2
4_DampedHarmonicOscillator.nb
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 in
amplitude. However, if there is some from of friction, then the amplitude will decrease as a
function of time
g
x
A0
t
A0
If the damping is sliding friction, Fsf = constant, then the work done by the frictional is equal
to the difference in the potential energy at the turning points. At the turning points, the
velocity is zero and therefore the energy is given by the potential energy.
HA0 + A1 L Fsf =
1
2
k IA20 - A21 M
(4.1)
or
Fsf =
1
2
Solving for A1
k HA0 - A1 L
(4.2)
4_DampedHarmonicOscillator.nb
A1 = A0 -
2 Fsf
3
(4.3)
k
damping coefficient
1.0
0.5
2
4
6
8
10
-0.5
-1.0
4.2 Velocity Dependent Friction
Another common form of friction is proportional to the velocity of the object. This includes, air
drag (at low velocities), viscous drag and magnetic drag.
Formally, the frictional force points in the opposite direction to the velocity
Ff = -b v
(4.4)
4
4_DampedHarmonicOscillator.nb
t
1
The total force on the mass is then F = -k x - b v. Therefore the equation of motion becomes
d2 x
F =m
d t2
= -k x - b
dx
(4.5)
dt
or
m
d2 x
d t2
+b
dx
dt
Let’s try our previous trick to find the energy.
d2 x
dv
=
2
dt
dt
mv
dv
dt
(4.6)
+kx = 0
Multiply the equation by v and rewrite
+ b v2 + k v x = 0
(4.7)
or
mv
dv
dt
+kx
dx
dt
= -b v2
(4.8)
The quantity on the left hand side is just the time rate of change of the energy
d
m v2
2
+
dt
k x2
2
=
dE
dt
= -b v2
(4.9)
Previously we had b = 0 and consequently the energy was a constant in time. Now we see that
4_DampedHarmonicOscillator.nb
5
dE
< 0 i.e. the energy decreases with time. To estimate the rate of the decrease of E, we can
dt
rewrite the above equation as
dE
dt
=-
2b 1
m
2
m v2
(4.10)
and replace both sides of the equation by the average value of the quanties. This is not precisely correct as we are neglecting some terms.
dE
dt
=-
2 b m v2
m
2
=-
2b E
m
2
=-
b
m
where we have used the fact that for the undamped harmonic oscillator
can solve the equation
-
bt
-
(4.11)
E
1
E
m v2 = . Now we
2
2
t
(4.12)
E = E0 ‰ m = E0 ‰-t g = E0 ‰ t
The average energy decreases exponentially with a characteristic time t = 1 ê g where g = b ê m.
From this equation, we see that the energy will fall by 1 ê ‰ of its initial value in time t
g
x
E0
E0
‰
1êg
For an undamped harmonic oscillator,
t
1
E
k x2 = . Therefor one would expect that
2
2
6
4_DampedHarmonicOscillator.nb
-
x∂
1
E ∂‰ 2
gt
(4.13)
4.3 Formal Solution
Dividing Equation 4.6 by m, we have that
d2 x
d t2
+
b
dx
m dt
+
k
m
x=
d2 x
d t2
+g
dx
dt
+ w20 x = 0
(4.14)
Substitute a solution of the form A ‰Â w t
 A ‰Â t w g w + w20 A ‰Â t w = 0
