18 Chapter Solutions
18.1 The resultant wave function has the form y = 2A0 cos (a)
sin kx t + 2 2
( /4) = 2(5.00) cos = 9.24 m 2 2
A = 2A0 cos f=
(b) *18.2
1200 = = 600 Hz 2 2
We write the second wave function as y2 = A sin(kx t + ) y2 = (0.0800 m) sin[2 (0.100x 80.0t) + ] Then y1 + y2 = (0.0800 m) sin[2 (0.100x 80.0t)] + (0.0800 m) sin[2 (0.100x 80.0t) + ] = 2(0.0800 m) cos
sin 2 (0.100x 80.0t) + 2 2
3
We require 2(0.0800 m) cos 3 = 2 2
= 0.0800 2
cos
= 60.0 =
3
Then the second wave function is y2 = (0.0800 m) sin 2 0.100x 80.0t + 1 6
y2 = (0.0800 m) sin [2(0.100x 80.0t + 0.167)]
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2
Chapter 18 Solutions
18.3
Suppose the waves are sinusoidal. The sum is (4.00 cm) sin(kx t) + (4.00 cm) sin(kx t + 90.0) 2(4.00 cm) sin(kx t + 45.0) cos 45.0 So the amplitude is (8.00 cm) cos 45.0 = 5.66 cm 2A0 cos 1 1 = A0, so 2 = cos 2 = 60.0 = 3 2 2 3 T 1 = = 3 3f 3v
18.4
Thus, the phase difference is = 120 =
This phase difference results if the time delay is 3.00 m = 0.500 s 3(2.00 m/s)
Time delay = 18.5
Waves reflecting from the near end travel 28.0 m (14.0 m down and 14.0 m back), while waves reflecting from the far end travel 66.0 m. The path difference for the two waves is: r = 66.0 m 28.0 m = 38.0 m Since = or, v r (r)f (38.0 m)(246 Hz) , then = = = 27.254 f v 343 m/s
r = 27.254
The phase difference between the two reflected waves is then
= 0.254(1 cycle) = 0.254(2 rad) = 91.3
18.6 (a) First we calculate the wavelength: = v 344 m/s = = 16.0 m f 21.5 Hz 1 2
Then we note that the path difference equals 9.00 m 1.00 m =
Therefore, the receiver will record a minimum in sound intensity.
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Chapter 18 Solutions
(b) If the receiver is located at point (x, y), then we must solve: (x + 5.00)2 + y2 (x 5.00)2 + y2 = 1 2 1 2
3
Then,
(x + 5.00)2 + y2 =
(x 5.00)2 + y2 +
Square both sides and simplify to get: 20.0x Upon squaring again, this reduces to: 400x2 10.02x +
2 = (x 5.00) 2 + y 2 4
4 = 2(x 5.00)2 + 2y2 16.0
Substituting = 16.0 m, and reducing, we have: 9.00x2 16.0y2 = 144 or x2 y2 =1 16.0 9.00
(When plotted this yields a curve called a hyperbola.) 18.7 We suppose the man's ears are at the same level as the lower speaker. Sound from the upper speaker is delayed by traveling the extra distance L 2 + d2 L.
He hears a minimum when this is (2n 1) /2 with n = 1, 2, 3, . . . Then, L2 + d2 L = (n 1/2)v/f L2 + d2 = (n 1/2)v/f + L L2 + d2 = (n 1/2)2v2/f2 + L2 + 2(n 1/2)vL/f L= d2 (n 1/2)2v2/f2 2(n 1/2)v/f n = 1, 2, 3, . . .
This will give us the answer to (b). The path difference starts from nearly zero when the man is very far away and increases to d when L = 0. The number of minima he hears is the greatest integer solution to d (n 1/2) v/f n = greatest integer df/v + 1/2 (a) df/v + 1 1 = (4.00 m)(200/s)/330 m/s + = 2.92 2 2
He hears two minima.
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4
Chapter 18 Solutions
(b)
With n = 1, L= d2 (1/2)2v2/f2 (4.00 m)2 (330 m/s)2/4(200/s)2 = 2(1/2)v/f (330 m/s)/200/s
L = 9.28 m with n = 2 L= 18.8 d2 (3/2)2v2/f2 = 1.99 m 2(3/2)v/f
Suppose the man's ears are at the same level as the lower speaker. Sound from the upper speaker is delayed by traveling the extra distance r = He hears a minimum when L 2 + d2 L.
