Physics 325
Lecture 5
So now, Equation 4.10 becomes
()
23
00
11
!
xx
dU
Ux
x
x
dx
dx
==
=+
L
+
(5.1)
We write leading term in the expansion as
2
1
2
k
x
≈
with
2
kdU
d
x
=
2
evaluated at
x=0
.
The
stability
of the motion depends on the sign of
the second derivative, that is the sign of
k
.
If
k>0
, the equilibrium is stable and if
k<0
the
equilibrium is unstable.
Let’s look at the stable case,
k>0
:
We assume that we have known initial conditions
x
0
and
v
0
that determine
E
0
22
0
E
mv
kx
We plug this in to Equation 4.8 to find
21
1
2
2
vx
E
k
x
E
k
x
mm
⎛⎞
=−= −
⎜⎟
⎝⎠
With the following substitutions
0
2
km
A
E k
ω
we can write
A
x
=
−
And with Equation 4.9
0
dx
tt
A
x
−=
−
∫
The integral can be done with the clever substitution
x=A
sin
θ
:
0
2
cos
1s
i
n
dx
A
d
d
Ax
A
θθ
=
−
−−
∫∫
∫
So we have
0
0
−=−
.
Or defining a phase angle
0
t
0
φ
=
+
we write
1
sin
x
t
A
ωφ
−
+==
Therefore
( ) ( )
sin
xt
A
t
=
+
(5.2)
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View Full DocumentThis is a simple harmonic oscillator with amplitude
A
and phase
φ
that are determined
from the initial conditions. For example, the amplitude
A
is determined by the total
energy
E
0
– the oscillator moves between the turning points.
We continue with our exploration of different types of forces.
e)
In one dimension:
()
FF
v
=
rr
r
dv
Fv
m
x m
dt
m
dv
dt
=
=→
=
∫∫
&&
(5.3)
The procedure is the same as we have done before.
Assuming we can integrate the left
hand side, we do so and then invert
t(v)
to find
v(t)
and integrate this a second time to
find
x(t)
.
There are many forces of this time, air resistance for example.
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 Fall '08
 Staff
 Physics, mechanics, Force, Drag force, Trigraph, terminal velocity

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