Morby, Grant – Homework 30 – Due: Apr 19 2006, noon – Inst: Drummond
1
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The due time is Central
time.
001
(part 1 of 1) 10 points
The equation of motion of a simple harmonic
oscillator is
d
2
x
dt
2
=

9
x ,
where
x
is displacement and
t
is time.
What is the period of oscillation?
1.
T
=
2
π
3
correct
2.
T
=
2
π
9
3.
T
= 6
π
4.
T
=
9
2
π
5.
T
=
3
2
π
Explanation:
For a simple harmonic oscillator, the equa
tion of motion can be written as
d
2
x
dt
2
=

ω
2
x
, where
ω
is the angular frequency, so
the period of oscillation is
T
=
2
π
ω
=
2
π
√
9
=
2
π
3
.
002
(part 1 of 1) 10 points
An oscillator is described by
1
2
3
4
5
6
7
8
9 10 11 12
1
2
x
(
t
)
t
(
sec
)
What
is
the
angular
frequency
ω
(in
rad/sec
) ?
1.
3.1
2.
2.0
3.
1.6
correct
4.
4.0
Explanation:
T
= 4
sec
ω
=
2
π
T
=
2
π
4
=
π
2
= 1
.
6
003
(part 1 of 1) 10 points
Given:
The oscillation of a massspring sys
tem
x
=
x
m
cos(
ω t
+
φ
)
,
where
x
m
is a positive number.
At time
t
= 0, the mass is at the equilibrium
point
x
= 0 and it is moving with positive
velocity
v
0
.
What is the phase angle
φ
?
1.
φ
=
3
2
π
correct
2.
φ
=
3
4
π
3.
φ
=
7
4
π
4.
φ
=
1
4
π
5.
φ
=
5
4
π
6.
φ
=
1
2
π
7.
φ
=
π
8.
φ
= 2
π
9.
φ
= 0
Explanation:
Since the initial position is
x
= 0, we know
that cos(
φ
) = 0, so
φ
=
1
2
π,
3
2
π,
5
2
π,
. . . .
The velocity can be found as
v
=
d x
dt
=

x
m
ω
sin(
ω t
+
φ
)
.
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Morby, Grant – Homework 30 – Due: Apr 19 2006, noon – Inst: Drummond
2
At
t
= 0 we have
v
0
=

x
m
ω
sin(
φ
)
>
0, so
the solution is
φ
=
3
2
π .
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 Spring '09
 KLEINMAN
 Physics, Simple Harmonic Motion, Work, kg, Simple Harmonic Oscillator, noon – Inst

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