r = (2n 1) with n = 1, 2, 3, . . . 2
Then, L2 + d2 L = (n 1/2)(v/f) L2 + d2 = (n 1/2)(v/f) + L L2 + d2 = (n 1/2)2(v/f)2 + 2(n 1/2)(v/f)L + L2 (1)
Equation 1 gives the distances from the lower speaker at which the man will hear a minimum. The path difference r starts from nearly zero when the man is very far away and increases to d when L = 0. (a) The number of minima he hears is the greatest integer value for which L 0. This is the same as the greatest integer solution to d (n 1/2)(v/f), or number of minima heard = nmax = greatest integer d(f/v) + 1/2 (b) From Equation 1, the distances at which minima occur are given by Ln = 18.9 d 2 (n 1/2) 2(v/f)2 where n = 1, 2, . . ., n max 2(n 1/2)(v/f)
y = (1.50 m) sin (0.400x) cos (200t) = 2A0 sin kx cos t Therefore, k= 2 rad = 0.400 m
=
2 = 15.7 m 0.400 rad/m
and = 2f, so f =
200 rad/s = = 31.8 Hz 2 2 rad
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Chapter 18 Solutions
The speed of waves in the medium is v = f = 200 rad/s 2 f = = = 500 m/s 2 k 0.400 rad/m x cos (40t) 2
5
18.10
y = 0.0300 m cos (a)
nodes occur where y = 0: x = (2n + 1) 2 2 so x = (2n + 1) = , 3 , 5 , . . . 0.400 = 0.0294 m 2
(b) 18.11
ymax = 0.0300 m cos
The facing speakers produce a standing wave in the space between them, with the spacing between nodes being dNN = 343 m/s v = = = 0.214 m 2 2f 2(800 s1)
If the speakers vibrate in phase, the point halfway between them is an antinode, at 1.25 m = 0.625 m from either speaker. 2
Then there is a node at 0.625 m 0.214 m = 0.518 m , a node at 2
0.518 m 0.214 m = 0.303 m , a node at 0.303 m 0.214 m = 0.0891 m , a node at 0.518 m + 0.214 m = 0.732 m , a node at 0.732 m + 0.214 m = 0.947 m , and a node at 0.947 m + 0.214 m = 1.16 m from either speaker
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6
Chapter 18 Solutions
*18.12
(a)
The resultant wave is y = 2A sin kx +
cos t 2 2
The nodes are located at kx +
= n 2
so
x=
n k 2k
which means that each node is shifted (b) The separation of nodes is x = (n + 1)
to the left. 2k
n k 2k k 2k
x =
= k 2
The nodes are still separated by half a wavelength. 18.13 y1 = 3.00 sin [(x + 0.600t)] cm y2 = 3.00 sin [(x 0.600t)] cm
y = y1 + y2 = [3.00 sin (x) cos (0.600 t) + 3.00 sin (x) cos (0.600t)] cm = (6.00 cm) sin (x) cos (0.600t) (a) We can take cos(0.600t) = 1 to get the maximum y. At x = 0.250 cm, ymax = (6.00 cm) sin(0.250) = 4.24 cm (b) (c) At x = 0.500 cm, ymax = (6.00 cm) sin(0.500) = 6.00 cm Now take cos (0.600t) = 1 to get ymax: At x = 1.50 cm, (d) ymax = (6.00 cm) sin(1.50)(1) = 6.00 cm
The antinodes occur when x = n/4 (n = 1, 3, 5, . . . ). But k = 2/ = , so = 2.00 cm, and x1 = /4 = 0.500 cm as in (b) x2 = 3/4 = 1.50 cm as in (c) x3 = 5/4 = 2.50 cm
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Chapter 18 Solutions
18.14 (a) Using the given parameters, the wave function is y = 2 sin
7
x cos (10t) 2 x =1 2
We need to find values of x for which sin
This condition requires that
x 1 = n + ; n = 0, 1, 2, . . . 2 2
For n = 0, x = 1.00 cm and for n = 1, x = 3.00 cm Therefore, the distance between antinodes, x = 2.00 cm A = 2 sin
(b) 18.15
x ; when x = 0.250 cm, A = 2.40 cm 2
y = 2A0 sin kx cos t
2y = 2A0k2 sin kx cos t x2 2y = 2A02 sin kx cos t t2
Substitution into the wave equation gives 2A0k2 sin kx cos t = 1 (2A02 sin kx cos t) v 2
This is satisfied, provided that v = 0.100 kg = 0.0500 kg/m 2.00 m T = (20.0 kg m/s2) 0.0500 kg/m
k
18.16
=
v=
= 20.0 m/s
For the simplest vibration possibility, NAN, dNN = 2.00 m =
2
= 4.00 m
f=
v (20.0 m/s) = = 5.00 Hz 4.00 m
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8