- A ‰Â t w w 2 +
A ‰Â t w I g w - w2 + w20 M = 0
(4.15)
Because ‰Â w t ∫ 0 or A ∫ 0, A ‰Â w t will be a solution if
-Â g w + w2 - w20 = 0
(4.16)
or
w1 =
w2 =
1
2
1
2
Âg+
-g2 + 4 w20
(4.17)
Âg-
-g2 + 4 w20
The complete solution then is
x@tD = A1 ‰Â t w1 + A2 ‰Â t w2
= A1 ‰
-
tg 1
+ Ât
2 2
-g2 +4 w2
0
+ A2 ‰
-
tg 1
- Ât
2 2
-g2 +4 w2
0
(4.18)
4.3.1 No Damping g = 0
For b = 0, we obtain our previous solution
x@tD = A1 ‰Â t w0 + A2 ‰- t w0
For intial condition at t = 0, x@0D = x0 and v@0D = v0 , we have that
(4.19)
4_DampedHarmonicOscillator.nb
x@0D = A1 + A2 = x0
7
(4.20)
v@0D = Â A1 w0 - Â A2 w0 = v0
Solving for A1 and A2 , we obtain
A1 =
A2 =
x0
2
x0
2
+
 v0
2 w0
(4.21)
 v0
2 w0
Clearly A1 is the complex conjugate of A2 . Rewritting them in exponential form
v20 + x02 w20
A1 =
‰
2 w0
-Â f
=
1
2
x02 +
v20
w20
‰-Â f
(4.22)
v20 + x02 w20
A2 =
where Tan@fD =
‰Â f =
2 w0
1
2
v20
x02 +
w20
‰Â f
v0
x0 w0
The displacement then becomes
x@tD = A1 ‰Â t w0 + A2 ‰- t w0 =
v20 + x02 w20
2 w0
‰Â f- t w0 + ‰- f+ t w0
(4.23)
Cos@f - t w0 D
=
4.3.2 Small Damping
For
wD =
v20 + x02 w20
w0
g
2
< w0
g
< w0 , the term in the exponent,
2
1
2
4 w20 - g2 =
w20 -
g2
, we obtain
4
4 w20 - g2 will be real. Defining a new frequency,
8
4_DampedHarmonicOscillator.nb
tg
1
-g2 +4 w20
x@tD = A1 ‰- 2 + 2 Â t
= ‰
-
tg
2
IA1 ‰
 t wD
+ A2 ‰
tg
1
+ A 2 ‰- 2 - 2 Â t
-Â t wD
-g2 +4 w20
M
(4.24)
We can now identify wD as the frequency of oscillations of the damped harmonic oscillator. As
before we can rewrite the exponentials in terms of Cosine function with an arbitrary phase.
-
tg
x@tD = A ‰ 2 Cos@wD t - fD
(4.25)
For intial condition at t = 0, x@0D = x0 and v@0D = v0 , we have that
x@0D = A Cos@fD = x0
v@0D = -
1
2
A g Cos@fD + A Sin@fD wD = v0
(4.26)
Substituting, for Cos[f] from the first equation into the second, we have that
-
g x0
2
+ A Sin@fD wD = v0
(4.27)
Therefore
A Cos@fD = x0
A Sin@fD =
2 v0 + g x0
(4.28)
2 wD
Solving for A and f
A
=
Tan@fD =
x02 +
H2 v0 + g x0 L2
4 w2D
(4.29)
2 v0 + g x 0
2 x0 w D
One can check this answer by taking the limit as g Ø 0 (no damping), wD Ø w0 and the expressions for A and f should reduce to our previous result.
A
=
Tan@fD =
x02 +
v0
x0 w0
v20
w20
(4.30)
4_DampedHarmonicOscillator.nb
9
initial angle HradiansL
g
w
2
initial angular velocity
°
q0 q 0
1.22 0
7.5 3.5
1.0
0.5
2
4
6
8
10
-0.5
-1.0
4.3.3 Critical Damping
At
g
2
= w0
g
= w0 , the frequency wD vanishes and the expression in the exponent reduces to
2
-
g
2
≤
1
2
-g2 + 4 w20 Ø -
Â
g
2
(4.31)
The solution no longer has an oscillatory part. In addition one no longer has two solutions
that can be used to fit arbitrary initial conditions.
