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Damped Harmonic Motion

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Damped Harmonic Motion

The ideal simple harmonic oscillator of M06-01 obeys

     2
¨x + ω0x = 0
(1)

and oscillates forever.

Real oscillators lose mechanical energy.

A standard model adds a resistive force proportional to velocity:

|----------|
-Fd-=-−-b˙x,-
(2)

where

b > 0
(3)

is the viscous damping coefficient.

For a mass attached to a spring,

mx¨=  − kx − b˙x.
(4)

Therefore

|------------------|
m-x¨+-bx˙+-kx--=-0.-
(5)

Dividing by

m
(6)

gives

|-------------------|
¨x + 2γ ˙x + ω20x = 0, |
---------------------
(7)

where

|--------|
|    -b--|
γ =  2m ,|
----------
(8)

and

|-----∘----|
|       k  |
ω0 =    --.|
--------m---
(9)

The parameter

γ
(10)

sets the damping time scale.

The relation between

γ
(11)

and

ω0
(12)

determines whether the system oscillates while decaying or returns to equilibrium without oscillation.

PIC

Figure 1. A damped spring-mass oscillator combines the restoring force −kx with a velocity-dependent damping force −bẋ.

1 Viscous damping force

The damping force

Fd = − bv
(13)

opposes the instantaneous velocity.

Its dimensions are

      -force--      − 1
[b] = velocity = M  T  .
(14)

Therefore

[   ]
  b-      −1
  m   = T   .
(15)

The damping rate

     b
γ = ----
    2m
(16)

has the same dimensions as an angular frequency.

This allows a direct comparison between damping and the undamped natural frequency

ω0.
(17)

2 State-space form

As in M06-01, define

v = ˙x.
(18)

Then

ẋ = v, (19)
v = −2γv − ω02x. (20)

The complete state is still

|------|
|(x,v).|
--------
(21)

In matrix form,

|---[-]---[-----------]-[-]--|
| d  x       0     1     x   |
|--  v  =   − ω2  − 2γ   v  .|
-dt------------0-------------
(22)

The damping regimes are therefore also eigenvalue regimes of this state matrix.

3 Characteristic equation

Seek a solution of the form

        rt
x(t) = e .
(23)

Then

x˙=  rert,     ¨x = r2ert.
(24)

Substitution into

x¨+  2γ ˙x + ω2x = 0
            0
(25)

gives

r2 + 2γr + ω20 = 0.
(26)

Thus

|----------------------|
|           ∘ --2----2 |
|r± = − γ ±   γ  − ω 0.|
-----------------------
(27)

Everything depends on the sign of

 2    2
γ −  ω0.
(28)

PIC

Figure 2. The damping regime is determined by the characteristic roots. Complex conjugate roots produce decaying oscillation, a repeated real root gives critical damping, and two distinct negative real roots give overdamping.

4 Damping ratio

Define the dimensionless damping ratio

|------------------|
|     γ       b    |
|ζ = --- = -√-----.|
-----ω0----2--mk---
(29)

Then:

  • underdamped:
    0 ≤ ζ < 1,
    (30)

  • critically damped:
    ζ = 1,
    (31)

  • overdamped:
    ζ > 1.
    (32)

The undamped limit is

ζ = 0.
(33)

5 Underdamped motion

If

γ < ω0,
(34)

the roots are complex:

r± =  − γ ± iωd,
(35)

where

|-----∘----------|
|         2    2 |
-ωd-=---ω-0 −-γ-.|
(36)

The real solution is

|--------−γt-----------------------|
-x(t) =-e---[C-cosωdt-+-D-sin-ωdt].|
(37)

Equivalently,

|----------−γt-------------|
-x(t) =-A0e---cos(ωdt-−-δ).-
(38)

The oscillation survives, but its envelope decays exponentially.

6 Damped period

The underdamped period is

|---------|
|     2π  |
Td =  --. |
------ωd---
(39)

Since

ωd <  ω0,
(40)

the damped period is slightly longer than the undamped period:

|--------|
-Td >-T0.-
(41)

For weak damping,

γ ≪  ω0,
(42)

the difference is small.