Chapter 18 Solutions
For the second state NANAN, dNN = 1.00 m f=
= 2.00 m
(20.0 m/s) = 10.0 Hz 2.00 m
For the third resonance, NANANAN, dNN = 2.00 m 3
= 1.33 m
f = 15.0 Hz
The mode mentioned in the problem has dNN = 0.400 m
= 0.800 m
f = 25.0 Hz
It is the fifth allowed state . 18.17 L = 30.0 m f1 = v 2L T 1/2 = 47.1 m/s = 9.00 103 kg/m T = 20.0 N
where v =
so
f1 =
47.1 = 0.786 Hz 60.0 f3 = 3f1 = 2.36 Hz f4 = 4f1= 3.14 Hz
f2 = 2f1 = 1.57 Hz
Goal Solution G: The string described in the problem is very long, loose, and somewhat massive, so it should have a very low fundamental frequency, maybe only a few vibrations per second. O: The tension and linear density of the string can be used to find the wave speed, which can then be used along with the required wavelength to find the fundamental frequency. A: The wave speed is v = F = 20 N = 47.1 m/s 9.0 103 kg/m
For a vibrating string of length L fixed at both ends, the wavelength of the fundamental frequency is = 2L = 60.0 m; and the frequency is f1 = v v 47.1 m/s = = = 0.786 Hz 2L 60 m
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Chapter 18 Solutions
9
The next three harmonics are f2 = 2f1 = 1.57 Hz f3 = 3f1 = 2.36 Hz f4 = 4f1 = 3.14 Hz L: The fundamental frequency is even lower than expected, less than 1 Hz. In fact, all 4 of the lowest resonant frequencies are below the normal human hearing range (20 to 17 000 Hz), so these harmonics are not even audible.
18.18
L = 120 cm (a)
f = 120 Hz
For four segments, L = 2 or
= 60.0 cm = 0.600 m
(b)
v = f = 72.0 m/s f1 = v 72.0 = = 30.0 Hz 2L 2(1.20)
18.19
dNN = 0.700 m
= 1.40 m
f = v = 308 m/s = (a) (b) T = 163 N f3 = 660 Hz T (1.20 103)/(0.700)
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10
Chapter 18 Solutions
Goal Solution G: The tension should be less than 100 lbs. (~500 N) since excessive force on the 4 cello strings would break the neck of the instrument. If the string vibrates in three segments, there will be three antinodes (instead of one for the fundamental mode), so the frequency should be three times greater than the fundamental. O: The length of the string can be used to find the wavelength, which can be used with the fundamental frequency to find the wave speed. The tension can then be found from the wave speed and linear mass density of the string. When the string vibrates in the lowest frequency mode, the length of string forms a standing wave where L = /2 (see Figure 18.2b), so the fundamental harmonic wavelength is
A:
= 2L = 2(0.700 m) = 1.40 m
and the velocity is v = f = (220 s1)(1.40 m) = 308 m/s From the tension equation v = T = T m/L we get
(a) T =
v2m (308 m/s)2(1.20 103 kg) = = 163 N L 0.700 m
(b) For the third harmonic, the tension, linear density, and speed are the same. However, the string vibrates in three segments so that the wavelength is one third as long as in the fundamental (see Figure 18.2d).
3 = /3
From the equation f = v , we find that the frequency is three times as high: f3 = L: v v = 3 = 3f = 660 Hz 3
The tension seems reasonable, and the third harmonic is three times the fundamental frequency as expected. Related to part (b), some stringed instrument players use a technique to double the frequency of a note by "cutting" a vibrating string in half. When the string is suddenly held at its midpoint to form a node, the second harmonic is formed, and the resulting note is one octave higher (twice the original fundamental frequency).
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Chapter 18 Solutions
v T 1/2 , where v = 2L If L is doubled, then f1 L1 will be reduced by a factor 1 . 2 1 2 2 . .
