hopkins (tlh982) – HW01 – criss – (4908)
1
This printout should have 19 questions.
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beFore answering.
001
10.0 points
At
t
= 0, a transverse wave pulse in a wire is
described by the Function
y
= 5
e

x
2
/
2
,
where
x
and
t
are given in SI units.
Which Function describes this wave iF it is
traveling in the negative
x
direction with a
speed oF 4
.
5 m
/
s?
1.
y
= 5
e

(
x
2

4
.
5
t
)
/
2
2.
y
= 5
e

x
2
/
2
+ 4
.
5
t
3.
y
= 5
e

(
x

4
.
5
t
)
2
/
2
4.
y
= 5
e

(
x
2
+4
.
5
t
)
/
2
5.
y
= 5
e

x
2
/
2

4
.
5
t
6.
y
= 5
e

(
x
+4
.
5
t
)
2
/
2
correct
7.
y
= 5
e

x
2
/
2
Explanation:
±or the wave traveling in the negative di
rection with the speed oF 4
.
5 m
/
s
,
we need to
replace
x
with
x
+ 4
.
5
t
which gives
y
= 5
e

(
x
+4
.
5
t
)
2
/
2
.
002
10.0 points
A carbon steel piano wire 1 m long with a
crosssectional area oF 2
×
10

6
m
2
and mass
0
.
1 kg is stretched 30 mm.
±or carbon steel, the Young’s modulus is
2
×
10
11
N
/
m
2
.
Determine the speed oF transverse waves on
the string.
Correct answer: 346
.
41 m
/
s.
Explanation:
Let :
L
= 1 m
,
Δ
L
= 30 mm
,
A
= 2
×
10

6
m
2
,
m
= 0
.
1 kg
,
and
Y
= 2
×
10
11
N
/
m
2
.
F
=
Y A
Δ
L
L
μ
=
m
L
v
=
r
F
μ
=
r
Y A
Δ
L / L
m / L
=
R
Y A
Δ
L
m
=
b
(2
×
10
11
N
/
m
2
) (2
×
10

6
m
2
)
0
.
1 kg
×
(0
.
03 m)
B
1
/
2
=
346
.
41 m
/
s
.
003
(part 1 oF 4) 10.0 points
A traveling wave propagates according to the
expression
y
= (5
.
55 cm) sin[(1
.
98 cm

1
)
x

(2
.
79 s

1
)
t
]
,
where
x
is in centimeters, and
t
is in seconds.
Determine the amplitude oF the wave.
Correct answer: 5
.
55 cm.
Explanation:
Given a wave
y
=
A
sin(
k x

ω t
), where
the amplitude oF the wave is
A
. In our case,
A
= 5
.
55 cm
.
004
(part 2 oF 4) 10.0 points
Determine the wavelength oF the wave.
Correct answer: 3
.
17333 cm.
Explanation:
The angular wave number oF the wave is
k
=
1
.
98 cm

1
and its relation to the wavelength
λ
is
λ
=
2
π
k
= 3
.
17333 cm
.
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2
005
(part 3 of 4) 10.0 points
Determine the frequency of the wave.
Correct answer: 0
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 Spring '08
 CRISS
 Correct Answer

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