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strong (ams4684) – HW12 – ditmire – (58216)
1
This printout should have 22 questions.
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beFore answering.
001
(part 1 oF 4) 10.0 points
A3
.
9 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
+
φ
0
)
.
±ind the angular Frequency oF the oscillat
ing mass.
Correct answer: 160
.
128 s

1
.
Explanation:
Let :
M
=3
.
9 kg
and
k
=1
×
10
5
N
/
m
.
When the mass moves out oF equilibrium,
it su²ers a net restoring Force
F
net
y
=
F
spring

Mg
=

k
(
y

y
eq

ky ,
and accelerates back towards the equilibrium
position at the rate
a
y
=
F
net
y
M
=

k
M
y.
ThereFore, the mass oscillates harmonically
with angular Frequency
ω
=
±

a
y
y
=
±
k
M
=
²
1
×
10
5
N
/
m
3
.
9 kg
=
160
.
128 s

1
.
002
(part 2 oF 4) 10.0 points
At time
t
0
= 0 the mass is at 17
.
5 cm and
moving upward at velocity +44
.
9m
/
s.
±ind the amplitude oF the oscillating mass.
Correct answer: 33
.
0529 cm.
Explanation:
Let :
y
0
= 17
.
5 cm
and
v
0
= 44
.
/
s
.
The mass oscillates according to the SHM
equation
y
(
t
A
sin(
+
φ
0
)
,
so its velocity is
v
y
(
t
dy
dt
=
A ω
cos(
+
φ
0
)
.
At time
t
0
= 0, we have
y
0
≡
y
(
t
= 0) =
A
sin
φ
0
and
v
0
=
v
y
(
t
= 0) =
cos
φ
0
,
so
A
sin
φ
0
=
y
0
and
A
cos
φ
0
=
v
0
ω
.
(1)
Consequently,
A
2
=(
A
sin
φ
0
)
2
+(
A
cos
φ
0
)
2
y
0
)
2
+
³
v
0
ω
´
2
A
=
±
(
y
0
)
2
+
³
v
0
ω
´
2
=
²
(17
.
5 cm)
2
+
µ
44
.
/
s
160
.
128 s

1
¶
2
=
33
.
0529 cm
.
003
(part 3 oF 4) 10.0 points
±ind the initial phase
φ
0
in the range 0
≤
φ
0
<
360
◦
.
Correct answer: 31
.
9686
◦
.
Explanation:
According to eq. (1),
tan
φ
0
=
A
sin
φ
0
A
cos
φ
0
=
ω y
0
v
0
=0
.
624108
.
y
0
>
0 and
v
0
>
0 imply sin
φ
0
>
0
,
cos
φ
0
>
0
,
and 0
< φ
0
<
90
◦
.
Consequently, the initial phase is
φ
0
= arctan(0
.
624108) =
31
.
9686
◦
.
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View Full Documentstrong (ams4684) – HW12 – ditmire – (58216)
2
004
(part 4 of 4) 10.0 points
Calculate the position of the oscillating mass
at the time
t
=0
.
0371 s
.
Correct answer: 7
.
06875 cm.
Explanation:
Let :
t
.
0371 s
.
y
=
A
sin(
ωt
+
φ
0
)
= (33
.
0529 cm)
×
sin[(160
.
128 s

1
)(0
.
0371 s) + 31
.
9686
◦
]
=
7
.
06875 cm
.
005
(part 1 of 2) 10.0 points
A6
.
3 kg mass slides on a frictionless surface
and is attached to two springs with spring
constants 23 N
/
m and 65 N
/
m as shown in
the Fgure.
23 N
/
m
65 N
/
m
6
.
3 kg
±ind the frequency of oscillation.
Correct answer: 0
.
594828 Hz.
Explanation:
Let :
k
1
= 23 N
/
m
,
k
2
= 65 N
/
m
,
and
m
=6
.
3 kg
.
By Hooke’s law,
F
=

kx
=
ma
=
m
d
2
x
dt
2
and
d
2
x
dt
2
+
k
m
x
,
(1)
whose integral form has a sine function
x
(
t
)=
A
sin(
ω t
+
δ
)
,
where
ω
=
±
k
m
, the square root of the coe²
cient of
x
in Eq. 1.
Call the displacement of the mass
x
, and
choose the positive direction to be to the right.
Then, the forces from the springs on the mass
m
are to the left:
F
1
=

k
1
x
and
F
2
=

k
2
x
,
so that force equilibrium is
F
=

k
parallel
x
F
1
+
F
2
=

k
parallel
x

k
1
x

k
2
x
=

k
parallel
x
k
parallel
=
k
1
+
k
2
.
ω
=
±
k
m
,
so
ω
parallel
=
±
k
parallel
m
=
±
k
1
+
k
2
m
and the frequency of oscillation is
f
=
ω
parallel
2
π
=
±
k
1
+
k
2
m
2
π
=
±
23 N
/
m + 65 N
/
m
6
.
3 kg
2
π
=
0
.
594828 Hz
.
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 SHIH
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