11
*18.20
f1 =
(a)
(b)
If is doubled, then f1 1/2 will be reduced by a factor If T is doubled, then f1
(c) 18.21
T will increase by a factor of
L = 60.0 cm = 0.600 m fn = where v= nv 2L
T = 50.0 N
= 0.100 g/cm = 0.0100 kg/m
T 1/2 = 70.7 m/s 70.7 = 58.9n = 20,000 Hz 1.20
fn = n
Largest n = 339 f = 19.976 kHz v = T 1 = T4 2 d 2 L
18.22
f=
since =
M V AL = = L L L 4Told4 2 old (2d old)2 Lold/2 Told4 2 2 = 2 fold = 800 Hz Lold
fnew =
=
old
2 dold
18.23
G = 2(0.350 m) = A = 2LA =
v fA fG f A
v fG
LG LA = LG
LG = LG Error! ) = (0.350 m) Error! = 0.0382 m
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12
Chapter 18 Solutions
Thus, LA = LG 0.0382 m = 0.350 m 0.0382 m = 0.312 m, or the finger should be placed 31.2 cm from the bridge . v 1 = 2fA 2 f A dT 4 fA T T
LA =
dLA =
dL A 1 dT = LA 2 T dLA dT 0.600 cm =2 =2 = 3.84% T LA (35.0 3.82) cm 18.24 In the fundamental mode, the string above the rod has only two nodes, at A and B, with an anti-node halfway between A and B. Thus, L -- = AB = 2 cos or
A
=
2L cos
L
B
Since the fundamental frequency is f, the wave speed in this segment of string is v = f = 2L f cos T = -- m/AB TL m cos
M
Also, v =
T =
T
where T is the tension in this part of the string. Thus, 2L f = cos TL m cos or 4L 2 f 2 TL = 2 m cos cos
F
and the mass of string above the rod is: m= T cos 4Lf 2 [Equation 1]
Mg
Now, consider the tension in the string. The light rod would rotate about point P if the string exerted any vertical force on it. Therefore, recalling Newton's third law, the rod must exert only a horizontal force on the string. Consider a free-body diagram of the string segment in contact with the end of the rod. Fy = T sin Mg = 0 T = Mg sin
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Chapter 18 Solutions
Then, from Equation 1, the mass of string above the rod is m= Mg cos Mg = 2 sin 4Lf 2 4Lf tan
13
18.25
(a)
Let n be the number of nodes in the standing wave resulting from the 25.0-kg mass. Then n + 1 is the number of nodes for the standing wave resulting from the 16.0-kg mass. For standing waves, = 2L/n, and the frequency is f = v/ . n 2L Tn n+1 , and also f = 2L Tn = Tn + 1 Tn + 1
Thus, f =
Thus,
n+1 = n
(25.0 kg)g 5 = (16.0 kg)g 4
Therefore, 4n + 4 = 5n, or n = 4 4 2(2.00 m) (25.0 kg)(9.80 m/s2) = 350 Hz 0.00200 kg/m
Then, f = (b)
The largest mass will correspond to a standing wave of 1 loop 1 2(2.00 m) m(9.80 m/s2) 0.00200 kg/m
(n = 1), so 350 Hz = yielding m = 400 kg *18.26
Using the frets does not change the speed of the wave. Therefore, if dNN is the distance between adjacent nodes,
1f1 = 2f2 = 2dNN1f1 = 2dNN2f2
dNN2 = dNN1
or
f1 2349 Hz = 21.4 cm = 22.7 cm f 2 2217 Hz
Thus, the distance between frets is dNN2 dNN1 = 22.7 cm 21.4 cm = 1.27 cm *18.27 The natural frequency is f= 1 1 = T 2 g 1 = L 2 9.80 m/s2 = 0.352 Hz 2.00 m
The big brother must push at this same frequency of 0.352 Hz to produce resonance.
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14
Chapter 18 Solutions
*18.28
The distance between adjacent nodes is one-quarter of the circumference. dNN = dAA = so
20.0 cm = = 5.00 cm 2 4
v 900 m/s = = 9000 Hz = 9.00 kHz 0.100 m
= 10.0 cm and f =
The singer must match this frequency quite precisely for some interval of time to feed enough energy into the glass to crack it. *18.29 (a) (b) The wave speed is v = 9.15 m = 3.66 m/s 2.50 s
From Figure P18.29, there are antinodes at both ends, so the distance between adjacent antinodes is dAA =
= 9.15 m, and the wavelength is = 18.3 m 2
v 3.66 m/s = = 0.200 Hz 18.3 m
The frequency is then f =
We have assumed the wave speed is the same for all wavelengths. *18.30 The wave speed is v = gd = (9.80 m/s2)(36.1 m) = 18.8 m/s
The bay has one end open and one end closed, so its simplest resonance is with a node (of velocity, antinode of displacement) at the head of the bay and an antinode (of velocity, node of displacement) at the mouth. Then, dNA = 210 103 m = Therefore, the period is T= 1 840 103 m = = = 4.47 104 s = 12 h 24 min f v 18.8 m/s
4
and
= 840 103 m
This agrees precisely with the period of the lunar excitation , so we identify the extrahigh tides as amplified by resonance. (a) 18.31 For the fundamental mode in a closed pipe, = 4L. (see Figure 18.3b) But v = f, therefore L = So, L = (b) v 4f
343 m/s = 0.357 m 4(240/s)
For an open pipe, = 2L. (see Figure 18.3a) So, L = v 343 m/s = = 0.715 m 2f 2(240/s)
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Chapter 18 Solutions
L = dAA = 2 n Since = n 2
15
*18.32
or
L=
for
n = 1, 2, 3, . . .
