damped harmonic oscillator

Chapter 4

Damped Harmonic Oscillator

4.1 Friction

In the absence of any form of friction, the system will continue to oscillate with no decrease inamplitude. However, if there is some from of friction, then the amplitude will decrease as afunction of time

g

t

A0

A0

x

If the damping is sliding friction, Fsf = constant, then the work done by the frictional is equalto the difference in the potential energy at the turning points. At the turning points, thevelocity is zero and therefore the energy is given by the potential energy.

(4.1)HA0 + A1LFsf = 1

2k IA0

2 - A12M

or

(4.2)Fsf =1

2k HA0 - A1L

Solving for A1

2 4_DampedHarmonicOscillator.nb

(4.3)A1 = A0 -2Fsfk

damping coefficient

2 4 6 8 10

-1.0

-0.5

0.5

1.0

4.2 Velocity Dependent Friction

Another common form of friction is proportional to the velocity of the object. This includes, airdrag (at low velocities), viscous drag and magnetic drag.

Formally, the frictional force points in the opposite direction to the velocity

(4.4)Ff = -b v

4_DampedHarmonicOscillator.nb 3

t

1

The total force on the mass is then F = -k x - b v. Therefore the equation of motion becomes

(4.5)F = md2 x

d t2= -k x - b

d x

d t

or

(4.6)md2 x

d t2+ b

d x

d t+ k x = 0

Let’s try our previous trick to find the energy. Multiply the equation by v and rewrited2 xd t2

= d vd t

(4.7)mvd v

d t+ b v2 + k v x = 0

or

(4.8)mvd v

d t+ k x

d x

d t= -b v2

The quantity on the left hand side is just the time rate of change of the energy

(4.9)d m v2

2+ k x2

2

d t=dE

d t= -b v2

Previously we had b = 0 and consequently the energy was a constant in time. Now we see that


dEd t

< 0 i.e. the energy decreases with time. To estimate the rate of the decrease of E, we can

rewrite the above equation as

(4.10)dE

d t= -

2 b

m

1

2mv2

and replace both sides of the equation by the average value of the quanties. This is not pre-cisely correct as we are neglecting some terms.

(4.11)dE

d t= -

2 b

m

mv2

2= -

2 b

m

E

2= -

b

mE

where we have used the fact that for the undamped harmonic oscillator 12mv2 = E

2. Now we

can solve the equation

(4.12)E = E0 ‰-b tm = E0 ‰

-t g = E0 ‰-tt

The average energy decreases exponentially with a characteristic time t = 1 ê g where g = b êm.From this equation, we see that the energy will fall by 1 ê ‰ of its initial value in time t

g

1 êg t

E0‰

E0

x

For an undamped harmonic oscillator, 12k x2 = E

2. Therefor one would expect that


(4.13)x ∂ E ∂ ‰-

12 g t

4.3 Formal Solution

Dividing Equation 4.6 by m, we have that

(4.14)d2 x

d t2+

b

m

d x

d t+

k

mx =

d2 x

d t2+ g

d x

d t+w0

2 x = 0

Substitute a solution of the form A ‰Â w t

(4.15)-A ‰Â tw w2 + Â A ‰Â tw g w +w0

2 A ‰Â tw = 0

A ‰Â tw IÂ g w -w2 +w02M = 0

Because ‰Â w t ∫ 0 or A ∫ 0, A ‰Â w t will be a solution if

(4.16)-Â g w+w2 -w02 = 0

or

(4.17)

w1 =1

2Â g + -g2 + 4w0

2

w2 =1

2Â g - -g2 + 4w0

2

The complete solution then is

(4.18)

x@tD = A1 ‰Â tw1 + A2 ‰

Â tw2

= A1 ‰-t g2 +

12 Â t -g2+4w0

2+ A2 ‰

-t g2 -

12 Â t -g2+4w0

2

4.3.1 No Damping g = 0

For b = 0, we obtain our previous solution

(4.19)x@tD = A1 ‰Â tw0 +A2 ‰

-Â tw0

For intial condition at t = 0, x@0D = x0 and v@0D = v0, we have that


(4.20)x@0D = A1 + A2 = x0

v@0D = Â A1 w0 - Â A2 w0 = v0

Solving for A1 and A2, we obtain

(4.21)

A1 =x0

2-Â v0

2w0

A2 =x0

2+Â v0

2w0

Clearly A1 is the complex conjugate of A2. Rewritting them in exponential form

(4.22)