The general solution as a function of time becomes
x@tD = ‰ 2 HA + B tL
-
tg
(4.32)
The second term is necessary to satisfy all possible initial conditions. Differentiating
v@tD = -
1 -tg
‰ 2 HA g + B H-2 + t gLL
2
1 -tg
‰ 2 g HA g + B H-4 + t gLL
a@tD =
4
(4.33)
10
4_DampedHarmonicOscillator.nb
Substituting in the differential equation
d2 x
d t2
dx
+g
dt
+ w20 x
= 0
1 -tg
‰ 2 g HA g + B H-4 + t gLL
4
-
tg
1 -tg
2 - 2
2
+ - ‰
g HA g + B H-2 + t gLL + w0 ‰
HA + B t (4.34)
2
1 -tg
‰ 2 HA + B tL Ig2 - 4 w20 M = 0
4
which by the definition, g2 = 4 w20 , is identically equal to zero
To determine A and B from the initial conditions, equate
‰ 2 HA + B tL
-
x@0D =
tg
=A
t=0
1 -tg
1
- ‰ 2 HA g + B H-2 + t gLL
= H2 B - A gL
2
2
t=0
v@tD =
(4.35)
Solving for A and B
A = x0
B =
1
2
H2 v0 + g x0 L
(4.36)
and therefore the general solution becomes
-
tg
x@tD = ‰ 2 x0 +
4.3.4 Overdamping
g
2
1
2
t H2 v0 + g x0 L
(4.37)
> w0
g
> w0 , the argument of the square root becomes negative. Taking the -1 out of the
2
squareroot so that its argument is now positive, the frequency, w, becomes completely
imaginary
For
w1 =
Âg
2
+
-
g2
4
+ w20
->
Âg
2
+Â
g2
4
- w20
(4.38)
w1 =
Âg
2
-
-
g2
4
+ w20 ->
Âg
2
-Â
g2
4
- w20
4_DampedHarmonicOscillator.nb
11
Again, there are no oscillatory terms. Both solutions are dying exponentials
g2
-w2
0
4
g
t - 2
x@tD = A1 ‰
g
t - +
2
g2
-w2
0
4
(4.39)
+ A2 ‰
Although the second term contains a growing exponential,
gD
=
2
g 2
g
J N - w20 <
and so the
2
2
overal function is still a dying exponential.
As before, to determine A1 and A2 from the initial conditions, equate
g g
t J- - D N
2
2
‰
x@0D =
g g
t J- + D N
2
2
A1 + ‰
A2
t=0
= A1 + A2
(4.40)
and
1 - 1 t Ig+g M t g
D I‰ D A H-g + g L - A Hg + g LM
‰ 2
2
D
1
D
2
t=0
v@0D =
=
1
2
HA2 H-g + gD L - A1 Hg + gD LL
(4.41)
Solving for A1 and A2
A1 =
A2 =
-2 v0 + x0 H-g + gD L
2 gD
2 v0 + x0 Hg + gD L
(4.42)
2 gD
and therefore the general solution becomes
x@tD =
1
2 gD
-
1
‰ 2
t Ig+gD M
I2 I-1 + ‰t gD M v0 + x0 I-g + gD + ‰t gD Hg + gD LMM
(4.43)
Note that this solution looks very much like our original solution for the underdamping case.
g2 - 4 w20 Ø Â
In particular if we let
x@tD =
1
2 Â wD
-
1
‰ 2
Ht gL
CoshB
2
2
4 w20 - g2 = 2 Â wD , then the above solution becomes
t  wD F 2  wD x0 + SinhB
1
2
t 2 Â wD F H2 v0 + g x0 L
(4.44)
‰Â y + ‰- y
‰Â y - ‰- y
= Cos@yD and Sinh@Â yD =
= Â Sin@yD so that the above
2
2
expression reduces to
Now, Cosh@Â yD =
12
4_DampedHarmonicOscillator.nb
x@tD = -
=
‰
-
tg
‰ 2
tg
2
HÂ Sin@t wD D H2 v0 + g x0 L + 2 Â Cos@t wD D x0 wD L
2 wD
Cos@t wD D x0 +
Equating x0 = Cos@fD and
v0 +
Sin@t wD D Jv0 +
g x0
2
N
wD
g
x
2 0 = Sin@fD, then the above solution becomes the solution for
wD
an underdamped harmonic oscillator
x@tD = Cos@f + t wD D
x02 +
Jv0 +
N
g x0 2
2
(4.46)
w2D
4.4 Quality Factor or Q
A dimensionless quantity used to measure the amount of damping is given by for a harmonic
oscillator is defined as
Q = H2 pHaverage energy storedLL ê Henergy lost per cycleL
(4.47)
As defined the Q will be large if the damping is small. We know that the average energy is
given approximately by E@tD º E0 ‰-g tê2 . The change is the energy in one period t is therefore
DE = E@t0 D - E@t + t0 D = ‰-g t0 H1 - ‰-g t L E0
(4.48)
For small damping g t << 1, one can expand the exponential, ‰-g t º 1 - g t + ... so that
DE = ‰-g t0 g t E0
(4.49)
Substituting this into the expression for the Q
Q=
2 ‰-g t0 p E0
‰-g t0 g t E0
=
2p
gt
=
w
g
where we have used the fact that w =
values of the Q.
