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Physics 105
Solutions to Problem Set 6
By Stephen Glassman, Fall 2007
8.22
If
F
=
kr

3
, then
U
=
1
2
kr

2
. We substitute this potential into equation
(8.30).
U
eﬀ
(
r
) =
U
(
r
) +
`
2
2
μr
2
=
1
r
2
±
1
2
k
+
`
2
2
μ
²
So
U
eﬀ
(
r
) =
αr

2
where
α
=
1
2
(
k
+
`
2
μ

1
) can be positive, negative, or zero depend
ing on
k
. Such graphs are simple enough that I won’t include them here.
For
α
= 0, every
r
is a critical point, and
r
satisﬁes ¨
r
= 0. For
α
positive, there
are no critical points. The potential is inﬁnite at the origin, so every orbit heads to
inﬁnity. For
α
negative, again there are no critical points. The potential at inﬁnity
is zero, so whether the orbit is bounded depends on whether the eﬀective 1D motion
has positive energy.
(b)
Since
u
=
r

1
,
F
=
ku
3
, and equation (8.41) becomes:
u
00
(
φ
) =

u
(
φ
)

μ
`
2
u
(
φ
)
2
F
=

u
(
φ
)

kμ
`
2
u
(
φ
)
=

±
1 +
kμ
`
2
²
u
(
φ
)
=

μ
`
2
αu
(
φ
)
For
α
= 0, the solution is
u
(
φ
) =
A
+
Bφ
. So
r
goes to inﬁnity at some (possibly
negative)
φ
. Also,
r
goes to 0 at large positive and negative
φ
.
For
α >
0, this is the usual diﬀerential equation for an harmonic oscillator (in
φ
not
t
). Its solutions are
u
(
φ
) =
A
cos
(p
1 +
μ
`
2
φ
+
δ
)
. Note that
u
has zeros but no
inﬁnities, so
r
goes to inﬁnity but not zero just as predicted.
For
α <
0, the solutions are exponential growth and decay.
u
(
φ
) =
Ae
βφ
+
Be

βφ
where
β
=
p

μ
`
2
α
. This function only has zeros if
A
and
B
have opposite signs.
Therefore,
r
deﬁnitely
has zeroes and
might
reach inﬁnity.
1
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View Full Document8.23
(a)
If
F
=

ku
2
+
λu
3
, then equation (8.41) becomes:
u
00
(
φ
) =

u
(
φ
)

m
`
2
u
(
φ
)
2
F
=

u
(
φ
)

λm
`
2
u
(
φ
) +
km
`
2
=

±
1 +
λm
`
2
²
u
(
φ
) +
km
`
2
This is the same diﬀerential equation we encountered in our discussion of a mass
on a spring in gravity. (See midterm solutions.) The homogenous solutions are the
harmonic oscillator solutions, plus there is a
φ
independent particular solution. Our
most general solution is then:
u
(
φ
) =
A
cos(
r
1 +
λm
`
2
φ
+
δ
) +
km
`
2
±
1 +
λm
`
2
²

1
=
A
cos(
r
1 +
λm
`
2
φ
+
δ
) +
±
`
2
km
+
λ
k
²

1
Now in the manner of
§
8.6, we’ll choose coordinates so that
δ
= 0.
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 Fall '07
 EdgarKnobloch
 Physics

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