1.6
Equilibrium, stability and bounded motion
Consider a potential energy that is only a function of the position
x
in onedimension:
U
=
U
(
x
). The energy conservation equation is
E
=
1
2
mv
2
+
U
(
x
)
implying
(56)
v
(
t
)=
dx
dt
=
±
±
2
m
[
E

U
(
x
)]
(57)
which implies periodic oscillations for bounded motion. We can integrate
t
2

t
1
=
±
m
2
²
x
2
x
1
dx
³
E

U
(
x
)
(58)
Looking at this equation, it is clear that only those values of
x
with
E

U
(
x
) positive are
possible.
example:
A particle of mass 2 moves on the positive
x
axis under the force
F
=4
/x
2

1.
Initially, the particle is released from rest at the point
x
= 4. Find the turning points and
the period of motion.
solution:
The potential is
U
(
x
) = 4
/x
+
x
, so that energy conservation states:
1
2
(2)
v
2
+
4
x
+
x
=
E
(59)
The initial condition
v
= 0 when
x
= 4 gives
E
= 5, so that
v
2
=5

4
x

x
(60)
The turning points occur when
v
= 0 , or
x
=1
,
4. To ±nd the period, write
v
=
dx/dt
in
the last equation and take the square roots:
dx
dt
=
±
±
(
x

1)(4

x
)
x
(61)
The period is the time it takes to get from
x
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 Spring '10
 RogerMelko
 Equilibrium, Conservation Of Energy, Energy, Kinetic Energy, Potential Energy

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