v v , L = n for n = 1, 2, 3, . . . f 2f
With v = 343 m/s, and f = 680 Hz, L = n 343 m/s = n(0.252 m) for n = 1, 2, 3, . . . 2(680 Hz)
Possible lengths for resonance are: L = 0.252 m, 0.504 m, 0.757 m, . . ., n(0.252) m 18.33 dAA = 0.320 m (a) (b) *18.34 f=
= 0.640 m
v = 531 Hz dAA = 42.5 mm
= 0.0850 m
The wavelength is
=
v 343 m/s = = 1.31 m f 261.6/s
so the length of the open pipe vibrating in its simplest (A-N-A) mode is dA to A = 1 = 0.656 m 2
A closed pipe has (N-A) for its simplest resonance, (N-A-N-A) for the second, and (N-A-N-AN-A) for the third. Here, the pipe length is 5dN to A = *18.35 5 5 = (1.31 m) = 1.64 m 4 4
The air in the auditory canal, about 3 cm long, can vibrate with a node at the closed end and antinode at the open end, with dN to A = 3 cm = so
4
= 0.12 m
v 343 m/s = 3 kHz 0.12 m
and f =
A small-amplitude external excitation at this frequency can, over time, feed energy into a larger-amplitude resonance vibration of the air in the canal, making it audible.
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16
Chapter 18 Solutions
18.36
=
v 343 m/s = = 0.780 m f 440/s
dN to A =
= 0.195 m = length of resonant air column 4
Water height = 0.400 m 0.195 m = 0.205 m m = V = Ah = (1000 kg/m3)(0.100 m2)(0.205 m) = 20.5 kg 18.37 For a closed box, the resonant frequencies will have nodes at both sides, so the permitted n wavelengths will be L = , (n = 1, 2, 3, . . . ). 2 i.e., L = n nv = 2 2f nv 2L
and f =
Therefore, with L = 0.860 m and L' = 2.10 m, the resonant frequencies are fn = n(206 Hz) and f n = n(84.5 Hz) ' 18.38 for L = 0.860 m for each n from 1 to 9 for L' = 2.10 m for each n from 2 to 23
We suppose these are the lowest resonances of the enclosed air columns. For one,
=
v (343 m/s) = = 1.34 m f 256/s
length = dAA = For the other,
= 0.670 m 2
=
v 343 m/s = = 0.780 m f 440/s
length = 0.390 m So, (b) original length = 1.06 m
= 2dAA = 2.12 m
(a) f= (343 m/s) = 162 Hz 2.12 m
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Chapter 18 Solutions
v f
17
18.39
The fork radiates sound with =
200 Hz
The distance between successive water levels at resonance is dNN = v 2f
So
Rt =
r 2v 2f
t=
r 2v 2R f (4.00 102 m)2(343 m/s) = 239 s 2(18.0 106 m3/s)(200/s)
v f
18.0 cm 3/s
t=
18.40
The wavelength of sound is =
The distance between water levels at resonance is d= v 2f
Rt = r 2d =
r 2v 2f
and t =
r 2v 2R f
v , giving 4L
18.41
The length corresponding to the fundamental satisfies f = v 343 = = 0.167 m 4f 4(512)
L1 =
Since L > 20.0 cm, the next two modes will be observed, corresponding to f= 3v 4L 2 and f= 5v 4L 3 and L3 5v = 0.837 m 4f
or
L2 =
3v = 0.502 m 4f
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18
Chapter 18 Solutions
18.42
Call L the depth of the well and v the speed of sound. Then for some integer n L = (2n 1)
1 v (2n 1)(343 m/s) = (2n 1) = 4 4f 1 4(51.5 s1)
and for the next resonance L = [2(n + 1) 1]
2 v (2n + 1)(343 m/s) = (2n + 1) = 4 4f 2 4(60.0 s1)
Thus,
(2n 1)(343 m/s) (2n + 1)(343 m/s) = , 4(51.5 s1) 4(60.0 s1) 2n + 1 2n 1 = 60.0 51.5
and we require an integer solution to
The equation gives n =
111.5 = 6.56, so the best fitting integer is n = 7. 17
Then L =
[2(7) 1](343 m/s) = 21.6 m 4(51.5 s1) [2(7) + 1](343 m/s) = 21.4 m 4(60.0 s1)
and L =
suggest the best value for the depth of the well is 21.5 m . 18.43 For resonance in a tube open at one end, v f = n (n = 1, 3, 5, . . .) 4L (a) Assuming n = 1 and n = 3, v 384 = 4(0.228) and 3v 384 = 4(0.683) Equation 18.12
22.8 cm 68.3 cm
In either case, v = 350 m/s (b) For the next resonance, n = 5, and L= 5v 5(350 m/s) = = 1.14 m 4f 4(384 s1)
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Chapter 18 Solutions
v v 331.5 m/s = = = 17.0 Hz 4L 4(4.88 m) v v = = 34.0 Hz 2L v(20.0C) f = v(0C) 1 1+ 1+ 20.0 f = 17.6 Hz 273 1
19
18.44
(a)
f1 =
(b)
f1 =
(c)
For the closed pipe, f =
For the open pipe, f = *18.45 (a)
20.0 f = 35.2 Hz 273 1
For the fundamental mode of an open tube, L= 343 m/s v = = = 0.195 m 2 2f 2(880 s1) 1+ (5.00) = 328 m/s 273
