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16 CHAPTER WAVES AND SOUND
CONCEPTUAL QUESTIONS
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1.
REASONING AND SOLUTION As Figure 16.4 shows, in a water wave, the wave motion of the water includes both transverse and longitudinal components. The water at the surface moves on nearly circular paths. When the wave passes beneath a fishing float, the float will simultaneously bob up and down, as well as move back and forth horizontally. Thus, the float will move in a nearly circular path in the vertical plane. It is not really correct, therefore, to say that the float bobs straight "up and down." REASONING AND SOLUTION "Domino Toppling" is an event that consists of lining up an incredible number of dominoes and then letting them topple, one after another. As the dominoes topple, their displacements contain both vertical and horizontal components. Therefore, the disturbance that propagates along the line of dominoes has both longitudinal (horizontal) and transverse (vertical) components. REASONING AND SOLUTION A longitudinal wave moves along a Slinky at a speed of 5 m/s. We cannot conclude that one coil of the Slinky moves through a distance of 5 m in one second. The quantity 5 m/s is the longitudinal wave speed, vspeed; it specifies how fast the disturbance travels along the spring. The wave speed depends on the properties of the spring. Like the transverse wave speed, the longitudinal wave speed depends upon the tension F in the spring and its linear mass density m/L. As long as the tension and the linear mass density remain the same, the disturbance will travel along the spring at constant speed. The particles in the Slinky oscillate longitudinally in simple harmonic motion with the same amplitude and frequency as the source. As with all particles in simple harmonic motion, the particle speed is not constant. The particle speed is a maximum as the particle passes through its equilibrium position and reaches zero when the particle has reached its maximum displacement from the equilibrium position. The particle speed depends upon the amplitude and frequency of the particle's motion. Thus, the particle speed, and therefore the longitudinal speed of a single coil, depends upon the properties of the source that causes the disturbance.
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2.
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3.
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4.
SSM REASONING AND SOLUTION A wave moves on a string with constant velocity. It is not correct to conclude that the particles of the string always have zero acceleration. As Conceptual Example 3 discusses, it is important to distinguish between the speed of the waves on the string, vwave, and the speed of the particles in the string, vparticle. The wave speed vwave is determined by the properties of the string; namely, the tension in the string and the linear mass density of the string. These properties determine the speed with which the disturbance travels along the string. The wave speed will remain constant as long as these properties remain unchanged.
808 WAVES AND SOUND
The particles in the string oscillate transversely in simple harmonic motion with the same amplitude and frequency as the source of the disturbance. Like all particles in simple harmonic motion, the acceleration of the particles continually changes. It is zero when the particles pass through their equilibrium positions and is a maximum when the particles are at their maximum displacements from their equilibrium positions.
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5.
REASONING AND SOLUTION A wire is strung tightly between two immovable posts. The speed of a transverse wave on the wire is given by Equation 16.2: vwave = F /(m / L) . The wire will expand because of the increase in temperature. Since the length of the wire increases slightly, it will sag, and the tension in the wire will decrease. From Equation 16.2, we see that the speed of the wave is directly proportional to the square root of the tension in the wire. If we ignore any change in the mass per unit length of the wire, then we can conclude that, when the temperature is increased, the speed of waves on the wire will decrease. ______________________________________________________________________________ 6. REASONING AND SOLUTION A rope of mass m is hanging down from the ceiling. Nothing is attached to the loose end of the rope. A transverse wave is traveling up the rope. The tension in the rope is not constant. The lower portion of the rope pulls down on the higher portions of the rope. If we imagine that the rope is divided into small segments, we see that the segments near the top of the rope are being pulled down by more weight than the segments near the bottom. Therefore, the tension in the rope increases as we move up the rope. The speed of a transverse wave on the rope is given by Equation 16.2: vwave = F /(m / L) . From Equation 16.2 we see that, as the tension F in the rope increases, the speed of the wave increases. Therefore, as the transverse wave travels up the rope, the speed of the wave increases. REASONING AND SOLUTION One end of each of two identical strings is attached to a wall. Each string is being pulled tightly by someone at the other end. A transverse pulse is sent traveling along one of the strings. A bit later, an identical pulse is sent traveling along the other string. In order for the second pulse to catch up with the first pulse, the speed of the pulse in the second string must be increased. The speed of the transverse pulse on the second string is given by Equation 16.2: vwave = F /(m / L) . This equation indicates that we can increase the speed of the second pulse by increasing the tension F in the string. Thus, the second string must be pulled more tightly. REASONING AND SOLUTION In Section 4.10 the concept of a "massless" rope is discussed. For a truly massless rope, the linear density of the rope, m/L, is zero. From Equation 16.2, vwave = F /(m / L) , the wave speed would be infinite if m/L were zero. Therefore, if the rope were really massless, the speed of transverse waves on the rope would be infinite, and a transverse wave would be instantaneously transmitted from one end of the rope to the other. It would not take any time for a transverse wave to travel the length of a massless rope.
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7.
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8.
Chapter 16 Conceptual Questions
809
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9.
REASONING AND SOLUTION As the disturbance moves outward when a sound wave is produced, it compresses the air directly in front of it. This compression causes the air pressure to rise slightly, resulting in a condensation that travels outward. The condensation is followed by a region of decreased pressure, called a rarefaction. Both the condensation and rarefaction travel away from the speaker at the speed of sound. As the condensations and rarefactions of the sound wave move away from the disturbance, the individual air molecules are not carried with the wave. Each molecule executes simple harmonic motion about a fixed equilibrium position. As the wave passes by, all the particles in the region of the disturbance participate in this motion. There are no particles that are always at rest.
