
------------------
(175)

with mode shape

[----]-|
|  1  .|
| − 1  |
--------
(176)

Normal coordinates decouple both the equations of motion and the quadratic energy.

For distinct modes of a general conservative system,

|-T----------|
-ar Mas-=--0.-
(177)

The same eigenvalue structure extends from a few coupled masses to the normal modes of continuous wave systems.

References

References

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

[2]   PhysicsLibrary, M06-02, Damped Harmonic Motion.

[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.


"Coupled Oscillators and Normal Modes" is owned by bloftin.
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Other names:  M06-04
Also defines:  coupled oscillator, normal mode, normal coordinate, mode shape, generalized eigenvalue problem, zero mode, mass orthogonality
Keywords:  coupled oscillators, normal modes, normal coordinates, mode shapes, eigenvalues, eigenvectors, mass matrix, stiffness matrix, generalized eigenvalue problem, beats, energy exchange, small oscillations

Attachments:
GRE Physics Companion: Coupled Oscillators and Normal Modes (Example) by bloftin

Cross-references: relation, determinant, external force, mechanics, particle, coupled oscillators, Standing Waves, generalized coordinates, inner product, isolated system, vectors, dot product, potential energy, kinetic energy, magnitude, energy, velocities, M06-01, phase, Normal, equilibrium, frequency, force, matrix, mass, vector, amplitudes, angular frequency, degrees of freedom, system, natural frequency, displacement, oscillator
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This is version 1 of Coupled Oscillators and Normal Modes, born on 2026-10-18.
Object id is 1494, canonical name is CoupledOscillatorsAndNormalModes.
Accessed 5 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)

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