wolz (cmw2833) – HW #12 – Antoniewicz – (57420)
1
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printout
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have
19
questions.
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before answering.
001
10.0points
A 62
.
9 g mass is attached to a horizontal
spring with a spring constant of 15
.
4 N
/
m
and released from rest with an amplitude of
24
.
6 cm.
What is the velocity of the mass when it
is halfway to the equilibrium position if the
surface is frictionless?
Correct answer: 3
.
3335 m
/
s.
Explanation:
Let :
m
= 62
.
9 g
,
k
= 15
.
4 N
/
m
,
and
A
= 24
.
6 cm
.
x
=
A
2
= 0
.
123 m
v
=
radicalbigg
k
m
(
A
2

x
2
)
=
radicalBigg
15
.
4 N
/
m
0
.
0629 kg
[(0
.
246 m)
2

(0
.
123 m)
2
]
=
3
.
3335 m
/
s
.
002
10.0points
The mass of the deuterium molecule D
2
is
twice that of the hydrogen molecule H
2
.
If the vibrational frequency of H
2
is 1
.
38
×
10
14
Hz, what is the vibrational frequency
of D
2
, assuming that the “spring constant”
of attracting forces is the same for the two
species?
Correct answer: 9
.
75807
×
10
13
Hz.
Explanation:
Let :
M
D
= 2
M
H
.
The
angular
frequencies
depend
only
on
spring constant and mass:
ω
=
radicalbigg
k
M
∝
radicalbigg
1
M
ω
D
=
radicalBigg
k
M
D
ω
H
=
radicalBigg
k
M
H
The spring constants
k
are the same, so
ω
D
ω
H
=
radicalBigg
M
H
M
D
=
radicalBigg
M
H
2
M
H
=
radicalbigg
1
2
=
1
√
2
.
The linear frequency is
f
=
ω
2
π
∝
ω,
so
f
D
f
H
=
ω
D
ω
H
=
1
√
2
f
D
=
f
H
√
2
=
1
.
38
×
10
14
Hz
√
2
=
9
.
75807
×
10
13
Hz
.
003
10.0points
At an outdoor market, a bunch of bananas is
set on a spring scale to measure the weight.
The spring sets the full bunch of bananas
into vertical oscillatory motion, which is har
monic with an amplitude 0
.
084 m. The max
imum speed of the bananas is observed to be
0
.
636 m
/
s.
What is the mass of the bananas?
The
spring
of
the
scale
has
a
force
constant
74
.
5 N
/
m.
Correct answer: 1
.
29957 kg.
Explanation:
Let :
A
= 0
.
084 m
,
k
= 74
.
5 N
/
m
,
and
v
max
= 0
.
636 m
/
s
.
A mass on a spring oscillates harmonically
according to
x
(
t
) =
A
sin(
ω t
+
φ
0
)
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wolz (cmw2833) – HW #12 – Antoniewicz – (57420)
2
so the velocity of the oscillating mass is
v
(
t
) =
d x
dt
=
A ω
cos(
ω t
+
φ
0
)
and the maximal speed is

v

max
=
A ω .
Thus the oscillation frequency is
ω
=

v

max
A
=
radicalbigg
k
M
M
=
k A
2

v

2
max
=
(74
.
5 N
/
m)(0
.
084 m)
2
(
v
)
2
=
1
.
29957 kg
.
004(part1of4)3.0points
An 11 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
)
.
Find the angular frequency of the oscillat
ing mass.
Correct answer: 95
.
3463 s
−
1
.
Explanation:
Let :
M
= 11 kg
and
k
= 1
×
10
5
N
/
m
.
When the mass moves out of equilibrium,
it suffers 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 .
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 Spring '08
 Turner
 Physics, Mass, Simple Harmonic Motion, Correct Answer

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