A1 =

v02 + x0

2 w02

2w0‰-Â f =

1

2x0

2 +v0

2

w02‰-Â f

A2 =

v02 + x0

2 w02

2w0‰Â f =

1

2x0

2 +v0

2

w02‰Â f

where Tan@fD = v0x0 w0

The displacement then becomes

(4.23)

x@tD = A1 ‰Â tw0 + A2 ‰

-Â tw0 =

v02 + x0

2 w02

2w0‰Â f-Â tw0 + ‰-Â f+Â tw0

=

Cos@f - tw0D v02 + x0

2 w02

w0

4.3.2 Small Damping g2< w0

For g2 < w0, the term in the exponent, 4w02 - g2 will be real. Defining a new frequency,

wD =12 4w0

2 - g2 = w02 -

g2

4 , we obtain


(4.24)x@tD = A1 ‰

-t g2 +

12 Â t -g2+4w0

2+ A2 ‰

-t g2 -

12 Â t -g2+4w0

2

= ‰-t g2 IA1 ‰

Â twD + A2 ‰-Â twD M

We can now identify wD as the frequency of oscillations of the damped harmonic oscillator. Asbefore we can rewrite the exponentials in terms of Cosine function with an arbitrary phase.

(4.25)x@tD = A ‰- t g2 Cos@wD t- fDFor intial condition at t = 0, x@0D = x0 and v@0D = v0, we have that

(4.26)x@0D = ACos@fD = x0

v@0D = -1

2A gCos@fD+ ASin@fDwD = v0

Substituting, for Cos[f] from the first equation into the second, we have that

(4.27)-g x0

2+ASin@fDwD = v0

Therefore

(4.28)

ACos@fD = x0

ASin@fD =2 v0 + g x0

2wD

Solving for A and f

(4.29)

A = x02 +

H2 v0 +g x0L24wD

2

Tan@fD =2 v0 + g x0

2 x0 wD

One can check this answer by taking the limit as g Ø 0 (no damping), wD Ø w0 and the expres-sions for A and f should reduce to our previous result.

(4.30)A = x0

2 +v0

2

w02

Tan@fD =v0

x0 w0


g

2w

7.5 3.5q0 q

°0

1.22 0

2 4 6 8 10

-1.0

-0.5

0.5

1.0

initial angle HradiansLinitial angular velocity

4.3.3 Critical Damping g2= w0

At g2= w0, the frequency wD vanishes and the expression in the exponent reduces to

(4.31)-g

2≤

1

2Â -g2 + 4w0

2 Ø -g

2

The solution no longer has an oscillatory part. In addition one no longer has two solutionsthat can be used to fit arbitrary initial conditions.

The general solution as a function of time becomes

(4.32)x@tD = ‰- t g2 HA+B tLThe second term is necessary to satisfy all possible initial conditions. Differentiating

(4.33)v@tD = -

1

2‰-t g2 HA g +B H-2+ t gLL

a@tD =1

4‰-t g2 g HA g +B H-4+ t gLL


Substituting in the differential equation

(4.34)

d2 x

d t2+ g

d x

d t+w0

2 x = 0

1

4‰-t g2 g HA g +B H-4+ t gLL + -

1

2‰-t g2 g HA g +B H-2+ t gLL+w0

2 ‰-t g2 HA+B t

-1

4‰-t g2 HA+B tL Ig2 - 4w0

2M = 0

which by the definition, g2 = 4w02, is identically equal to zero

To determine A andB from the initial conditions, equate

(4.35)

x@0D = ‰-t g2 HA+B tL

t=0= A

v@tD = -1

2‰-t g2 HA g +B H-2+ t gLL

t=0=

1

2H2B-A gL

Solving for A andB

(4.36)

A = x0

B =1

2H2 v0 +g x0L

and therefore the general solution becomes

(4.37)x@tD = ‰- t g2 x0 +1

2t H2 v0 +g x0L

4.3.4 Overdamping g2> w0

For g2> w0, the argument of the square root becomes negative. Taking the -1 out of the

squareroot so that its argument is now positive, the frequency, w, becomes completelyimaginary

(4.38)

w1 =Â g

2+ -

g2

4+w0

2 ->Â g

2+ Â

g2

4-w0

2

w1 =Â g

2- -

g2

4+w0

2 ->Â g

2- Â

g2

4-w0

2


Again, there are no oscillatory terms. Both solutions are dying exponentials

(4.39)x@tD = A1 ‰

t -g2-g24 -w0

2

+A2 ‰

t -g2+g24 -w0

2

Although the second term contains a growing exponential, gD2

= J g2

N2 - w02 <

g

2 and so the

overal function is still a dying exponential.

As before, to determine A1 and A2 from the initial conditions, equate

(4.40)x@0D = ‰t J-g2-gD2 N A1 + ‰

t J-g2+gD2 N A2t=0

= A1 +A2

and

(4.41)

v@0D =1

2‰-

12 t Ig+gDM I‰t gD A2 H-g + gDL- A1 Hg + gDLM

t=0

=1

2HA2 H-g + gDL- A1 Hg + gDLL

Solving for A1 and A2

(4.42)

A1 =-2 v0 + x0 H-g + gDL

2 gD

A2 =2 v0 + x0 Hg + gDL

2 gD

and therefore the general solution becomes

(4.43)x@tD = 1

2 gD‰-

12 t Ig+gDM I2 I-1+‰t gD M v0 + x0 I-g + gD +‰t gD Hg + gDLMM

Note that this solution looks very much like our original solution for the underdamping case.