(b)
v = 331 m/s
We ignore the thermal expansion of the metal. f= v v 328 m/s = = = 841 Hz 2L 2(0.195 m)
The flute is flat by a semitone. 18.46 When the rod is clamped at one-quarter of its length, the fundamental frequency corresponds to a mode of vibration in which L = . Therefore, L = v 5100 m/s = = 1.16 m f 4400 Hz
18.47
(a) (b)
f=
v 5100 = = 1.59 kHz 2L (2)(1.60)
Since it is held in the center, there must be a node in the center as well as antinodes at the ends. The even harmonics have an antinode at the center so only the odd harmonics are present.
(c) f = 18.48
v' 3560 = = 1.11 kHz 2L (2)(1.60)
v = 4500 m/s so, f1 =
1 = 4L = 240 cm = 2.40 m
v v 4500 = = = 1.88 kHz 1 4L 2.40
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20
Chapter 18 Solutions
18.49
fv
T 540 = 104.4 Hz 600
fnew = 110
f = 5.64 beats/s
Goal Solution G: Beat frequencies are usually only a few Hertz, so we should not expect a frequency much greater than this. O: As in previous problems, the two wave speed equations can be used together to find the frequency of vibration that corresponds to a certain tension. The beat frequency is then just the difference in the two resulting frequencies from the two strings with different tensions. A: Combining the velocity equation v = f/ and the tension equation v = f= T 2 f2 = f1 T2 T1 540 N = 104.4 Hz 600 N T we find that
and since and are constant, we can divide to get
With f1 = 110 Hz, T1 = 600 N, and T2 = 540 N: f2 = (110 Hz)
The beat frequency is: fb = f 1 f 2 = 110 Hz 104.4 Hz = 5.64 Hz L: As expected, the beat frequency is only a few cycles per second. This result from the interference of the two sound waves with slightly different frequencies has a tone that varies in amplitude over time, similar to the sound made by saying "wa-wa-wa..." Note: The beat frequency above is written with three significant figures on the assumption that the data and known precisely enough to warrant them. This assumption implies that the original frequency is known more precisely than to the three significant digits quoted in "110 Hz." For example, if the original frequency of the strings were 109.6 Hz, the beat frequency would be 5.62 Hz.
*18.50
(a) (b)
The string could be tuned to either 521 Hz or 525 Hz from this evidence. Tightening the string raises the wave speed and frequency. If the frequency were originally 521 Hz, the beats would slow down. Instead, the frequency must have started at 525 Hz to become 526 Hz .
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Chapter 18 Solutions
21
(c)
From f =
v T/ 1 = = 2L 2L T2 T1
T , T2 = f2 2 523 Hz 2 T1 = T = 0.989T1 f 1 526 Hz 1
f2 = f1
and
The fractional change that should be made in the tension is then fractional change = T1 T2 = 1 0.989 = 0.0114 = 1.14% lower T1
The tension should be reduced by 1.14% . (v + vs) For an echo f ' = f (v v s ) the beat frequency is f b = f ' f (2v s ) when approaching wall. (v v s ) beat frequency
18.51
Solving for fb gives fb = f
(a) (b)
fb = (256)
(2)(1.33) = 1.99 Hz (343 1.33)
When moving away from wall, vs changes sign. Solving for vs gives vs = fbv (5)(343) = = 3.38 m/s 2f f b (2)(256) 5
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22
Chapter 18 Solutions
*18.52
We evaluate s = 100 sin + 157 sin 2 + 62.9 sin 3 + 105 sin 4 + 51.9 sin 5 + 29.5 sin 6 + 25.3 sin 7 where s represents particle displacement in nanometers and represents the phase of the wave in radians. As advances by 2, time advances by (1/523) s. Here is the result:
Flute Waveform
400
200
s (nm)
0
0
2
4
6
8
10
12
--200
--400
Phase (rad)
*18.53
We list the frequencies of the harmonics of each note in Hz: Harmonic 3 1320.0 1663.1 1977.8
Note A C# E
1 440.00 554.37 659.26
2 880.00 1108.7 1318.5
4 1760.0 2217.5 2637.0
5 2200.0 2771.9 3296.3
The second harmonic of E is close to the third harmonic of A, and the fourth harmonic of C# is close to the fifth harmonic of A.
