Real Oscillators
… constant forces  integrate EOM  parabolic trajectories.
… linear restoring force  guess EOM solution  SHM
… nonlinear restoring forces  ?
nonlinear spring?
linear spring
F  c 1  e x 
F   kx
F
F
x
x
 kx  mx
c 1  e x   mx
F   mg  APatm  AP  mx
The spring of air :
use Ideal Gas Law: PV=NRT
NRT
 mg  APatm  A
 mx
V
Patm
chamber volume: V=Ax
NRT
 mg  APatm 
 mx
x
m
A
+x
Stable Equilibrium at
P, V
Force
xeq = NRT / (mg + APatm)
0
0
0
X
EOM
Taylor Series Expansions:
f x 


n 0
dn f
a 
n
dx
 x  a n
n!
Turns a function into a polynomial near x = a
Example:
0.3
f x 
f(x)
0.2
1  cos x 
1 x2  ex
0.1
0
-0.1
-6
-4
-2
0
x
2
4
6
Expand NRT/x around xeq:


NRT NRT
NRT
2
x  xeq   3 x  xeq   ...  mx

 mg  APatm 
2
xeq
xeq
xeq


 NRT

NRT
2
x  xeq   3 x  xeq   ...  mx
0 
2
xeq
xeq


Is it safe to linearize it? Better check a unitless ratio. How about:
 x  xeq 


 x 
 eq 
(Yes, excellent choice Dr. Hafner!)
2


 x  xeq   x  xeq 
NRT

  ...  mx
0  


xeq   xeq   xeq 


Displacement 5% of xeq:
0

.05
.0025
NRT
x  xeq   mx
2
xeq
Perhaps you would prefer….
..
NRT
x  xeq   x  xeq 

2
mxeq
SHM with
NRT m
o 
xeq
….
Simple Pendulum:
Length: L
Mass: m
Stable Equilibrium:
Q
Fx  0
Fy  T  mg  0
Displace by Q:
Fx   mg cosQ  sin Q 
T
mg cosQ sinQ
-x
mg cosQ
L2  x 2 x
  mg
L
L
EOM:
mg
mg
L2  x 2 x
 mg
 mx
L
L
Expand it!
g
 2 x L2  x 2  x
L
Derivatives:
f  x L2  x 2
f   L  x  x L  x
2
2
2
2

f   3 xL  x
1
2 2
f   3L  x

2
2
1
2 2

1
2 2
 x L  x
3
2
 6 x L  x
2
2

3
2 2

3
2 2
 3 x L  x
4
2

5
2 2

g
3 3 
0

Lx

0

x ...  x
2 
L 
6L

Now express as a unitless ratio of the dependent variable and some
parameter of the system:
3
  x
1 x  
 g 0     0    ...  x
2 L 
  L
Displacement 5% of length:
0
.05
g
 x  x
L
0
.0000625 …
SHM with
g
o 
L
The world is not linear. However, one can use a
Taylor expansion to linearize an EOM by
assuming only small perturbations around a
point of stable equilibrium (which may not be
the origin).