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10. REASONING AND SOLUTION Assuming that we can treat air as an ideal gas, then the speed of sound in air is given by Equation 16.5, v = kT / m , where is the ratio of the specific heats c P / c V , k is Boltzmann's constant, T is the Kelvin temperature, and m is the mass of a molecule of the gas. We can see from Equation 16.5 that the speed of sound in air is proportional to the square root of the Kelvin temperature of the gas. Therefore, on a hot day, the speed of sound in air is greater than it is on a cold day. Hence, we would expect an echo to return to us more quickly on a hot day as compared to a cold day, other things being equal.
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11. SSM REASONING AND SOLUTION A loudspeaker produces a sound wave. The sound wave travels from air into water. As indicated in Table 16.1, the speed of sound in water is approximately four times greater than it is in air. We are told in the hint that the frequency of the sound wave does not change as the sound enters the water. The relationship between the frequency f, the wavelength , and the speed v of a wave is given by Equation 16.1: v = f . Since the wave speed increases and the frequency remains the same as the sound enters the water, the wavelength of the sound must increase.
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12. REASONING AND SOLUTION A person is making JELL-O. It starts out as a liquid and then sets into a gel. In general, sound travels slowest in gases, faster in liquids, and fastest in solids. When the JELL-O sets into a gel, it is more "solid;" therefore, the speed of sound should be larger in the set JELL-O than when the JELL-O is in the liquid state.
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13. REASONING AND SOLUTION Animals that rely on an acute sense of hearing for survival often have relatively large external ear parts. Sound intensity is defined as the sound power P that passes perpendicularly through a surface divided by the area A of that surface: I = P / A . For low intensity sounds, the power per unit area is small. Relatively large outer ears have a greater area than smaller outer ears. Hence, large outer ears intercept and direct more sound power into the auditory system than smaller outer ears do.
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810 WAVES AND SOUND
14. REASONING AND SOLUTION A source is emitting sound uniformly in all directions. According to Equation 16.9, I = P /(4 r 2 ) , the intensity of such a source varies as 1/r2. Thus, the intensity I at a point in space depends on the distance of that point from the source. A flat surface faces the source. As suggested in the following figure, the distance between the source and the flat sheet varies, in general, from point to point on the sheet. The figure indicates that, as we move up the screen, the distance between the source and the screen increases. Therefore, the sound intensity is not the same at all points on the screen.
uniformly emitting source
flat screen
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15. REASONING AND SOLUTION Two people talk simultaneously and each creates an intensity level of 65 dB at a certain point. The total intensity level at this point does not equal 130 dB. Intensity levels are defined in terms of logarithms. According to Equation 16.10, the sound intensity level in decibels is given by = (10 dB ) log( I / I 0 ) , where I0 is the threshold of human hearing. We can show, using Equation 16.10, that an intensity level of 65 dB corresponds to a sound intensity of 3.2 10 6 W/m 2 . If two people simultaneously talk and each creates an intensity of 3.2 10 6 W/m 2 at a certain point, then the total intensity at that point is 6.4 10 6 W/m 2 . However, since the intensity level is defined in terms of logarithms, the intensity level at that point is (from Equation 16.10) 68 dB. While the second person doubles the intensity at the point in question, he only increases the intensity level (or the loudness) by 3 dB.
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16. REASONING AND SOLUTION Two cars, one behind the other, are traveling in the same direction at the same speed. The velocity of either car relative to the other one is zero. Since there is no relative motion between the cars, either driver will hear the other's horn at exactly the same frequency that would be heard if both cars were at rest.
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17. REASONING AND SOLUTION According to Table 16.1, the speed of sound in air at 20 C is v = 343 m/s, while its value in water at the same temperature is v = 1482 m/s. These values influence the Doppler effect, because of which an observer hears a frequency fo that is different from the frequency fs that is emitted by the source of sound. For purposes of this question, we assume that fs = 1000 Hz and that the speed at which the source moves is vs = 25 m/s. Our conclusions, however, will be valid for any values of fs and vs .
Chapter 16 Conceptual Questions
811
a. The Doppler-shifted frequency when the source approaches the observer is given by Equation 16.11 as f o = fs 1 - (vs / v) . Applying this equation for air and water, we find
Air
fo = fo =
1000 Hz = 1079 Hz 1 - (25 m s )/(343 m s )
Water
1000 Hz = 1017 Hz 1 - (25 m s )/(1482 m s )
The change in frequency due to the Doppler effect in air is greater. b. The Doppler-shifted frequency when the source moves away from the observer is given by Equation 16.12 as f o = fs 1 + (vs / v) . Applying this equation for air and water, we find
Air
fo = fo =
1000 Hz = 932 Hz 1 + (25 m s )/(343 m s ) 1000 Hz = 983 Hz 1 + (25 m s )/(1482 m s )
Water
The change in frequency due to the Doppler effect in air is greater.
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18. REASONING AND SOLUTION A music fan at a swimming pool is listening to a radio on a diving platform. The radio is playing a constant frequency tone when this fellow, clutching his radio, jumps. a. The frequency heard by a person left behind on the platform is given by Equation 16.12: f o = fs 1 + (vs / v) . Since the denominator, 1 + (vs / v) , is necessarily greater than one, the frequency fo is less than fs. Thus, a person left behind on the platform will hear a frequency that is smaller than the frequency produced by the radio. As the radio falls, it accelerates, so its speed increases. As vs increases, the quantity 1 + (vs / v) also increases, so that fo decreases with time. Therefore, the observed frequency is not constant and decreases during the fall. b. The frequency heard by a person down below floating on a rubber raft is given by Equation 16.11: f o = fs 1 - (vs / v) . We can assume that during the fall vs < v at all times so that the quantity vs / v is less than one, and the denominator, 1 - (vs / v) , is necessarily less than one. Therefore, the frequency fo is greater than fs. A person down below floating on a rubber raft will hear a frequency that is greater than the frequency produced by the radio. As the radio falls, vs increases as discussed in part (a), so that the
812 WAVES AND SOUND
quantity 1 - (vs / v) decreases. As a result, the frequency fo increases with time. Therefore, the observed frequency is not constant and increases during the fall.
