Solutions to Assignment 2
Physics 214, Spring 2008
February 8, 2008
Problem 1: Damped Oscillators
In the previous assignment we studied the damped oscillator in the under
damped regime. In this problem we look at the general case. The nature of
the solution depends on the dimensionless parameter
ω
0
τ
.
(a) Free Oscillations
The equation of motion is
m
d
2
x
dt
2
+
b
dx
dt
+
kx
= 0
(1)
As usual we obtain the solution of eqn.[1] by ﬁnding the real part of a complex
function
z
(
t
) that satisﬁes it. This is done so by substituting
z
(
t
) =
e
αt
and
solving for
α
.
mα
2
+
bα
+
mω
2
0
= 0
⇒
α
±
=

b
±
√
b
2

4
km
2
m
=

1
τ
±
r
1
τ
2

ω
2
0
(2)
as
τ
=
2
m
b
.
1) Underdamped case;
ω
0
τ >
1
⇒
1
τ
2

ω
2
0
<
0
⇒
α
±
=

1
τ
±
i
q
ω
2
0

1
τ
2
⇒
z
(
t
) =
Be
α
+
t
+
Ce
α

t
⇒
x
(
t
) =
Re
[
z
(
t
)] =
Ae

t
τ
cos(
r
ω
2
0

1
τ
2
t
)
(3)
2) Critically damped case;
ω
0
τ
= 1
⇒
α
=

1
τ
⇒
x
(
t
) =
Re
[
z
(
t
)] =
Ae

t
τ
(4)
1
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View Full Document3) Overdamped case;
ω
0
τ <
1
⇒
1
τ
2

ω
2
0
>
0
⇒
α
±
=

1
τ
±
q
1
τ
2

ω
2
0
⇒
z
(
t
) =
Ce
α
+
t
+
De
α

t
⇒
x
(
t
) =
Ae
(

1
τ
+
p
1
τ
2

ω
2
0
)
+
Be
(

1
τ

p
1
τ
2

ω
2
0
)
(5)
Notice that both
α
+
and
α

are real and negative and that
A
and
B
are the
real parts of
C
and
D
.
(b) Plots of x(t) for the 3 damping regimes: (See end of document) (c) The
frequency of the oscillations in the underdamped case;
ω
=
ω
0
q
ω
2
0

1
(
ω
0
τ
)
2
We now calculate the values of
ω
0
τ
required to produce various percentage
changes in natural frequency.
ω
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 Spring '08
 THORNE
 Physics, Cos, ω ω, critically damped oscillator

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