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Version One – Homework 10 – savrasov – 39822 – Dec 05, 2007
1
This printout should have 9 questions.
Multiplechoice questions may continue on
the next column or page – fnd all choices
beFore answering.
The due time is Central
time.
Tipler PSE5 14 104
13:01, calculus, numeric,
>
1 min, normal.
001
(part 1 oF 1) 10 points
A wooden cube with edge length 5 cm and
mass 62
.
5 g ±oats in water with one oF its
Faces parallel to the water surFace.
²ind the period oF oscillation in the vertical
direction iF the cube is pushed down slightly.
The density oF the water is 1 g
/
cc, and the
acceleration oF gravity is 9
.
81 m
/
s
2
.
Correct answer: 0
.
317187 s.
Explanation:
Let :
m
= 62
.
5 g
,
a
= 5 cm
,
ρ
= 1 g
/
cc
,
and
g
= 9
.
81 m
/
s
2
.
The density oF the wood is
ρ
wood
=
m
a
3
=
62
.
5 g
(5 cm)
3
= 0
.
5 g
/
cc
.
Applying
X
F
y
= 0 to the cube when it is
±oating in the water,
m g

F
B
= 0
.
Applying
X
T
y
=
m a
y
to the cube when it
is pushed down a small distance
y
,
m g

F
0
B
=
m a
y
.
Subtracting ,
F
B

F
0
B
=
m a
y
= Δ
F
B
.
²or
y
¿
1
,
Δ
F
B
=
dF
B
=

ρ V g
=

a
2
ρ g y
=
m
d
2
y
dt
2
.
The equation oF motion is
m
d
2
y
dt
2
=

a
2
ρ g y
d
2
y
dt
2
=

a
2
ρ g
m
y
=

ω
2
y
ω
=
a
r
ρ g
m
,
so the period oF the oscillation is
T
=
2
π
ω
=
2
π
a
r
m
ρ g
=
2
π
5 cm
s
62
.
5 g
(1 g
/
cc)(9
.
81 m
/
s
2
)
×
r
1m
100cm
=
0
.
317187 s
.
keywords:
Vibrating D2 Molecule
13:02, trigonometry, numeric,
>
1 min, nor
mal.
002
(part 1 oF 1) 10 points
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
.
3
×
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
.
19239
×
10
13
Hz.
Explanation:
Let :
M
D
= 2
M
H
.
The angular Frequencies depend only on
spring constant and mass:
ω
=
r
k
M
∝
r
1
M
ω
D
=
s
k
M
D
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View Full DocumentVersion One – Homework 10 – savrasov – 39822 – Dec 05, 2007
2
ω
H
=
s
k
M
H
The spring constants
k
are the same, so
ω
D
ω
H
=
s
M
H
M
D
=
s
M
H
2
M
H
=
r
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
.
3
×
10
14
Hz
√
2
=
9
.
19239
×
10
13
Hz
.
keywords:
Uniform Rod as a Pendulum 05
13:06, calculus, numeric,
>
1 min, normal.
003
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 Fall '09
 RichardScalettar
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