Transcript Duffing Oscillator - Northern Illinois University

Duffing Oscillator
Two Springs

k
A mass is held between two
springs.
• Spring constant k
• Natural length l
l
m

k
l
Springs are on a horizontal
surface.
• Frictionless
• No gravity
Transverse Displacement

s
The force for a displacement
is due to both springs.
• Only transverse component
• Looks like its harmonic
x
q
s
s  l 2  x2

F  2k  l

 l
F  2k l 2  x 2  l sin q
2
 x2
x
l 2  x2


1


F  2kx 1 
2 2 

1

x
l 

Purely Nonlinear

The force can be expanded
as a power series near
equilibrium.
• Expand in x/l

x 
1

F  2kl 1 
2 2 
l
1

x
l 

3

The lowest order term is
non-linear.
 x
F  kl   
l
• F(0) = F’(0) = F’’(0) = 0
• F’’’(0) = 3

Quartic potential
• Not just a perturbation
V
k 4
x 
2
4l
Mixed Potential

l+d
Typical springs are not at
natural length.
s
• Approximation includes a
linear term
s
F 
x
l+d
V
2kd
k l  d  3
x
x 
3
l
l
kd 2 k l  d  4
x 
x 
3
l
4l
Quartic Potentials

The sign of the forces influence the shape of the
potential.
hard
V
double well
k 2 k 4
x 
x
2
4
V 
k 2 k 4
x 
x
2
4
soft
V
k 2 k 4
x 
x
2
4
Driven System

Assume a more complete,
realistic system.
• Damping term
• Driving force

x  2
b
x  w02 x  bx 3  f cos wt
m
Rescale the problem:
• Set t such that w02 = k/m = 1
• Set x such that b = k/m = 1

mx  2bx  kx  kx3  f coswt
This is the Duffing equation
x  2x  x  x3  f coswt
Steady State Solution

Try a solution, match terms
x(t )  A(w ) cos[wt  q (w )]
x  2x  x  x3  f coswt
A(1  w 2 ) cos(wt  q )  2 Aw sin(wt  q )  A3 cos3 (wt  q )  f coswt
trig
3
3
1
cos
(
w
t

q
)

cos(
w
t

q
)

4
4 cos3(wt  q )
identities
f coswt  f cosq cos(wt  q )  f sin q sin(wt  q )
[ A(1  w 2  34 A2 )  f coswt ] cos(wt  q )
f coswt  A(1  w 2  34 A2 )
 [2 Aw  f sin wt ] sin(wt  q )
f sin wt  2 Aw
 14 A3 cos3(wt  q )
 14 A3 cos3(wt  q )  0
0
Amplitude Dependence

Find the amplitudefrequency relationship.
• Reduces to forced harmonic
oscillator for A  0
f 2 cos2 wt  A2 (1  w 2  34 A2 ) 2
f 2 sin 2 wt  4 A2 2w 2
f 2  A2 [(1  w 2  34 A2 ) 2  4 2w 2 ]
f 2  A2[(1  w 2 )2  (2w)2 ]

Find the case for minimal
damping and driving force.
• f,  both near zero
• Defines resonance condition
0  A2 [(1  w 2  34 A2 ) 2  0]
0  1  w 2  34 A2
A(w )   43 (w 2  1)
Resonant Frequency
A
Linear
oscillator
The resonant frequency of a
linear oscillator is
independent of amplitude.

The resonant frequency of a
Duffing oscillator increases
with amplitude.
A   43 (w 2  1)
Duffing
oscillator
w1

w
Hysteresis

A
A Duffing oscillator behaves
differently for increasing and
decreasing frequencies.
• Increasing frequency has a
jump in amplitude at w2
• Decreasing frequency has a
jump in amplitude at w1
w1
w2

This is hysteresis.
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