7 Underdamped initial-value solution

Let

x (0) = x0,    v(0) = v0.
(43)

Begin with

x(t) = e−γt[C cosω  t + D sin ω t].
                   d          d
(44)

At

t = 0,
(45)

C =  x0.
(46)

Differentiate:

v(t) = e−γt[− γ(C cos ωdt + D sin ωdt) (47)
− Cωd sin ωdt + Dωd cos ωdt]. (48)

At

t = 0,
(49)

v0 = − γC + ωdD.
(50)

Thus

|--------------|
|     v0 +-γx0-|
|D =     ω    .|
----------d-----
(51)

Therefore

|-----------[---------------------------]--|
|x(t) = e−γt x0 cosωdt + v0-+-γx0-sin ωdt  .|
----------------------------ωd-------------|
(52)

8 Underdamped amplitude envelope

In amplitude-phase form,

x(t) = A0e −γtcos(ωdt − δ).
(53)

The positive and negative envelopes are

|------------------|
|xenv,± = ±A0e  −γt.|
-------------------
(54)

The amplitude envelope therefore has time constant

|--------|
|     1  |
|τA = --.|
------γ--
(55)

After one amplitude time constant,

         A0
A (τA) = ---.
          e
(56)

PIC

Figure 4. An underdamped oscillator remains sinusoidal while its amplitude is bounded by the exponential envelopes ±A0e−γt.

9 Example 1: underdamped oscillator

Let

m = 1.0 kg, (57)
k = 100 N∕m, (58)
b = 4.0 kg∕s. (59)

Then

     ∘  ----
ω0 =    100-=  10rad∕s,
         1
(60)

and

γ = 4-= 2 s−1.
    2
(61)

Because

γ < ω0,
(62)

the motion is underdamped.

The damped angular frequency is

ωd = √ --2----2
  10  − 2 (63)
= √ ---
  96 (64)
≈ 9.80 rad∕s. (65)

The amplitude time constant is

|------------|
-τA-=-0.50-s.|
(66)

10 Critical damping

If

γ = ω0,
(67)

the characteristic equation has one repeated root:

|---------|
r = − ω0. |
-----------
(68)

A repeated root requires a second independent solution proportional to

te− ω0t.
(69)

Therefore

|----------------------|
|x(t) = (C + Dt )e−ω0t.|
-----------------------
(70)

Critical damping separates oscillatory decay from nonoscillatory decay.

It is also the fastest possible return to equilibrium within this linear model without oscillation.

11 Critical-damping initial-value solution

At

t = 0,
(71)

C =  x0.
(72)

Differentiate:

v(t) = [D − ω0(C + Dt )]e−ω0t.
(73)

At

t = 0,
(74)

v0 = D  − ω0C.
(75)

Thus

|---------------|
D--=-v0-+-ω0x0.--
(76)

Hence

|--------------------------−ω0t|
x-(t)-=-[x0 +-(v0 +-ω0x0-)t]e---.-
(77)

12 Overdamped motion

If

γ > ω0,
(78)

define

|----∘----------|
α =    γ2 − ω20. |
-----------------
(79)

The two roots are real and negative:

|--------------|
-r±-=-−-γ-±-α.-|
(80)

Therefore

|----------------------|
|x(t) = Cer+t + Der −t.|
-----------------------
(81)

There is no sinusoidal oscillation.

One exponential mode decays slowly and the other rapidly.

At sufficiently long times, the slower mode dominates.

13 Overdamped initial-value solution

At

t = 0,
(82)

x0 =  C + D.
(83)

Also,

v0 = r+C  + r− D.
(84)

Solving,

|----------------|
|C =  v0-−-r− x0-,|
|      r+ − r−   |
-----------------
(85)

and

|----------------|
|     r+x0-−-v0- |
|D  =  r+ − r−  .|
-----------------
(86)

These coefficients completely determine the overdamped trajectory.

14 The three regimes compared

The three forms are:

  • underdamped:
          −γt
x =  e   (C  cosωdt + D sinωdt),
    (87)

  • critically damped:
                   −ω0t
x =  (C  + Dt )e   ,
    (88)

  • overdamped:
           r+t      r− t
x =  Ce   +  De   .
    (89)

All three return toward equilibrium when

b > 0.
(90)

Only the underdamped case crosses equilibrium repeatedly.

PIC

Figure 3. Underdamped motion oscillates through equilibrium, critical damping gives the fastest nonoscillatory return, and overdamped motion returns more slowly as a sum of two decaying exponentials.