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19. SSM REASONING AND SOLUTION When a car is at rest, its horn emits a frequency of 600 Hz. A person standing in the middle of the street hears the horn with a frequency of 580 Hz. As shown in Figure 16.28, when a vehicle is moving, a stationary observer in front of the vehicle will hear sound of a shorter wavelength, while an observer behind the vehicle will hear sound of a longer wavelength. We can deduce, therefore, from Equation 16.1, v = f , that the person standing in front of the vehicle will hear sound of a higher frequency than when the car is at rest, while a person standing behind the vehicle will hear sound of a lower frequency than when the car is at rest. Since the person standing in the middle of the street hears a frequency of 580 Hz, which is less than the stationary frequency of 600 Hz, the person is behind the car. Therefore, the person need not jump out of the way.
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20. REASONING AND SOLUTION A car is speeding toward a large wall and sounds the horn. The sound of the horn in front of the car will have a shorter wavelength and a higher frequency than the wavelength and frequency of the sound of the horn when the car is at rest. This Doppler shift occurs because the source is moving. The echo is produced when this higher-frequency sound is reflected from the wall. The wall now acts as a stationary source of this higher-frequency sound. The driver is an observer who is moving toward a stationary source. Since the driver is moving toward the echo, he will hear a frequency that is greater than the sound heard by a stationary observer. Thus, the echo heard by the driver is Doppler-shifted twice. The sound is shifted once to a higher frequency because the source of the sound is moving. Then the sound is Doppler shifted to a higher frequency again because the driver is moving relative to the stationary source (of the echo).
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Chapter 16 Problems
813
CHAPTER 16 WAVES AND SOUND
PROBLEMS
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SSM REASONING Since light behaves as a wave, its speed v, frequency f, and wavelength are related to according to v = f (Equation 16.1). We can solve this equation for the frequency in terms of the speed and the wavelength. SOLUTION Solving Equation 16.1 for the frequency, we find that
f = v = 3.00 108 m/s = 5.50 1014 Hz 5.45 10-7 m
2.
REASONING a. The period is the time required for one complete cycle of the wave to pass. The period is also the time for two successive crests to pass the person.
b. The frequency is the reciprocal of the period, according to Equation 10.5. c. The wavelength is the horizontal length of one cycle of the wave, or the horizontal distance between two successive crests. d. The speed of the wave is equal to its frequency times its wavelength (see Equation 16.1). e. The amplitude A of a wave is the maximum excursion of a water particle from the particle's undisturbed position.
SOLUTION a. After the initial crest passes, 5 additional crests pass in a time of 50.0 s. The period T of the wave is 50.0 s T= = 10.0 s 5
b. Since the frequency f and period T are related by f = 1/T (Equation 10.5), we have f = 1 1 = = 0.100 Hz T 10.0 s
c. The horizontal distance between two successive crests is given as 32 m. This is also the wavelength of the wave, so = 32 m
814 WAVES AND SOUND
d. According to Equation 16.1, the speed v of the wave is v = f = ( 0.100 Hz )( 32 m ) = 3.2 m/s e. There is no information given, either directly or indirectly, about the amplitude of the wave. Therefore, it is not possible to determine the amplitude.
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3.
REASONING AND SOLUTION From Equation 16.1, we have = v/f. But v = x/t, so we find x v 2.5 m = = = = 0.49 m f t f (1.7 s )(3.0 Hz ) ______________________________________________________________________________
4.
REASONING The speed of a Tsunamis is equal to the distance x it travels divided by the time t it takes for the wave to travel that distance. The frequency f of the wave is equal to its speed divided by the wavelength , f = v/ (Equation 16.1). The period T of the wave is related to its frequency by Equation 10.5, T = 1/f. SOLUTION a. The speed of the wave is (in m/s)
v= x 3700 103 m 1 h = = 190 m/s t 5.3 h 3600 s
b. The frequency of the wave is
f = v
=
190 m/s = 2.5 10-4 Hz 3 750 10 m
(16.1)
c. The period of any wave is the reciprocal of its frequency:
1 1 = = 4.0 103 s (10.5) f 2.5 10-4 Hz ______________________________________________________________________________ T=
5.
SSM REASONING When the end of the Slinky is moved up and down continuously, a transverse wave is produced. The distance between two adjacent crests on the wave, is, by definition, one wavelength. The wavelength l is related to the speed and frequency of a periodic wave by Equation 16.1, = v / f . In order to use Equation 16.1, we must first determine the frequency of the wave. The wave on the Slinky will have the same frequency
Chapter 16 Problems
815
as the simple harmonic motion of the hand. According to the data given in the problem statement, the frequency is f = (2.00 cycles)/(1 s) = 2.00 Hz .
SOLUTION Substituting the values for and f, we find that the distance between crests is
v 0.50 m/s = = 0.25 m f 2.00 Hz ______________________________________________________________________________
=
6.
REASONING AND SOLUTION The period of the wave is the same as the period of the person, so T = 5.00 s.
a.
f = 1/T = 0.200 Hz
(10.5)
b. v = l f = (20.0 m)(0.200 Hz) = 4.00 m/s (16.1) ______________________________________________________________________________ 7.
