sanchez (ds28677) – homework 29 – Turner – (58220)
1
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12
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001
(part 1 of 3) 10.0 points
A weight suspended from a spring is seen to
bob up and down over a distance of 27 cm
twice each second.
What is its frequency?
Correct answer: 2 Hz.
Explanation:
Let :
n
= 2
.
freq =
2 bobs
1 second
=
2 Hz
.
002
(part 2 of 3) 10.0 points
What is its period?
Correct answer: 0
.
5 s.
Explanation:
period =
1
freq
=
1
2 Hz
=
0
.
5 s
.
003
(part 3 of 3) 10.0 points
What is its amplitude?
Correct answer: 13
.
5 cm.
Explanation:
Let :
d
= 27 cm
.
The amplitude is the distance from the
equilibrium position to maximum displace
ment:
1
2
d
=
1
2
(27 cm) =
13
.
5 cm
.
004
10.0 points
The oscillation of a massspring system
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
=
1
2
π
2.
φ
= 2
π
3.
φ
=
1
4
π
4.
φ
=
π
5.
φ
=
5
4
π
6.
φ
=
3
2
π
correct
7.
φ
=
3
4
π
8.
φ
=
7
4
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
)
.
At
t
= 0 we have
v
0
=

x
m
ω
sin(
φ
)
>
0, so
the solution is
φ
=
3
2
π
.
005
(part 1 of 4) 10.0 points
A 3
.
4 kg mass is suspended on a 1
×
10
5
N
/
m
spring. The mass oscillates up and down from
the equilibrium position
y
eq
= 0 according to
y
(
t
) =
A
sin(
ωt
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 Spring '08
 Turner
 Physics, Mass, Simple Harmonic Motion, Work, Hertz, Angular frequency

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