WAVE MOTION
E
XERCISES
Section 14.1 Waves and Their Properties
16.
Wave crests (adjacent wavefronts) take a time of one period to pass a fixed point, traveling at the wave speed (or
phase velocity) for a distance of one wavelength. Thus
/
18 m/(5.3 m/s)
3.40 s.
T
v
λ
=
=
=
17.
I
NTERPRET
This problem is about wave propagation. Given the speed and frequency of the ripples, we are asked to
compute the period and the wavelength.
D
EVELOP
Equation 14.1 relates the speed of the wave to its period, frequency, and wavelength:
v
f
T
λ
λ
=
=
This is the equation we shall use to solve the problem.
E
VALUATE
Equation 14.1 gives
(a)
1
1
5.2 Hz
0.192 s,
f
T
=
=
=
and
(b)
34 cm/s
5.2 Hz
6.54 cm.
v
f
λ
=
=
=
A
SSESS
The unit of frequency is Hz, with
1
1 Hz
1s
.

=
If the frequency is kept fixed, then increasing the
wavelength will increase the speed of propagation.
18.
From Equation 14.1,
8
6
/
(3
10
m/s)/(88.7
10
Hz)
3.38 m.
v f
λ
=
=
×
×
=
19.
I
NTERPRET
This problem is about wave propagation. Given the speed and frequency of various electromagnetic
waves, we are asked to compute their wavelength.
D
EVELOP
Equation 14.1 relates the speed of the wave to its period, frequency, and wavelength:
v
v
f
T
f
λ
λ
λ
=
=
→
=
This is the equation we shall use to solve the problem.
E
VALUATE
Since the speed of propagation of electromagnetic waves in vacuum is simply equal to the speed of
light,
8
3.0
10
m/s,
v
c
=
=
×
Equation 14.1 gives
(a)
8
6
3
10
m/s
10
Hz
300 m
c
f
λ
×
=
=
=
;
(b)
8
6
3
10
m/s
190
10
Hz
1.58 m
c
f
λ
×
×
=
=
=
;
(c)
8
10
3
10
m/s
10
Hz
0.03 m
3 cm
c
f
λ
×
=
=
=
=
;
(d)
8
13
3
10
m/s
6
4
10
Hz
7.5
10
m
7.5 m
c
f
λ
μ
×

×
=
=
=
×
=
;
(e)
8
14
3
10
m/s
7
6
10
Hz
5.0
10
m
500 nm
c
f
λ
×

×
=
=
=
×
=
;
(f )
8
18
o
3
10
m/s
10
1.0
10
Hz
3.0
10
m
3A
c
f
λ
×

×
=
=
=
×
=
(See Appendix C on units.)
A
SSESS
If the speed of propagation is kept fixed, then a higher frequency means a shorter wavelength.
20.
The wave speed can be calculated from the distance and the travel time, which, together with the frequency and
Equation 14.1, gives a wavelength of
/
(
/ )/
1200 km/(5
60
3.1)
1.29 km.
v f
d t
f
λ
=
=
=
×
×
=
Section 14.2 Wave Math
14.1
14
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14.2
Chapter 14
21.
I
NTERPRET
This problem is about the ultrasound wave. Given its frequency, and wavelength, we want to find its
angular frequency, wave number, and wave speed.
D
EVELOP
The relationships between the speed of the wave, its wave number, frequency, and wavelength are given
by Equations 13.6, 14.1, and 14.2:
2
,
2
f
v
f
k
T
ϖ
λ
π
λ
π
λ
=
=
=
=
E
VALUATE
(a)
Equation 13.6 gives
7
1
2
2
(4.8 MHz)
3.02
10
s
.
f
ϖ
π
π

=
=
=
×
(b)
Equation 14.2 gives
4
1
2
2
0.31 mm
2.03
10
m
.
k
π
π
λ

=
=
=
×
(c)
Using Equation 14.1, the speed of the ultrasound wave is
6
3
3
(4.8
10
Hz)(0.31
10
m)
1.49
10
m/s
v
f
λ

=
=
×
×
=
×
A
SSESS
The speed of the wave can also be computed as
7
1
3
4
1
3.02
10
s
1.49
10
m/s
2.03
10
m
v
k
ϖ


×
=
=
=
×
×
Thus, we see that the pairs
,
f
λ
and
,
k
ϖ
are equivalent ways to describe the same wave.
22.
From Equation 14.2,
(a)
1
2
/10.8 m
0.582 m
,
k
π

=
=
and
(b)
1
2
/(4.1 s)
1.53 s
.
ϖ
π

=
=
23.
I
NTERPRET
We are given a function that describes a traveling sinusoidal wave, and asked to compute various
physical quantities associated with the wave.
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 Spring '09
 Prof.Kim
 Physics, Frequency, Wavelength, shock waves

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