… constant forces integrate EOM parabolic trajectories.
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… constant forces integrate EOM parabolic trajectories … linear restoring force guess EOM solution SHM … nonlinear restoring forces ? linear spring x F nonlinear spring? x F Real Oscillators kx F x m kx x e c F 1 x m e c x 1
description
Real Oscillators. … constant forces integrate EOM parabolic trajectories. … linear restoring force guess EOM solution SHM. … nonlinear restoring forces ?. nonlinear spring?. linear spring. F. F. x. x. 0. 0. The spring of air :. use Ideal Gas Law: PV=NRT. - PowerPoint PPT Presentation
Transcript of … constant forces integrate EOM parabolic trajectories.
… constant forces integrate EOM parabolic trajectories.
… linear restoring force guess EOM solution SHM
… nonlinear restoring forces ?
linear spring
kxF
x
F
xmkx
nonlinear spring?
xecF 1
x
F
xmec x 1
Real Oscillators
The spring of air : xmAPAPmgF atm
P, V
m
A
Patm
+x
0
use Ideal Gas Law: PV=NRT
xmV
NRTAAPmg atm
xmx
NRTAPmg atm
chamber volume: V=Ax
Stable Equilibrium at
xeq = NRT / (mg + APatm)Force
X
0
0
EOMWTF!
(whoa there, fella)
Taylor Series Expansions:
xex
xxf
21
cos1
-0.1
0
0.1
0.2
0.3
-6 -4 -2 0 2 4 6
f(x)
x
0
!n
nn
n
axn
adx
fd
xf
Turns a function into a polynomial near x = a
Example:
... 3
2
3 3
1
3 1
13 22
2
xdxfd
xdxdf
fxf
-0.1
0
0.1
0.2
0.3
-6 -4 -2 0 2 4 6
f(x)
x
Expand around x = -3:
-0.1
0
0.1
0.2
0.3
-6 -4 -2 0 2 4 6
f(x)
x
-0.1
0
0.1
0.2
0.3
-6 -4 -2 0 2 4 6
f(x)
x
... 30.0083483 - 3.10320 19801.0 2 xx
-0.1
0
0.1
0.2
0.3
-6 -4 -2 0 2 4 6
f(x)
x
0th order1st order2nd order
... 2
2
2 2
1
2 1
12 22
2
xdxfd
xdxdf
fxf
-0.1
0
0.1
0.2
0.3
-6 -4 -2 0 2 4 6
f(x)
x
Expand around x = 2:
-0.1
0
0.1
0.2
0.3
-6 -4 -2 0 2 4 6
f(x)
x
-0.1
0
0.1
0.2
0.3
-6 -4 -2 0 2 4 6
f(x)
x
... 20.030981 2.031685 11431. 2 x-x
-0.1
0
0.1
0.2
0.3
-6 -4 -2 0 2 4 6
f(x)
x
0th order
1st order
2nd order
Expand NRT/x around xeq:
xmxxx
NRTxx
x
NRTx
NRTAPmg eq
eq
eq
eqeqatm
...2
32
xmxxx
NRTxx
x
NRTeq
eq
eq
eq
...0 2
32
Is it safe to linearize it? Better check a unitless ratio. How about:
eq
eq
x
xx
(Yes, excellent choice Dr. Hafner!)
xmx
xx
x
xx
xNRT
eq
eq
eq
eq
eq
...0
2
Displacement 5% of xeq: 0 .05 .0025 ….
xmxxx
NRTeq
eq
2
SHM witheq
o x
mNRT
Perhaps you would prefer….
eqeq
eq
xxxxmx
NRT 2
..
Simple Pendulum:
Length: LMass: m Q
mg
cosQ
T
mg cosQ sinQ
mg
mg
0 xF
Stable Equilibrium:
0 mgTFy
Displace by :Q
sincosmgFx
Lx
LxL
mg22
mg cosQ
-x
xmLx
LxL
mg 22
EOM:
Expand it!
xxLxLg
222
25
23
21
23
21
21
22422222
22322
22222
22
363
3
xLxxLxxLf
xLxxLxf
xLxxLf
xLxf
Derivatives:
xxL
LxLg
...
63
00 32
Now express as a unitless ratio of the dependent variable and some parameter of the system:
xLx
Lx
g
...
21
003
xxLg
SHM withL
go
Displacement 5% of length: 0 .05 0 .0000625 …
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).