15 Why critical damping is special

For a specified

m
(91)

and

k,
(92)

critical damping occurs when

|------√-----|
|b =  2  mk. |
--c----------
(93)

If

b < bc,
(94)

the response oscillates.

If

b > bc,
(95)

the response is nonoscillatory but contains a slow exponential mode.

At

b = b ,
     c
(96)

the system sits at the boundary and returns without oscillation as rapidly as possible in the linear viscous model.

PIC

Figure 7. Critical damping lies at the boundary between oscillatory and overdamped behavior and gives the fastest nonoscillatory return to equilibrium for the standard linear model.

16 Phase-plane behavior

The undamped SHO of M06-01 follows closed ellipses in the

x − v
(97)

phase plane.

Damping changes the geometry.

For the underdamped oscillator, the trajectory spirals toward

(x,v) = (0,0).
(98)

For critical and overdamped motion, trajectories approach the origin without repeated revolutions.

Thus damping turns the equilibrium from a conservative center into an attracting equilibrium.

PIC

Figure 5. Damping causes phase-space trajectories to approach the equilibrium state (0,0). Underdamped trajectories spiral inward, while critical and overdamped trajectories approach without sustained rotation.

17 State-space eigenvalues and damping regimes

The state matrix is

     [           ]
A  =    0 2   1   .
      − ω 0  − 2γ
(99)

Its eigenvalues satisfy

det(A  − r1) = 0.
(100)

This gives

r2 + 2γr + ω20 = 0,
(101)

the same characteristic equation obtained from the scalar differential equation.

Therefore:

  • complex conjugate eigenvalues correspond to a stable spiral,
  • a repeated real eigenvalue corresponds to critical damping,
  • two negative real eigenvalues correspond to a stable node.

This is the dynamical-systems interpretation of the three damping regimes.

18 Mechanical energy decreases

Define the usual mechanical energy

|-------------------|
|    1-   2   1- 2  |
E  = 2 mv  +  2kx . |
---------------------
(102)

Differentiate:

dE
---
dt = mvv + kxẋ (103)
= v(mẍ + kx). (104)

From the damped equation of motion,

m ¨x + kx = − bv.
(105)

Therefore

|------------|
|dE-       2 |
|dt  = − bv .|
--------------
(106)

Since

b > 0,
(107)

|--------|
|dE- ≤ 0.|
-dt-------
(108)

Mechanical energy decreases monotonically.

The lost mechanical energy is transferred to thermal or other internal degrees of freedom represented by the damping model.

PIC

Figure 6. The viscous damping force removes mechanical energy at the instantaneous rate dE∕dt = −bv2. Energy decreases most rapidly when speed is large.

19 Energy decay for light damping

For weak underdamping,

          −γt
x(t) ≈ A0e   cos(ωdt − δ).
(109)

Because oscillator energy scales approximately with amplitude squared,

      2
E ∝ A  ,
(110)

the slowly varying energy envelope obeys

|----------------|
|           −2γt |
-Eenv(t) ∝-e----.
(111)

Thus the characteristic energy-decay time is

|---------|
|     1-- |
τE =  2γ. |
-----------
(112)

The exact instantaneous mechanical energy contains oscillatory modulation.

The simple exponential statement refers to the slow envelope or cycle-averaged energy in the lightly damped regime.

20 Logarithmic decrement

For an underdamped oscillator, consider two same-sign displacement maxima separated by one damped period.

Their envelope amplitudes satisfy

An+1  = Ane −γTd.
(113)

Therefore

  (  An  )
ln   -----  = γTd.
    An+1
(114)

Define the logarithmic decrement:

|------(------)--|
|        -An--   |
Λ  = ln  A      .|
-----------n+1----
(115)

Then

|----------------|
|           2πγ- |
|Λ = γTd =   ω  .|
--------------d---
(116)

More generally, using peaks

N
(117)

periods apart,

|---------(-------)--|
|     1       An     |
|Λ = ---ln  ------  .|
-----N------An+N------
(118)

PIC

Figure 8. The logarithmic decrement measures damping from the ratio of successive same-sign peak amplitudes in an underdamped free decay.

