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Chapter 15
Page 1
CHAPTER 15 – Wave Motion
1.
The speed of the wave is
v
=
f
λ
=
/
T
= (9.0 m)/(4.0 s) =
2.3 m/s
.
2.
For AM we find the wavelengths from
AMhigher
=
v
/
f
AMlower
= (3.00
×
10
8
m/s)/(550
×
10
3
Hz) =
545 m
;
AMlower
=
v
/
f
AMhigher
= (3.00
×
10
8
m/s)/(1600
×
10
3
Hz) =
188 m
.
For FM we have
FMhigher
=
v
/
f
FMlower
= (3.00
×
10
8
m/s)/(88
×
10
6
Hz) =
3.4 m
;
FMlower
=
v
/
f
FMhigher
= (3.00
×
10
8
m/s)/(108
×
10
6
Hz) =
2.78 m
.
3.
We find the wavelength from
v
=
f
;
330 m/s = (262 Hz)
, which gives
=
1.26 m
.
4.
We find the speed of the longitudinal (compression) wave from
v
= (
B
/
ρ
)
1/2
for fluids and
v
= (
E
/
)
1/2
for solids.
(
a
)
For water we have
v
= (
B
/
)
1/2
= [(2.0
×
10
9
N/m
2
)/(1.00
×
10
3
kg/m
3
)]
1/2
=
1.4
×
10
3
m/s
.
(
b
)
For granite we have
v
= (
E
/
)
1/2
= [(45
×
10
9
N/m
2
)/(2.7
×
10
3
kg/m
3
)]
1/2
=
4.1
×
10
3
m/s
.
5.
The speed of the longitudinal (compression) wave is
v
= (
E
/
)
1/2
,
so the wavelength is
=
v
/
f
= (
E
/
)
1/2
/
f
= [(100
×
10
9
N/m
2
)/(7.8
×
10
3
kg/m
3
)]
1/2
/(5000 Hz) =
0.72 m
.
6.
We find the speed of the wave from
v
= [
F
T
/(
m
/
L
)]
1/2
= {(120 N)/[(0.65 kg)/(30 m)]}
1/2
= 74.4 m/s.
We find the time from
t
=
L
/
v
= (30 m)/(74.4 m/s) =
0.40
s.
7.
We find the tension from the speed of the wave:
v
= [
F
T
/(
m
/
L
)]
1/2
;
(4.8 m)/(0.85 s) = {
F
T
/[(0.40 kg)/(4.8 m)]}
1/2
, which gives
F
T
=
2.7 N
.
8.
The speed of the longitudinal wave is
v
= (
B
/
)
1/2
,
so the distance down and back that the wave traveled is
2
D
=
vt
= (
B
/
)
1/2
t
;
2
D
= [(2.0
×
10
9
N/m
2
)/(1.00
×
10
3
kg/m
3
)]
1/2
(3.5 s), which gives
D
= 2.5
×
10
3
m =
2.5 km
.
9. (
a
) Because the pulse travels up and back, the speed is
v
= 2
L
/
t
= 2(600 m)/(16 s) =
75 m/s
.
(
b
) The mass density of the cable is
µ
=
m
/
L
=
AL
/
L
=
A
.
We find the tension from
v
= (
F
T
/
)
1/2
= (
F
T
/
A
)
1/2
;
75 m/s = [
F
T
/(7.8
×
10
3
kg/m
3
)p(0.75
×
10
–2
m)
2
]
1/2
, which gives
F
T
=
7.8
×
10
3
N
.
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View Full DocumentChapter 15
Page 2
10. (
a
)
The shape is maintained
and moves 1.80 m in 1.00 s.
(
b
)
At the instant shown, point A is moving down.
We use the slope of the string to estimate the
vertical speed.
In the time to move vertically from 1 cm to – 1 cm, we estimate the string moves
30 cm, so the time is
?
t
= (0.30 m)/(1.80 m/s) = 0.17 s.
The vertical velocity is – (2 cm)/(0.17 s) =
– 0.1 m/s (down)
.
11. (
a
) Because both waves travel the same distance, we have
?
t
= (
d
/
v
S
) – (
d
/
v
P
) =
d
[(1/
v
S
) – (1/
v
P
)];
94 s =
d
{[1/(5.5 km/s)] – [1/(9.0 km/s)]}, which gives
d
=
1.3
×
10
3
km
.
(
b
) The direction of the waves is not known, thus the position of the epicenter
cannot be determined
.
Because two circles have two intersections, it would take at least two more stations.
12. Because the speed, frequency, and medium are the same for the two waves, the intensity depends on the
amplitude only:
I
∝
D
M
2
.
For the ratio of intensities we have
I
2
/
I
1
= (
D
M2
/
D
M1
)
2
;
2 = (
D
M2
/
D
M1
)
2
, which gives
D
M2
/
D
M1
=
1.41
.
13. We assume that the wave spreads out uniformly in all directions.
(
a
) The intensity will decrease as 1/
r
2
, so the ratio of intensities is
I
2
/
I
1
= (
r
1
/
r
2
)
2
= [(10 km)/(20 km)]
2
=
0.25
.
(
b
) Because the intensity depends on
D
M
2
, the amplitude will decrease as 1/
r
, so the ratio of
amplitudes is
D
M2
/
D
M1
=
r
1
/
r
2
= (10 km)/(20 km) =
0.50
.
14. We assume that the wave spreads out uniformly in all directions.
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 Spring '09
 Fielding
 Physics

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