In particular if we let g2 - 4w02 Ø Â 4w0

2 - g2 = 2 Â wD, then the above solution becomes

(4.44)x@tD = 1

2 Â wD‰-

12 Ht gL CoshB 2

2t Â wDF 2 Â wD x0 +SinhB 1

2t 2 Â wDF H2 v0 + g x0L

Now, Cosh@Â yD = ‰Â y + ‰-Â y

2= Cos@yD and Sinh@Â yD = ‰Â y - ‰-Â y

2= ÂSin@yD so that the above

expression reduces to


x@tD = -Â ‰-t g2 HÂSin@twDD H2 v0 + g x0L+ 2 ÂCos@twDD x0 wDL

2wD

= ‰-t g2 Cos@twDD x0 +

Sin@twDD Jv0 +g x0

2N

wD

Equating x0 = Cos@fD and v0 +

g2x0

wD= Sin@fD, then the above solution becomes the solution for

an underdamped harmonic oscillator

(4.46)x@tD = Cos@f + twDD x02 +

Jv0 +g x0

2N2

wD2

4.4 Quality Factor or Q

A dimensionless quantity used to measure the amount of damping is given by for a harmonicoscillator is defined as

(4.47)Q = H2 pHaverage energy storedLL ê Henergy lost per cycleLAs defined the Q will be large if the damping is small. We know that the average energy isgiven approximately by E@tD º E0 ‰

-g tê2. The change is the energy in one period t is therefore

(4.48)DE = E@t0D-E@t + t0D = ‰-g t0 H1- ‰-g tL E0

For small damping g t << 1, one can expand the exponential, ‰-g t º 1- g t + ... so that

(4.49)DE = ‰-g t0 g t E0

Substituting this into the expression for the Q

(4.50)Q =2 ‰-g t0 p E0

‰-g t0 g t E0=

2 p

g t=w

g

where we have used the fact that w = 2 pt

. The following illustrates the behavior for several

values of the Q.


4 6 8 10

= 0.0125664 Q = 500

0 2 4 6 8 10

-1.0

-0.5

0.0

0.5

1.0 g = 0.0314159 Q = 200

0 2 4 6

-1.0

-0.5

0.0

0.5

1.0 g = 0.0628319 Q =

4 6 8 10

g = 0.125664 Q = 50

0 2 4 6 8 10

-1.0

-0.5

0.0

0.5

1.0 g = 0.314159 Q = 20

0 2 4 6

-1.0

-0.5

0.0

0.5

1.0 g = 0.628319 Q =

4 6 8 10

g = 1.25664 Q = 5

0 2 4 6 8 10

-1.0

-0.5

0.0

0.5

1.0 g = 3.14159 Q = 2

0 2 4 6

-1.0

-0.5

0.0

0.5

1.0 g = 6.28319 Q =

4 6 8 10

g = 12.5664 Q = 0.5

0 2 4 6 8 10

-1.0

-0.5

0.0

0.5

1.0 g = 31.4159 Q = 0.2

0 2 4 6

-1.0

-0.5

0.0

0.5

1.0 g = 62.8319 Q =

4.5 Energy of Damped Harmonic Oscillator

Using the expression for an underdamped harmonic oscillator, x@tD = ACos@wD t+ fD e-g tê2, thepotential energy becomes

(4.51)PE =1

2k x@tD2 = 1

2mw2 x@tD2 = 1

2A2 ‰-t gmw2 Cos@f+ twDD2

To calculate the kinetic energy, first calculate the velocity


(4.52)v@tD = - 1

2A ‰-

t g2 gCos@f+ twDD- A ‰- t g2 Sin@f + twDDwD

and the kinetic energy becomes

(4.53)PE =

1

2mv@tD2 = 1

2m -

1

2A ‰-

t g2 gCos@f+ twDD-A ‰- t g2 Sin@f + twDDwD 2

=1

8A2 ‰-t g k HgCos@f + twDD+ 2Sin@f + twDDwDL2

The total energy is then

(4.54)E@tD = 1

8A2 ‰-t gmIIg2 + 4w2MCos@f + twDD2 + 2 gSin@2 Hf + twDLDwD + 4Sin@f + twDD2 wD2 M

Substituting w2 =g2

4+wD

2

(4.55)

E@tD =1

4A2 ‰-t gm Ig2 Cos@f+ twDD2 + gSin@2 Hf+ twDLDwD + 2wD

2 ME@tD =

1

2A2 ‰-t gmwD

2 1+g2 Cos@f+ twDD2

2wD2

+gSin@2 Hf+ twDLD

2wD

Plotting the total energies assuming f = 0.


tmax

g

w

0.0 0.2 0.4 0.6 0.8 1.00.0

0.2

0.4

0.6

0.8

1.0

1.2

............................................................


damped harmonic oscillator

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