REASONING AND SOLUTION Using Equation 16.1, we find that
= v/f = (343 m/s)/(4185.6 Hz) = 8.19 10 2 m
(16.1)
______________________________________________________________________________ 8. REASONING The wave is moving with a speed v (unknown), and the jetskier is moving in the same direction with a speed vjetskier (known). To find v, we proceed as follows. If the jetskier were traveling with the same speed as the wave, his speed relative to the wave would be zero and he would never experience any "bumps." He is traveling faster than the wave, however, so his speed relative to the wave is vjetskier - v. This relative speed times the time between "bumps" must, then, equal the wavelength of the water wave, because the distance between adjacent crests is the wavelength. Moreover, the time between "bumps" or crests is the reciprocal of the given "bump" frequency. In this fashion, we can obtain an equation that can be solved for the wave speed.
SOLUTION The speed of the jetskier relative to the wave times the time t between "bumps" is the wavelength of the water wave, so that we have
( v jetskier - v ) t =
Speed of jetskier relative to wave
The times between bumps is the reciprocal of the "bump" frequency f or t = 1/f. With this substitution we find 1 v jetskier - v = or v jetskier - v = f f
(
)
816 WAVES AND SOUND
Solving for the speed v of the wave gives
v = v jetskier - f = 8.4 m/s - (1.2 Hz )( 5.8 m ) = 1.4 m/s
9.
REASONING During each cycle of the wave, a particle of the string moves through a total distance that equals 4A, where A is the amplitude of the wave. The number of wave cycles per second is the frequency f of the wave. Therefore, the distance moved per second by a string particle is 4Af. The time to move through a total distance L, then, is t = L/(4Af). According to Equation 16.1, however, we have that f = v/, so that
t= L L = 4 Af 4 A ( v / )
SOLUTION Using the result obtained above, we find that
1.0 103 m ( 0.18 m ) L t= = = 5.0 101 s 3 4 Av 4 2.0 10 m ( 450 m/s )
( (
)
)
______________________________________________________________________________ 10. REASONING AND SOLUTION The maximum acceleration of the dot occurs at the extreme positions and is
amax = ( 2 f ) A = 2 ( 4.0 Hz )
2 2
( 5.4 10-3 m ) = 3.4 m/s2
(10.10)
______________________________________________________________________________ 11. REASONING AND SOLUTION When traveling with the waves, the skier "sees" the waves to be traveling with a velocity of vw vs and with a period of T1 = 0.600 s. Suppressing the units for convenience, we find that the distance between crests is (in meters)
l = (vs vw)T1 = (12.0 - vw)(0.600) = 7.20 0.600 vw
Similarly, for the skier traveling opposite the waves
(1)
l = (vs + vw)T2 = 6.00 + 0.500 vw
a. Subtracting Equation (2) from Equation(1) and solving for vw gives vw = 1.1 m/s .
(2)
b. Substituting the value for vw into Equation (1) gives l = 6.55 m . ______________________________________________________________________________
Chapter 16 Problems
817
12. REASONING The speed v of a transverse wave on a wire is given by v = F / ( m /L ) (Equation 16.2), where F is the tension and m/L is the mass per unit length (or linear density) of the wire. We are given that F and m are the same for the two wires, and that one is twice as long as the other. This information, along with knowledge of the wave speed on the shorter wire, will allow us to determine the speed of the wave on the longer wire.
SOLUTION The speeds on the longer and shorter wires are: [Longer wire]
vlonger = F m /L longer
[Shorter wire]
vshorter =
F m /L shorter
Dividing the expression for vlonger by that for vshorter gives
F m /L longer F m /L shorter
vlonger vshorter
=
=
L longer L shorter
Noting that vshorter = 240 m/s and that Llonger = 2Lshorter, the speed of the wave on the longer wire is vlonger = vshorter L longer L shorter = ( 240 m/s ) 2L shorter L shorter = ( 240 m/s ) 2 = 340 m/s
______________________________________________________________________________ 13. SSM WWW REASONING According to Equation 16.2, the linear density of the string is given by (m / L) = F / v 2 , where the speed v of waves on the middle C string is given by Equation 16.1, v = f = / T .
SOLUTION Combining Equations 16.2 and 16.1 and using the given data, we obtain
F FT 2 (944 N)(3.82 10 3 s) 2 = 2 = = 8.68 10 3 kg/m 2 2 v (1.26 m) ______________________________________________________________________________ m/ L =
818 WAVES AND SOUND
14. REASONING The length L of the string is one of the factors that affects the speed of a wave traveling on it, in so far as the speed v depends on the mass per unit length m/L according to F (Equation 16.2). The other factor affecting the speed is the tension F. The speed v= m/ L is not directly given here. However, the frequency f and the wavelength are given, and the speed is related to them according to v = f (Equation 16.1). Substituting Equation 16.1 into Equation 16.2 will give us an equation that can be solved for the length L.
SOLUTION Substituting Equation 16.1 into Equation 16.2 gives
v= f = F m/ L
Solving for the length L, we find that
-3 f 2 2 m ( 260 Hz ) ( 0.60 m ) 5.0 10 kg L= = = 0.68 m F 180 N
2
2
(
)
15. REASONING Each pulse travels a distance that is given by vt, where v is the wave speed and t is the travel time up to the point when they pass each other. The sum of the distances traveled by each pulse must equal the 50.0-m length of the wire, since each pulse starts out from opposite ends of the wires.