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Chapter 18 Solutions
18.54 (a) For the block: Fx = T Mg sin 30.0 = 0 so (b) T = Mg sin 30.0 = 1 Mg 2
h
23
The length of the section of string parallel to the incline is h/sin 30.0 = 2h. The total length of the string is then 3 h .
M
(c)
The mass per unit length of the string is = m/3h . T =
(d) (e)
The speed of waves in the string is v =
Mg 3h = 2 m
3Mgh 2m
In the fundamental mode, the segment of length h vibrates as one loop. The distance between adjacent nodes is then dNN = /2 = h, so the wavelength is = 2h. v 1 = 2h 3Mgh = 2m 3Mg 8m h
The frequency is f =
(g)
When the vertical segment of string vibrates with 2 loops (i.e., 3 nodes), then h = 2 and the wavelength is = h .
2
(f)
The period of the standing wave of 3 nodes (or two loops) is 1 = =h f v 2m = 3Mgh 2m h 3Mg 3Mg 8m h
T=
(h)
fb = 1.02f f = (2.00 102)f = (2.00 10 2 ) x=
18.55
(a)
(9.00 + 4.00)
3.00 =
13.0 3.00 = 0.606 m
The wavelength is =
v 343 m/s = = 1.14 m f 300 Hz
Thus,
x 0.606 = = 0.530 1.14
of a wave, or = 2 (0.530) = 3.33 rad
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24
Chapter 18 Solutions
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Chapter 18 Solutions
(b) For destructive interference, we want x x = 0.500 = f v where x is a constant in this set up. f= 18.56 f = 87.0 Hz speed of sound in air: va = 340 m/s (a) v 343 = = 283 Hz 2 x (2)(0.606)
l = 0.400 m
25
b= l
v = f b = (87.0 s1)(0.400 m)
L
v = 34.8 m/s
(b)
a L = 4f = 4(87.0 s1) = 0.977 m va = a f
a = 4L
v
340 m/s
18.57
Moving away from station, frequency is depressed: f ' = 180 2.00 = 178 Hz 178 = 180 (343) (343 + v)
Solving for v gives v= (2.00)(343) 178
Therefore, v = 3.85 m/s away from station Moving towards the station, the frequency is enhanced: f ' = 180 + 2.00 = 182 Hz 182 = 180 (343) (343 v)
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26
Chapter 18 Solutions
Solving for v gives v= (2.00)(343) 182
Therefore, v = 3.77 m/s towards the station *18.58 Use the Doppler formula f'=f (v v 0 ) (v v s ) +
With f 1 = frequency of the speaker in front of student and ' f 2 = frequency of the speaker behind the student. ' f 1 = (456 Hz) ' (343 m/s + 1.50 m/s) = 458 Hz (343 m/s 0) (343 m/s 1.50 m/s) = 454 Hz (343 m/s + 0)
f 2 = (456 Hz) '
Therefore, fb = f 1 f 1 = 3.99 Hz ' ' 18.59 From the leading train she hears f1 = f ' 343 m/s v + 0 =f v + v s 343 m/s + 8.00 m/s
From the still-approaching train, f2 = f '
343 343 8.00
Then, 4.00 Hz = f 2 f 1 = 1.0239f 0.9772f ' ' f= 4.00 Hz = 85.7 Hz 0.0467
18.60
v=
(48.0)(2.00) = 141 m/s 4.80 103
dNN = 1.00 m
= 2.00 m
f=
v = 70.7 Hz
a =
va 343 m/s = = 4.85 m f 70.7 Hz
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Chapter 18 Solutions
*18.61 The second standing wave mode of the air in the pipe reads ANAN, with dNA =
27
1.75 m = 4 3
so
= 2.33 m and
f=
v 343 m/s = = 147 Hz 2.33 m
For the string, and v are different but f is the same. 0.400 m = dNN = 2 2 so
= 0.400 m
T/
v = f = (0.400 m)(147 Hz) = 58.8 m/s =
T = v2 = (9.00 103 kg/m)(58.8 m/s)2 = 31.1 N 18.62 (a) L= v 4f so L' f = L f' 4 2 1 : : 5 3 2
