homework 20 – RAHMAN, TARIQUE – Due: Apr 18 2008, 3:00 am
1
Question 1, chap 14, sect 4.
part 1 of 1
10 points
A steel cable with a crosssectional area of
1
.
5 cm
2
has a mass of 2
.
95 kg
/
m
.
Its Young’s
modulus is 1
.
27
×
10
11
N
/
m
2
.
The acceleration of gravity is 9
.
8 m
/
s
2
.
If 762 m of the cable is hung over a vertical
cliF, how much does the cable stretch under
its own weight?
Correct answer: 44
.
0588 cm (tolerance
±
1
%).
Explanation:
Basic Concepts:
Δ
L
L
=
F
Y A
Let :
A
= 1
.
5 cm
2
= 0
.
00015 m
2
,
Y
= 1
.
27
×
10
11
N
/
m
2
,
L
= 762 m
,
λ
= 2
.
95 kg
/
m
,
The tension in this cable is not uniform, so
this becomes a fairly di±cult problem. ²rom
the de³nition of Young’s modulus, we have
Δ
L
L
=
F
Y A
.
F
=
mg
=
λy g
weight of the cable at posi
tion
y .
Δ
y
=
i
L
0
0
Δ
L
L
dy
=
λg
Y A
i
L
0
0
y dy
=
1
2
λg L
2
0
Y A
=
1
2
(2
.
95 kg
/
m) (9
.
8 m
/
s
2
) (762 m)
2
(1
.
27
×
10
11
N
/
m
2
) (0
.
00015 m
2
)
= 0
.
440588 m =
44
.
0588 cm
.
Question 2, chap 14, sect 4.
part 1 of 1
10 points
Use bulk modulus
B
= 2
.
1
×
10
9
Pa . What
increase of pressure is required to change the
volume of a sample of water by 0
.
71 %?
Correct answer: 1
.
491
×
10
7
Pa (tolerance
±
1 %).
Explanation:
²rom the de³nition of bulk modulus, the
change in pressure is
Δ
P
=

Δ
V
V
B.
²or our situation,
Δ
V
V
= 0
.
0071
.
Then,
Δ
P
=

(0
.
0071)
B
=

(0
.
0071) 2
.
1
×
10
9
Pa
= 1
.
491
×
10
7
Pa
.
Question 3, chap 15, sect 1.
part 1 of 1
10 points
When an object oscillating in simple har
monic motion is at its maximum displacement
from the equilibrium position, which of the
following is true of the values of its speed and
the magnitude of the restoring force?
Magnitude of
Speed
Restoring Force
1.
Zero
Maximum
correct
2.
Maximum
Zero
3.
1
2
maximum
1
2
maximum
4.
Zero
Zero
5.
Maximum
1
2
maximum
Explanation:
The maximum displacement occurs at the
turning points, which are the points where the
velocity or speed is zero.
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View Full Documenthomework 20 – RAHMAN, TARIQUE – Due: Apr 18 2008, 3:00 am
2
The magnitude of the restoring force is
given by Hooke’s law
F
=

k x,
where
k
is the spring constant and
x
is the
displacement. Since
x
is a maximum,
F
is a
maximum.
From a di±erent perspective, the displace
ment from the equilibrium position can be
written as
y
=
A
sin
θ ,
where
θ
is the phase of the oscillation.
When the object is at its maximum dis
placement,
θ
=
π
2
.
The speed is then
v
=
ω A
cos
θ
=
ω A
cos
π
2
= 0
and the restoring force is
F
=
mA ω
2
sin
θ
=
mA ω
2
π
2
=
mA ω
2
,
at its maximum value.
Question 4, chap 15, sect 1.
part 1 of 1
10 points
A mass attached to a spring oscillates back
and forth as indicated in the position vs. time
plot below.
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