SOLUTION Using vA and vB to denote the speeds on either wire, we have
vA t + vB t = 50.0 m
Solving for the time t and using Equation 16.2 v = F , we find m/ L
t=
50.0 m = vA + v B
50.0 m FA m/L + FB m/L
=
50.0 m 6.00 10 N + 0.020 kg/m
2
3.00 10 2 N 0.020 kg/m
= 0.17 s
16. REASONING The speed v of a transverse wave on a string is given by v = F / ( m /L ) (Equation 16.2), where F is the tension and m/L is the mass per unit length (or linear density) of the string. The strings are identical, so they have the same mass per unit length. However, the tensions are different. In part (a) of the text drawing, the string supports the entire weight of the 26-N block, so the tension in the string is 26 N. In part (b), the block is supported by the part of the string on the left side of the middle pulley and the part of the string on the right side. Each part supports one-half of the block's weight, or 13 N. Thus, the tension in the string is 13 N.
Chapter 16 Problems
819
SOLUTION a. The speed of the transverse wave in part (a) of the text drawing is
v=
F 26 N = = 2.0 101 m/s m /L 0.065 kg/m
b. The speed of the transverse wave in part (b) of the drawing is v= F 13 N = = 1.4 101 m/s m /L 0.065 kg/m
17. REASONING The speed v of the transverse pulse on the wire is determined by the tension F F in the wire and the mass per unit length m/L of the wire, according to v = (Equation m/ L 16.2). The ball has a mass M. Since the wire supports the weight Mg of the ball and since the weight of the wire is negligible, it is only the ball's weight that determines the tension in the wire, F = Mg. Therefore, we can use Equation 16.2 with this value of the tension and solve it for the acceleration g due to gravity. The speed of the transverse pulse is not given, but we know that the pulse travels the length L of the wire in a time t and that the speed is v = L/t.
SOLUTION Substituting the tension F = Mg and the speed v = L/t into Equation 16.2 for the speed of the pulse on the string gives
v= F m/ L or L Mg = t m/ L
Solving for the acceleration g due to gravity, we obtain
L t g=
2
(m / L)
M
0.95 m -4 1.2 10 kg/m 0.016 s = = 7.7 m/s 2 0.055 kg
2
(
)
18. REASONING AND SOLUTION The speed of the wave on the cable is 1.00 104 N F = = 21.2 m/s (16.2) A (7860 kg/m3 )(2.83 10-3 m 2 ) ______________________________________________________________________________ v= F = (m / L)
820 WAVES AND SOUND
19. SSM REASONING Newton's second law can be used to analyze the motion of the blocks using the methods developed in Chapter 4. We can thus determine an expression that relates the magnitude P of the pulling force to the magnitude F of the tension in the wire. Equation 16.2 [ v = F /(m / L) ] can then be used to find the tension in the wire.
SOLUTION The following figures show a schematic of the situation described in the problem and the free-body diagrams for each block, where m1 = 42.0 kg and m2 = 19.0 kg . The pulling force is P, and the tension in the wire gives rise to the forces F and -F, which act on m1 and m2, respectively.
Tension in wire
m1 F m2 P FN1 FN2
F F
-FF
P
m1g
m2g
Newton's second law for block 1 is, taking forces that point to the right as positive, F = m1a , or a = F / m1 . For block 2, we obtain P - F = m2 a . Using the expression for a obtained from the equation for block 1, we have
m PF =F 2 m 1
or
m P = F 2 + F = F m 1
m2 m + 1 1
According to Equation 16.2, F = v 2 (m / L) , where m/L is the mass per unit length of the wire. Combining this expression for F with the expression for P, we have m 19.0 kg P = v 2 (m / L) 2 + 1 = (352 m/s)2 (8.50 10-4 kg/m) + 1 = 153 N m 42.0 kg 1 ______________________________________________________________________________ 20. REASONING AND SOLUTION Initially, the tension in the wire is F0 = (m/L)v2 = (7.0 103 kg/m)(46 m/s)2 As the temperature is lowered, the wire will attempt to shrink by an amount L = aLT, where a is the coefficient of thermal expansion. Since the wire cannot shrink, a stress will develop, according to Stress = YL/L = YaT where Y is Young's modulus. This stress corresponds to an additional tension:
Chapter 16 Problems
821
F = (Stress)A = YAaT = (1.1 1011 Pa)(1.1 106 m2)(17 106/C)(14 C) The total tension in the wire at the lower temperature is now F = F0 + F, so that the new speed of the waves on the wire is = 79 m/s m/ L ______________________________________________________________________________ 21. REASONING AND SOLUTION If the string has length L, the time required for a wave on the string to travel from the center of the circle to the ball is
t= L vwave
v=
F0 + F
(1)
The speed of the wave is given by text Equation 16.2
vwave = F mstring / L
(2)
The tension F in the string provides the centripetal force on the ball, so that F = mball 2 r = mball 2 L Eliminating the tension F from Equations (2) and (3) above yields vwave = mball 2 L mstring / L = mball 2 L2 mstring =L mball 2 mstring (3)
Substituting this expression for vwave into Equation (1) gives
t= L L mball mstring
2
=
mstring mball
2
=
0.0230 kg
(15.0 kg ) (12.0 rad/s )
2
= 3.26 10-3 s
______________________________________________________________________________ 22. REASONING AND SOLUTION a. Comparing with Equations 16.3 and 16.4, we see that the wave travels in the +x direction b. The displacement at x = 13 m and t = 38 s is
822 WAVES AND SOUND
y = ( 0.26 m ) sin ( 38 ) - ( 3.7 ) (13) = -0.080 m ______________________________________________________________________________ 23. SSM REASONING Since the wave is traveling in the +x direction, its form is given by Equation 16.3 as 2 x y = A sin 2 ft - 2 We are given that the amplitude is A = 0.35 m. However, we need to evaluate 2f and . Although the wavelength is not stated directly, it can be obtained from the values for the speed v and the frequency f, since we know that v = f (Equation 16.1).