21 Example 2: damping from successive peaks

Suppose successive positive maxima are

An = 8.0 cm,     An+1 =  6.0cm,
(119)

and the damped period is

Td = 0.50 s.
(120)

Then

Λ = ln (  )
  8-
  6 (121)
≈ 0.2877. (122)

Thus

γ = -Λ-
Td (123)
≈0.2877
-------
 0.50 (124)
≈ 0.575 s−1. (125)

22 Quality factor for weak damping

A common dimensionless measure of weak damping is the quality factor

|--------|
Q  ≈  ω0.|
|     2γ |
----------
(126)

Using

      b
γ =  ---,
     2m
(127)

|------------√------|
|    m-ω0-   --mk-- |
Q  ≈   b  =    b  . |
---------------------
(128)

For light damping:

  • large Q means slow decay,
  • small Q means faster decay.

A more precise energy-based definition of

Q
(129)

is especially useful for driven oscillators and Resonance and will be developed further in M06-03.

23 Relation between damping ratio and quality factor

For weak damping,

ζ = -γ-.
    ω0
(130)

Therefore

|--------|
|     1  |
|Q ≈  --.|
------2ζ--
(131)

This approximation is excellent when

ζ ≪  1.
(132)

24 Example 3: classify from damping ratio

Let

m = 2.0 kg,     k = 50N ∕m.
(133)

Then

      ∘ ---
        50-
ω0 =     2 =  5.0 rad∕s.
(134)

The critical damping coefficient is

bc = 2√mk--- (135)
= 2∘ -------
  (2)(50) (136)
= 20 kg∕s. (137)

Therefore:

  • b = 8 kg/s is underdamped,
  • b = 20 kg/s is critically damped,
  • b = 30 kg/s is overdamped.

25 What damping does to frequency

For underdamped motion,

     ∘ --------
ωd =   ω2 −  γ2.
         0
(138)

Thus damping reduces the oscillation frequency.

For weak damping, use

√ -----
  1 − 𝜖 ≈ 1 − 𝜖.
              2
(139)

Then

ωd = ω0∘ -------
      γ2-
  1 − ω2
        0 (140)
≈ ω0[         ]
     1-γ2-
 1 − 2 ω2
        0. (141)

Therefore the frequency shift is second order in weak damping.

This is why lightly damped oscillators often oscillate at nearly the undamped natural frequency even while their amplitude visibly decays.

26 Weak damping versus strong damping

A useful distinction is:

  • weak damping:
    γ ≪  ω0,
    (142)

    many oscillations occur before substantial decay;

  • moderate underdamping:
    γ <  ω
      0
    (143)

    but not much smaller;

  • critical or overdamping: no sustained oscillation occurs.

The word damped does not imply the motion is necessarily oscillatory.

27 The equilibrium state is an attractor

For

b > 0,
(144)

the origin

(x,v) = (0,0)
(145)

is an attracting equilibrium.

Mechanical energy satisfies

dE        2
--- = − bv ≤  0.
dt
(146)

Therefore trajectories lose energy until they approach equilibrium.

This contrasts with the ideal SHO, whose phase-plane trajectories remain on closed constant-energy ellipses.

28 Damping as a nonconservative process

The damping force is

Fd = − bv.
(147)

Its instantaneous power is

|----------------2-|
-Pd-=-Fdv-=--−-bv-.|
(148)

This is always nonpositive.

The damping force therefore removes mechanical energy regardless of the sign of

v.
(149)

The force reverses direction when the velocity reverses, but its power remains negative.

29 Example 4: instantaneous energy loss rate

Let

b = 3.0 kg∕s,
(150)

and suppose at one instant

v = − 2.0 m ∕s.
(151)

Then

dE-
dt = −bv2 (152)
= −(3.0)(4.0) (153)
= −12 W. (154)

The negative sign means mechanical energy is being removed.

30 Critical damping and engineering design

Critical damping is important when oscillation is undesirable but rapid settling is needed.

Examples include idealized models of:

  • door closers,
  • measuring instruments,
  • suspension components,
  • actuator position control,
  • vibration isolation systems.

Real devices can involve dry friction, nonlinear damping, hydraulic effects, or active control.

The linear viscous model remains the fundamental first approximation.

31 Beyond viscous damping

Not all damping forces are proportional to velocity.

Examples include:

  • Coulomb or dry friction,
  • quadratic fluid drag,
  • structural hysteresis,
  • radiation damping,
  • actively controlled damping.

For these systems, the equation

m ¨x + b˙x + kx = 0
(155)

may be only an approximation.

