14_InstSolManual_PC - WAVE MOTION 14 v= EXERCISES Section...

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

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.