SOLUTION Since the frequency is f = 14 Hz, we have
2f = 2(14 Hz) = 88 rad/s It follows from Equation 16.1 that 2 = 2 f 2 (14 Hz ) = = 17 m -1 v 5.2 m/s in Equation 16.3, we have
Using these values for 2f and
2
2 x y = A sin 2 ft - y = ( 0.35 m ) sin ( 88 rad/s ) t - 17 m -1 x ______________________________________________________________________________ 24. REASONING AND SOLUTION We find from the graph on the left that = 0.060 m 0.020 m = 0.040 m and = 0.010 m. From the graph on the right we find that = 0.30 s 0.10 s = 0.20 s. Then, f = 1/(0.20 s) = 5.0 Hz. Substituting these into Equation 16.3 we get 2 x y = A sin 2 f t and y = ( 0.010 m ) sin (10 t 50 x ) ______________________________________________________________________________
(
)
Chapter 16 Problems
823
25. REASONING The mathematical form for the displacement of a wave traveling in the x 2 x . direction is given by Equation 16.4: y = A sin 2 f t +
SOLUTION Using Equation 16.1 and the fact that f = 1/ T , we obtain the following numerical values for f and : f = 1/ T = 1.3 Hz , and = v / f = 9.2 m . Omitting units and substituting these values for f and into Equation 16.4 gives
y = ( 0.37 m ) sin ( 2.6 t + 0.22 x )
______________________________________________________________________________ 26. REASONING The speed of a wave on the string is given by Equation 16.2 as v =
F , m/ L where F is the tension in the string and m/L is the mass per unit length (or linear density) of the string. The wavelength is the speed of the wave divided by its frequency f (Equation 16.1).
SOLUTION a. The speed of the wave on the string is
v=
F 15 N = = 4.2 m/s ( m / L ) 0.85 kg/m
b. The wavelength is
=
v 4.2 m/s = = 0.35 m f 12 Hz
c. The amplitude of the wave is A = 3.6 cm = 3.6 10-2 m. Since the wave is moving along the -x direction, the mathematical expression for the wave is given by Equation 16.4 as 2 x y = A sin 2 f t + Substituting in the numbers for A, f, and , we have
2 x 2 x -2 y = A sin 2 f t + = 3.6 10 m sin 2 (12 Hz ) t + 0.35 m
(
)
= 3.6 10-2 m sin ( 75 s -1 ) t + (18 m -1 ) x ______________________________________________________________________________
(
)
824 WAVES AND SOUND
27. SSM REASONING According to Equation 16.2, the tension F in the string is given by F = v 2 (m / L) . Since v = f from Equation 16.1, the expression for F can be written m F = ( f )2 L (1)
where the quantity m / L is the linear density of the string. In order to use Equation (1), we must first obtain values for f and ; these quantities can be found by analyzing the expression for the displacement of a string particle.
SOLUTION The displacement is given by y = (0.021 m)sin(25t - 2.0x) . Inspection of this 2 x , gives equation and comparison with Equation 16.3, y = A sin 2 f t
2 f = 25 rad/s and
or
f =
25 Hz 2
2
= 2.0 m 1
or
=
2 m 2.0
Substituting these values f and into Equation (1) gives
25 Hz m 2 (1.6 10 2 kg/m) = 2.5 N F = ( f ) = m 2 L 2.0 ______________________________________________________________________________
2 2
28. REASONING Using either Equation 16.3 or 16.4 with x = 0 m, we obtain
y = A sin (2 f t )
Applying this equation with f = 175 Hz, we can find the times at which y = 0.10 m.
SOLUTION Using the given values for y and A, we obtain
0.10 m = (0.20 m )sin (2 f t )
or
0.10 m 2 f t = sin 1 = sin 1 (0.50 ) 0.20 m
The smallest two angles for which the sine function is 0.50 are 30 and 150. However, the values for 2ft must be expressed in radians. Thus, we find that the smallest two angles for which the sine function is 0.50 are expressed in radians as follows:
(30)
2 360
and
(150 )
2 360
Chapter 16 Problems
825
The times corresponding to these angles can be obtained as follows:
2 2 f t1 = (30 ) 360 t1 = 30 f (360) 2 2 f t 2 = (150 ) 360 t2 = 150 f (360 )
Subtracting and using f = 175 Hz gives
(150 30 ) 150 30 = = 1.9 10 3 s f (360) f (360 ) (175 Hz )(360 ) ______________________________________________________________________________
t2 t1 =
29. SSM REASONING AND SOLUTION The speed of sound in an ideal gas is given by Equation 16.5, v = kT / m . The ratio of the speed of sound v2 in the container (after the temperature change) to the speed v1 (before the temperature change) is
v2 = v1 T2 T1
Thus, the new speed is 405 K = 1730 m/s T1 201 K ______________________________________________________________________________
v 2 = v1 T2
= (1220 m/s)
30. REASONING The wavelength of a sound wave is equal to its speed v divided by its frequency f (Equation 16.1): v = (1) f The speed of sound in a gas is given by Equation 16.5 as v = kT / m , where T is the Kelvin temperature and m is the mass of a single air molecule. The Kelvin temperature is related to the Celsius temperature Tc by T = Tc + 273.15 (Equation 12.1), so the speed of sound can be expressed as
v=
k (Tc + 273.15 )
m
(2)
We know that the speed of sound is 343 m/s at 20.0 C, so that 343 m/s =
k ( 20.0 C + 273.15 )
m
(3)
826 WAVES AND SOUND
Dividing Equation (2) by (3) gives
v = 343 m/s
Tc + 273.15 m = 20.0 C + 273.15 k ( 20.0 C + 273.15 ) m
k (Tc + 273.15 )
(4)
SOLUTION Solving Equation (4) for v and substituting the result into Equation (1) gives
Tc + 273.15 ( 343 m/s ) 35 C + 273.15 v 20.0 C + 273.15 = 20.0 C + 273.15 = 3.9 10-3 m = = f f 91 103 Hz ______________________________________________________________________________
( 343 m/s )
31. REASONING AND SOLUTION a. The travel time is
t = x/v = (2.70 m)/(343 m/s) = 7.87 10 3 s
b. The wavelength is
l = v/f = (343 m/s)/(523 Hz) = 0.656 m
The number of wavelengths in 2.70 m is, therefore,
(16.1)
Number of wavelengths = (2.70 m)/(0.656 m) = 4.12 ______________________________________________________________________________ 32. REASONING A rail can be approximated as a long slender bar, so the speed of sound in the rail is given by Equation 16.7. With this equation and the given data for Young's modulus and the density of steel, we can determine the speed of sound in the rail. Then, we will be able to compare this speed to the speed of sound in air at 20 C, which is 343 m/s.