The three classical damping regimes apply specifically to the linear viscous oscillator.

32 Common mistakes

  1. Assuming every damped oscillator is underdamped.
  2. Confusing the undamped natural frequency ω0 with the damped frequency ωd.
  3. Forgetting the factor of 2 in 2γ = b∕m.
  4. Writing γ = b∕m when using the equation ẍ + 2γẋ + ω02x = 0.
  5. Using ωd = ∘ --------
  ω20 + γ2 instead of the minus sign.
  6. Forgetting the exponential envelope e−γt in the underdamped solution.
  7. Assuming the instantaneous mechanical energy is exactly proportional to e−2γt at every instant.
  8. Saying critical damping returns to equilibrium faster than every underdamped trajectory for every possible initial condition; the standard statement concerns fastest nonoscillatory return in the linear model.
  9. Forgetting the te−ω0t term for a repeated characteristic root.
  10. Treating the overdamped solution as sinusoidal.
  11. Confusing damping ratio ζ with damping rate γ.
  12. Forgetting that logarithmic decrement uses same-sign peaks separated by a full damped period.
  13. Forgetting that damping force always removes mechanical energy because Pd = −bv2.
  14. Assuming phase-space trajectories remain closed when damping is present.
  15. Applying viscous-damping formulas to dry-friction or quadratic-drag systems without justification.

33 Practice exercises

  1. Derive the damped oscillator equation from the spring force and viscous damping force.
  2. Derive the characteristic roots r± = −γ ±∘  --------
   γ2 − ω20.
  3. Express the damping ratio ζ in terms of m, b, and k.
  4. Derive the underdamped frequency ωd.
  5. Starting from x0 and v0, derive the underdamped initial-value solution.
  6. Derive the critical-damping solution and its initial-condition constants.
  7. Derive the overdamped initial-condition coefficients C and D.
  8. Show that the critical damping coefficient is bc = 2√ ----
  mk.
  9. Derive dE∕dt = −bv2.
  10. Explain why the underdamped phase-plane trajectory spirals toward equilibrium.
  11. Derive the logarithmic decrement formula Λ = 2πγ∕ωd.
  12. Use two peaks separated by several cycles to derive the multi-cycle logarithmic-decrement formula.
  13. Show that weak damping changes the oscillation frequency only at second order in γ∕ω0.
  14. For m = 2 kg and k = 50 N/m, verify the critical damping coefficient in Example 3.
  15. Explain the dynamical-systems meaning of the eigenvalue transition from complex to repeated real to distinct real roots.

34 Summary

The linearly damped harmonic oscillator obeys

|------------------|
m-x¨+-bx˙+-kx--=-0.-
(156)

With

                  ∘  ---
γ = -b-,     ω0 =    k-,
    2m               m
(157)

|-------------------|
¨x + 2γ ˙x + ω20x = 0. |
---------------------
(158)

The characteristic roots are

|-----------∘----------|
|               2    2 |
-r±-=-−-γ-±---γ--−-ω-0.|
(159)

For

γ < ω0,
(160)

|-----∘----------|
|         2    2 |
-ωd-=---ω-0 −-γ-,|
(161)

and the motion oscillates with exponentially decaying amplitude.

For

γ = ω0,
(162)

the system is critically damped.

For

γ > ω0,
(163)

it is overdamped.

Mechanical energy obeys

|----------------|
|dE              |
|--- = − bv2 ≤ 0.|
-dt---------------
(164)

The underdamped logarithmic decrement is

|-----2πγ--|
|Λ =  ----.|
------ωd---|
(165)

Damping converts the closed phase-space orbits of the ideal oscillator into trajectories attracted toward equilibrium.

M06-03 adds an external periodic force and develops driven motion, resonance, phase lag, and the deeper role of the quality factor.

References

References

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

[2]   PhysicsLibrary, simple harmonic oscillator, canonical object SimpleHarmonicOscillator.

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

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


"Damped Harmonic Motion" is owned by bloftin.
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Also defines:  damped harmonic motion, viscous damping, damping coefficient, damping rate, damping ratio, underdamped motion, critical damping, overdamped motion, damped angular frequency, logarithmic decrement
Keywords:  damped harmonic motion, damping, viscous damping, underdamped oscillator, critical damping, overdamped oscillator, damping ratio, logarithmic decrement, quality factor, phase plane, energy dissipation, transient response

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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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