Solutions for ch 17 and ch 24
Answers for ch. 17
33. REASONING A standing wave is composed of two oppositely traveling waves.
F
The speed v of these waves is given by v 
(Equation 16.2), where F is the
m/ L
tension in the string and m/L is its linear density (mass per unit length). Both F and
m/L are given in the statement of the problem. The wavelength  of the waves can be
obtained by visually inspecting the standing wave pattern. The frequency of the
waves is related to the speed of the waves and their wavelength by f = v/ (Equation
16.1).
SOLUTION
a. The speed of the waves is
v
F
280 N

 180 m/s
m/ L
8.5 103 kg/m
1.8 m
b. Two loops of any standing wave comprise one
wavelength. Since the string is 1.8 m long and consists
of three loops (see the drawing), the wavelength is

2
3
1.8 m   1.2 m



c. The frequency of the waves is
v 180 m/s
f  
 150 Hz

1.2 m
19. SSM REASONING The beat frequency of two sound waves is the difference
between the two sound frequencies. From the graphs, we see that the period of the
wave in the upper text figure is 0.020 s, so its frequency is
f1  1/ T1  1/(0.020 s) = 5.0 101 Hz . The frequency of the wave in the lower figure
is f 2  1/(0.024 s)=4.2 101 Hz .
SOLUTION The beat frequency of the two sound waves is
f beat  f1  f 2  5.0  101 Hz – 4.2  101 Hz = 8 Hz
57. REASONING The natural frequencies of a tube open at only one end are given by
 v 
Equation 17.5 as f n  n 
 , where n is any odd integer (n = 1, 3, 5, …), v is the
 4L 
speed of sound, and L is the length of the tube. We can use this relation to find the
value for n for the 450-Hz sound and to determine the length of the pipe.
SOLUTION
 v 
a. The frequency fn of the 450-Hz sound is given by 450 Hz  n 
 . Likewise,
 4L 
 v 
the frequency of the next higher harmonic is 750 Hz   n  2  
 , because n is an
 4L 
odd integer and this means that the value of n for the next higher harmonic must be n
+ 2. Taking the ratio of these two relations gives
 v 
n  2 


750 Hz
 4L   n  2

450 Hz
n
 v 
n

 4L 
Solving this equation for n gives n = 3 .
 v 
b. Solving the equation 450 Hz  n 
 for L and using n = 3, we find that the
 4L 
length of the tube is
 343 m/s 
 v 
L  n
 3

  0.57 m
4f 
 n
 4  450 Hz  
5.
SSM REASONING The tones from the two speakers will produce destructive
interference with the smallest frequency when the path length difference at C is
one-half of a wavelength. From Figure 17.7, we see that the path length difference is
s  sAC – sBC . From Example 1, we know that sAC  4.00 m , and from Figure
sBC  2.40 m .
17.7,
Therefore,
s  4.00 m – 2.40 m = 1.60 m .
the
path
length
SOLUTION Thus, destructive interference will occur when

2
 1.60 m
or
 = 3.20 m
This corresponds to a frequency of
v 343 m / s
f  
 107 Hz

3.20 m
difference
is
1.
(d) If we add pulses 1 and 4 as per the principle of linear superposition, the resultant
is a straight horizontal line that extends across the entire graph.
10. (d) According to the discussion in Section 17.5, one loop of a transverse standing
wave corresponds to one-half a wavelength. The two loops in the top picture mean
that the wavelength of 1.2 m is also the distance L between the walls, so L = 1.2 m.
The bottom picture contains three loops in a distance of 1.2 m, so its wavelength is
2 1.2 m  0.8 m.

3
11. (b) The frequency of a standing wave is directly proportional to the speed of the
traveling waves that form it (see Equation 17.3). The speed of the waves, on the
other hand, depends on the mass m of the string through the relation
v  F /  m / L  , so the smaller the mass, the greater is the speed and, hence, the
greater the frequency of the standing wave.
14. (c) One loop of a longitudinal standing wave corresponds to one-half a wavelength.
Since this standing wave has two loops, its wavelength is equal to the length of the
tube, or 0.80 m.
Answers for Ch 24
3. (b) The magnitudes of the electric and magnetic fields of the wave are proportional to
each other, according to E = cB (Equation 24.3). As Section 24.4 discusses, the wave
carries equal amounts of electric and magnetic energy.
 v 
6. (d) The observed frequency is f o  fs 1  rel  according to Equation 24.6. The
c 

frequency fs emitted by the source is the same in each case, so that only the relative
speed vrel and the direction of the relative motion determine the observed frequency.
In each case either the source or the observer is moving, so the relative speed is just
the magnitude of the velocity vector shown in the drawing. Since the velocity vector
has the same magnitude in each case, the relative speed is the same in each case.
Thus, it is only the direction of the relative motion that needs to be considered here.
In A and C the source and observer are moving apart at the same relative speed, the
minus sign applies in Equation 24.6, and the observed frequencies are the same. In B
and D they are coming together at the same relative speed, the plus sign applies in
Equation 24.6, and the observed frequencies are the same, but greater than that in A
and C.
SSM REASONING AND SOLUTION The energy is equal to the power P
27.
multiplied by the time t. The power, on the other hand, is equal to product of the
intensity S of the wave and the area A through which the wave passes.
Energy = P t = (SA)t = (1390 W/m2)(25 m  45 m)(3600 s)  5.6  10 J
9
________________________________________________________________________
______
29.
REASONING AND SOLUTION Since the sun emits radiation uniformly in all
directions, at a distance r from the sun's center, the energy spreads out over a sphere of
2
surface area 4 r . Therefore, according to S  P /(4 r 2 ) (Equation 16.9), the total
power radiated by the sun is
P  S (4 r 2 )  (1390 W/m 2 )(4 )(1.50 1011 m) 2  3.93 1026 W
35. REASONING The Doppler effect for electromagnetic radiation is given by
Equation 24.6;
v 

f o  fs 1  rel 
c 

if vrel  c
where f o is the observed frequency, f s is the frequency emitted by the source, and
vrel is the speed of the source relative to the observer. As discussed in the text, the
plus sign applies when the source and the observer are moving toward one another,
while the minus sign applies when they are moving apart. According to
Equation 16.1, the wavelength of these waves is  = c/f. Therefore, the Doppler shift
can be written in terms of wavelengths:
1
o

vrel 
1
1 

s 
c 
if vrel  c
SOLUTION
a. The wavelength o of the light observed on earth is greater than the wavelength s
of the light when it is emitted from the distant galaxy (the source). Therefore, the
frequency of the light observed on earth is less than the frequency of the light when it
is emitted from the distant galaxy. Thus, the quantity in the brackets in Equation 24.6
must be less than one; it must be equal to 1 – (vrel / c)  . Since the minus sign applies,
we can conclude that
the galaxy must be receding from the earth .
b. We can find the speed of the galaxy relative to the earth by solving the
wavelength version of Equation 24.6 for vrel:
  
 434.1 nm 
6
vrel  c 1 – s   (3.0 108 m/s) 1 
  3.110 m/s
  
 438.6 nm 
o