SOLUTION The speed of sound in the rail is
v= Y = 2.0 1011 N/m 2 7860 kg/m
3
= 5.0 10 3 m/s
This speed is greater than the speed of sound in air at 20 C by a factor of
v rail v air
5.0 10 3 m/s = = 15 343 m/s
Chapter 16 Problems
827
______________________________________________________________________________ 33. SSM REASONING If we treat the sample of argon atoms like an ideal monatomic gas ( = 1.67 ) at 298 K, Equation 14.6
(
1 2 mvrms 2
= kT can be solved for the root-mean-square
3 2
)
speed v rms of the argon atoms. The speed of sound in argon can be found from Equation 16.5: v = kT / m .
SOLUTION We first find the mass of an argon atom. Since the molecular mass of argon is 39.9 u, argon has a mass per mole of 39.9 10 3 kg /mol. Thus, the mass of a single argon atom is 39.910-3 kg/mol m= = 6.6310-26 kg 23 -1 6.02210 mol
a. Solving Equation 14.6 for v rms and substituting the data given in the problem statement, we find vrms 3 1.38 10-23 J/K ( 298 K ) 3kT = = = 431 m/s m 6.63 10-26 kg
(
)
b. The speed of sound in argon is, according to Equation 16.5,
= = 322 m/s m 6.63 10-26 kg ______________________________________________________________________________ 34. REASONING AND SOLUTION Since f = v/l and the frequency of the sound remains unchanged in water, we have va vw =
v=
kT
(1.67 ) (1.38 10-23 J/K ) ( 298 K )
a
w
Taking the speed of sound in fresh water from Table 16.1, we find that
1482 m/s = 11.8 m = ( 2.74 m ) 343 m/s ______________________________________________________________________________
w = a
vw va
35. REASONING AND SOLUTION The wheel must rotate f = (2200 Hz)/20 = 110 Hz. The angular speed of the wheel is
at
a
frequency
of
= 2 f = 2 (110 Hz) = 690 rad/s
828 WAVES AND SOUND
______________________________________________________________________________ 36. REASONING AND SOLUTION The sound wave will travel at constant speed through the water (speed = vwater) and the copper block (speed = vcopper). Thus, the times it takes for the sound wave to travel downward through the water a distance dwater and downward through the copper a distance dcopper are twater = d water vwater and tcopper = dcopper vcopper
The time required for the incident sound wave to reach the bottom of the copper block just before reflection is equal to the time required for the reflected sound wave to reach the surface of the water. Using values for the speed of sound in water and copper from Table 16.1 in the text, we find that the total time interval between when the sound enters and leaves the water is, then, dcopper d ttotal = 2(twater + tcopper ) = 2 water + vwater vcopper
0.15 m 0.45 m 4 ttotal = 2 + = 6.7 10 s 1482 m/s 5010 m/s ______________________________________________________________________________ 37. REASONING AND SOLUTION a. In order to determine the order of arrival of the three waves, we need to know the speeds of each wave. The speeds for air, water and the metal are va = 343 m/s, vw = 1482 m/s, vm = 5040 m/s The order of arrival is metal wave first, water wave second, air wave third . b. Calculate the length of time each wave takes to travel 125 m. tm = (125 m)/(5040 m/s) = 0.025 s tw = (125 m)/(1482 m/s) = 0.084 s ta = (125 m)/(343 m/s)= 0.364 s Therefore, the delay times are t12 = tw tm = 0.084 s 0.025 s = 0.059 s t13 = ta tm = 0.364 s 0.025 s = 0.339 s ______________________________________________________________________________
Chapter 16 Problems
829
38. REASONING Two facts allow us to solve this problem. First, the sound reaches the microphones at different times, because the distances between the microphones and the source of sound are different. Since sound travels at a speed v = 343 m/s and the sound arrives at microphone 2 later by an interval of Dt = 1.46 10-3 s, it follows that L2 - L1 = vt (1)
Second, the microphones and the source of sound are located at the corners of a right triangle. Therefore, the Pythagorean theorem applies:
2 L2 = D 2 + L1 2
(2)
Since v, Dt, and D are known, these two equations may be solved simultaneously for the distances L1 and L2.
