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CHAPTER
15
Wave Motion
1* ·
A rope hangs vertically from the ceiling. Do waves on the rope move faster, slower, or at the same speed as they
move from bottom to top? Explain.
They move faster as they move up because the tension increases due to the weight of the rope below.
2
·
(
a
)
The bulk modulus for water is 2.0
×
10
9
N/m
2
. Use it to find the speed of sound in water. (
b
)The speed of
sound in mercury is 1410 m/s. What is the bulk modulus for mercury (
r
= 13.6
×
10
3
kg/m
3
)?
(
a
) Use Equ. 154
(
b
) From Equ. 154,
B
=
v
2
r
10
1.41
=
m/s
10
/
10
2
3
3
9
×
×
=
v
m/s
B
= (1410
2
×
1.36
×
10
3
) N/m
2
= 2.70
×
10
10
N/m
2
3
·
Calculate the speed of sound waves in hydrogen gas at
T
= 300 K. (Take
M
= 2 g/mol and
g
= 1.4.)
Use Equ. 155;
M
= 2
×
10

3
kg/mol
10
300/2
8.314
1.4
3

×
×
×
=
v
m/s = 1320 m/s
4
·
A steel wire 7 m long has a mass of 100 g. It is under a tension of 900 N. What is the speed of a transverse wave
pulse on this wire?
Use Equ. 153
0.1/7.0
900
=
v
m/s = 251 m/s
5* ·
Transverse waves travel at 150 m/s on a wire of length 80 cm that is under a tension of 550 N. What is the mass
of the wire?
From Equ. 153,
m
=
F
/
v
2
;
m
=
m
L
=
FL
/
v
2
m
= (550
×
0.8/150
2
) kg = 0.0196 kg = 19.6 g
6
·
A wave pulse propagates along a wire in the positive
x
direction at 20 m/s. What will the pulse velocity be if we
(
a
) double the length of the wire but keep the tension and mass per unit length constant? (
b
) double the tension while
holding the length and mass per unit length constant? (
c
) double the mass per unit length while holding the other
variables constant?
See Equ. 153. (
a
) 20 m/s.
(
b
) 20
2
m/s = 28.8 m/s.
(
c
) 20/
2
m/s = 14.1 m/s.
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View Full DocumentChapter 15
Wave Motion
7
·
A steel piano wire is 0.7 m long and has a mass of 5 g. It is stretched with a tension of 500 N. (
a
) What is the
speed of transverse waves on the wire? (
b
) To reduce the wave speed by a factor of 2 without changing the tension,
what mass of copper wire would have to be wrapped around the steel wire?
(
a
) Use Equ. 153
(
b
) From Equ. 153,
m
f
= 4
m
i
0.005/0.7
500
=
v
m/s = 265 m/s
∆
m
= 3
m
i
= 15 g
8
·
The cable of a ski lift runs 400 m up a mountain and has a mass of 80 kg. When the cable is struck with a
transverse blow at one end, the return pulse is detected 12 s later. (
a
) What is the speed of the wave? (
b
) What is the
tension in the cable?
v
=
∆
x
/
∆
t
From Equ. 153,
F
=
v
2
m
/
L
v
= 800/12 m/s = 66.7 m/s
F
= (66.7
2
×
80/400) N = 889 N
9* ··
A common method for estimating the distance to a lightning flash is to begin counting when the flash is observed
and continue until the thunder clap is heard. The number of seconds counted is then divided by 3 to get the distance in
kilometers. (
a
) What is the velocity of sound in kilometers per second? (
b
) How accurate is this procedure? (
c
) Is a
correction for the time it takes for the light to reach you important? (The speed of light is
3
×
10
8
m/s.)
(
a
)
v
= 340 m/s = 0.340 km/s.
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 Spring '08
 staff

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