(4.50)
2p
. The following illustrates the behavior for several
t
4_DampedHarmonicOscillator.nb
= 0.0125664 Q = 500
4
6
8
6
8
6
-0.5
-0.5
-1.0
-1.0
2
8
6
8
10
0
0.5
0.0
0.0
-0.5
-0.5
-1.0
-1.0
2
4
6
8
10
0
g = 3.14159 Q = 2
0.5
0.0
0.0
-0.5
-0.5
-1.0
-1.0
0
2
4
6
8
10
0.5
0.0
0.0
-0.5
-0.5
-1.0
-1.0
0
2
4
6
8
10
6
4
6
2
4
6
g = 62.8319 Q =
1.0
0.5
4
g = 6.28319 Q =
0
g = 31.4159 Q = 0.2
1.0
2
1.0
0.5
10
2
g = 0.628319 Q =
1.0
0.5
0
10
4
g = 0.314159 Q = 20
1.0
8
6
0.0
10
g = 12.5664 Q = 0.5
4
0.0
1.0
g = 1.25664 Q = 5
4
0.5
0
g = 0.0628319 Q =
1.0
0.5
10
g = 0.125664 Q = 50
4
g = 0.0314159 Q = 200
1.0
13
0
2
4
6
4.5 Energy of Damped Harmonic Oscillator
Using the expression for an underdamped harmonic oscillator, x@tD = A Cos@wD t + fD e-g tê2 , the
potential energy becomes
PE =
1
2
k x@tD2 =
1
2
m w2 x@tD2 =
1
2
A2 ‰-t g m w2 Cos@f + t wD D2
To calculate the kinetic energy, first calculate the velocity
(4.51)
14
4_DampedHarmonicOscillator.nb
1
v@tD = -
2
A‰
-
tg
2
g Cos@f + t wD D - A ‰
-
tg
2
Sin@f + t wD D wD
(4.52)
and the kinetic energy becomes
PE =
=
1
2
1
8
m v@tD2 =
1
2
m -
1
2
A ‰ 2 g Cos@f + t wD D - A ‰ 2 Sin@f + t wD D wD
-
tg
-
tg
2
(4.53)
A2 ‰-t g k Hg Cos@f + t wD D + 2 Sin@f + t wD D wD L2
The total energy is then
E@tD =
1
8
A2 ‰-t g m
IIg + 4 w M Cos@f + t wD D + 2 g Sin@2 Hf + t wD LD wD + 4 Sin@f + t wD D
2
2
Substituting w2 =
E@tD =
E@tD =
1
4
1
2
2
2
w2D M
(4.54)
g2
+ w2D
4
A2 ‰-t g m Ig2 Cos@f + t wD D2 + g Sin@2 Hf + t wD LD wD + 2 w2D M
2 -t g
A ‰
m w2D
1+
g2 Cos@f + t wD D2
2 w2D
Plotting the total energies assuming f = 0.
+
g Sin@2 Hf + t wD LD
2 wD
(4.55)
4_DampedHarmonicOscillator.nb
tmax
g
w
1.2
1.0
0.8
0.6
0.4
0.2
0.0
0.0
0.2
0.4
0.6
0.8
............................................................
1.0
15