Letting the longest L be 1, the ratio is 1 : or in integers 30 : 24 : 20 : 15 343 = 33.5 cm (4)(256)
(b)
L=
This is the longest pipe, so using the ratios the lengths are: 33.5, 26.8, 22.3, 16.7 cm (c) The frequencies are using the ratio 256, 320, 384, and 512 Hz . These represent notes C, E, G, and C' on the physical pitch scale. 18.63 (a) Since the first node is at the weld, the wavelength in the thin wire is 2L or 80.0 cm. The frequency and tension are the same in both sections, so f= 1 2L T 1 = 2(0.400) 4.60 = 59.9 Hz 2.00 103
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28
Chapter 18 Solutions
(b)
As the thick wire is twice the diameter, the linear density is 4 times that of the thin wire. ' = 8.00 g/m so L' = 1 2f T
(4.60) (8.00 10-3)
half the length of the thin wire
L' =
1 (2)(59.9) = 20.0 cm
18.64
fB = fA
2
B =
1 3 A
vB =
1 v 3 A
vB =
1 2 v 9 A T
2
v=
TB vB = 2 = 0.111 TA vA 18.65 (a) f= n 2L T
so
f' L L 1 = = = f L' 2L 2
The frequency should be halved to get the same number of antinodes for twice the length. (b) n' = n so T T' T' n = T n'
2
=
n (n + 1)
2
The tension must be
T' =
n T (n + 1)
2
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Chapter 18 Solutions
29
(c)
f ' n' L = f n L' so
T' T
2
T' n f 'L ' = T n' f L T' 3 = T 2 2 T' 9 = T 16
2
to get twice as many antinodes.
18.66
For the wire, = 0.0100 kg = 5.00 103 kg/m 2.00 m T = (200 kg m/s2) 5.00 103 kg/m
v=
v = 200 m/s If it vibrates in its simplest state, dNN = 2.00 m =
2
f= (a)
v (200 m/s) = = 50.0 Hz 4.00 m
The tuning fork can have frequencies 45.0 Hz or 55.0 Hz
(b)
If f = 45.0 Hz, v = f = (45.0/s) 4.00 m = 180 m/s Then, T = v2 = (180 m/s)2(5.00 103 kg/m) = 162 N or if f = 55.0 Hz T = v2 = f 2 2 = (55.0/s)2(4 .00 m)2(5.00 103 kg/m) = 242 N
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30
Chapter 18 Solutions
18.67
The odd-numbered harmonics of the organ-pipe vibration are: 650 Hz, 550 Hz, 450 Hz, 350 Hz, 250 Hz, 150 Hz, 50.0 Hz Closed f1 = 50.0 Hz
= 6.80 m
L = 1.70 m
18.68
We look for a solution of the form 5.00 sin (2.00x 10.0t) + 10.0 cos(2.00x 10.0t) = A sin (2.00x 10.0t + ) = A sin (2.00x 10.0t)cos + A cos (2.00x 10.0t) sin This will be true if both 5.00 = A cos and 10.0 = A sin , requiring (5.00)2 + (10.0)2 = A2 A = 11.2 and
= 63.4
The resultant wave 11.2 sin(2.00x 10.0t + 63.4) is sinusoidal. *18.69 (a) With k = Error! and
= 2 f = Error!
2 x 2 v t cos
y(x, t) = 2A sin kx cos t = 2A sin (b) For the fundamental vibration,
1 = 2L
so y1(x, t) = 2A sin
x vt cos L L
(c)
For the second harmonic
2 = L
and y2(x, t) = 2A sin 2 x 2 vt cos L L
(d)
In general,
n =
2L n
and
yn(x, t) = 2A sin
n x n vt cos L L
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Chapter 18 Solutions
18.70 (a) In the diagram, observe that: sin = 1.00 m 2 = 1.50 m 3 or
31
d
d
= 41.8
Considering the mass, Fy = 0 or (b) T= gives 2T cos = mg
g
(12.0 kg)(9.80 m/s2) = 78.9 N 2 cos 41.8
m m (b) (a)
The speed of transverse waves in the string is v= T = 78.9 N = 281 m/s 0.00100 kg/m
For the standing wave pattern shown (3 loops), d =
3 , or 2
=
2(2.00 m) = 1.33 m 3
Thus, the required frequency is f= v 281 m/s = = 211 Hz 1.33 m
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