Physics 325, Fall 2006
Solutions to Homework #7
1) The amplitude of a damped, driven oscillator is
D
(
ω
f
) =
A
q
(
ω
2

ω
2
f
)
2
+ 4
ω
2
f
β
2
If
ω
f
≈
ω
and
β
±
ω
(that is, the driving frequency is close to the resonance frequency,
and the resonance is narrow), then
D
(
ω
f
) =
A
q
(
ω

ω
f
)
2
(
ω
+
ω
f
)
2
+ 4
ω
2
f
β
2
D
(
ω
f
)
≈
A
q
(
ω

ω
f
)
2
4
ω
2
+ 4
ω
2
β
2
=
A
2
ω
q
(
ω

ω
f
)
2
+
β
2
Since
D
R
=
A/
(2
βω
), we can write
D
(
ω
f
)
≈
βD
R
q
(
ω

ω
f
)
2
+
β
2
With
ω
f
=
ω
²
β
√
3, then (
ω

ω
f
)
2
= 3
β
2
. So
D
(
ω
f
) =
βD
R
√
3
β
2
+
β
2
=
D
R
2
δ
= tan

1
±
2
βω
f
ω
2

ω
2
f
!
= tan

1
±
2
βω
f
(
ω
+
ω
f
)(
ω

ω
f
)
!
But
ω
+
ω
f
≈
2
ω
≈
2
ω
f
. So
δ
≈
tan

1
±
β
ω

ω
f
!
= tan

1
±
β
²
β
√
3
!
=
²
tan

1
s
1
3
2)
F
ext
(
t
) = R
e
(
F
0
e
(

α
+
iω
f
)
t
)
Let’s solve the problem without the “Re” in there, adding in a ﬁctitious imaginary compo
nent
F
ext
(
t
). Since the resulting solution to the damped oscillator’s diﬀerential equation
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View Full Documentwill be the sum of a real and an imaginary part, we’ll just remember to take the real part
at the end of the problem. So let’s take
F
ext
(
t
) =
F
0
e
(

α
+
iω
f
)
t
and solve
¨
x
+ 2
β
˙
x
+
ω
2
x
=
F
ext
(
t
)
m
Let’s take the hint and try a solution of the form
x
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 Fall '09
 Lamb
 mechanics, Work

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