SOLUTION Solving Equation (1) for L2 gives
L2 = L1 + vt Substituting this result into Equation (2) gives
2 ( L1 + vt )2 = D 2 + L1 2 2 L1 + 2 L1vt + ( vt ) = D 2 + L1 2
2 L1vt + ( vt ) = D 2
2
Solving for L1, we find that
2 (1.50 m ) - ( 343 m/s ) 1.46 10-3 s D 2 - ( vt ) = L1 = 2vt 2 ( 343 m/s ) 1.46 10-3 s 2 2
(
(
)
)
2
= 2.00 m
Solving Equation (2) for L2 and substituting the result for L1 reveal that
2 L2 = D 2 + L1 =
(1.50 m )2 + ( 2.00 m )2
= 2.50 m
______________________________________________________________________________
830 WAVES AND SOUND
39. SSM REASONING The sound will spread out uniformly in all directions. For the purposes of counting the echoes, we will consider only the sound that travels in a straight line parallel to the ground and reflects from the vertical walls of the cliff. Let the distance between the hunter and the closer cliff be x1 and the distance from the hunter to the further cliff near wall far wall be x2 . The first echo arrives at the location of the hunter after x1 traveling a total distance 2x1 in a time t1, so that, if vs is x2 the speed of sound, t1 = 2x1 / vs . Similarly, the second echo arrives after reflection from the far wall and in an point of amount of time t2 after the firing of the gun. The firing quantity t2 is related to the distance x2 and the speed of sound vs according to t2 = 2x2 / vs . The time difference between the first and second echo is, therefore 2 t = t2 t1 = (x x1) vs 2 (1)
The third echo arrives in a time t3 after the second echo. It arises from the sound of the second echo that is reflected from the closer cliff wall. Thus, t3 = 2x1 / vs , or, solving for x1 , we have vt x1 = s 3 (2) 2 Combining Equations (1) and (2), we obtain
t = t 2 t 1 = vst 3 2 x2 vs 2
Solving for x2 , we have
v x2 = s (t + t3 ) 2
(3)
The distance between the cliffs can be found from d = x1 + x 2 , where x1 and x2 can be determined from Equations (2) and (3), respectively.
SOLUTION According to Equation (2), the distance x1 is
x1 =
(343 m/s)(1.1 s) = 190 m 2
According to Equation (3), the distance x2 is x2 =
(343 m/s )
2
(1.6 s + 1.1 s ) = 460 m
Chapter 16 Problems
831
Therefore, the distance between the cliffs is d = x1 + x 2 = 190 m + 460 m = 650 m ______________________________________________________________________________ 40. REASONING Since the sound wave travels twice as far in neon as in krypton in the same time, the speed of sound in neon must be twice that in krypton: vneon = 2vkrypton Furthermore, the speed of sound in an ideal gas is given by v = (1)
, according to m Equation 16.5. In this expression is the ratio of the specific heat capacities at constant pressure and constant volume and is the same for either gas (see Section 15.6), k is Boltzmann's constant, T is the Kelvin temperature, and m is the mass of an atom. This expression for the speed can be used for both gases in Equation (1) and the result solved for the temperature of the neon.
SOLUTION Using Equation 16.5 in Equation (1), we have
kT
kTneon
mneon
Speed in neon
=2
kTkrypton
mkrypton
Speed in krypton
Squaring this result and solving for the temperature of the neon give
kTneon
mneon
kTkrypton = 4 mkrypton
or
m Tneon = 4 neon mkrypton
Tkrypton
We note here that the mass m of an atom is proportional to its atomic mass in atomic mass units (u). As a result, the temperature of the neon is
m 20.2 u Tneon = 4 neon Tkrypton = 4 ( 293 K ) = 283 K mkrypton 83.8 u ______________________________________________________________________________
41. SSM REASONING Equation 16.7 relates the Young's modulus Y, the mass density r, and the speed of sound v in a long slender solid bar. According to Equation 16.7, the Young's modulus is given by Y = v 2 .The data given in the problem can be used to compute values for both r and v.
832 WAVES AND SOUND
SOLUTION Using the values of the data given in the problem statement, we find that the speed of sound in the bar is
v=
L 0.83 m = = 4.4 103 m/s -4 t 1.9 10 s
where L is the length of the rod and t is the time required for the wave to travel the length of the rod. The mass density of the bar is, from Equation 11.1, = m / V = m /( LA) , where m and A are, respectively, the mass and the cross-sectional area of the rod. The density of the rod is, therefore, m 2.1 kg = = = 1.9 104 kg/m3 LA (0.83 m)(1.3 10 4 m 2 ) Using these values, we find that the bulk modulus for the rod is Y = v 2 = 1.9 104 kg/m3 4.4 103 m/s
(
)(
)
2
= 3.7 1011 N/m 2
Comparing this value to those given in Table 10.1, we conclude that the bar is most likely made of tungsten . ______________________________________________________________________________ 42. REASONING Let vP represent the speed of the primary wave and vS the speed of the secondary wave. The travel times for the primary and secondary waves are tP and tS, respectively. If x is the distance from the earthquake to the seismograph, then tP = x/ vP and tS = x/ vS. The difference in the arrival times is tS - tP = 1 x x 1 - = x - v vS vP S vP
We can use this equation to find the distance from the earthquake to the seismograph.
SOLUTION Solving the equation above for x gives
tS - tP 78 s = = 8.0 105 m 1 1 1 1 - - vS vP 4.5 103 m/s 8.0 103 m/s ______________________________________________________________________________ x= 43. REASONING AND SOLUTION The speed of sound in an ideal gas is given by text Equation 16.5: kT v= m
Chapter 16 Problems
833
where = cp/cv, k is Boltzmann's constant, T is the Kelvin temperature of the gas, and m is the mass of a single gas molecule. If mTOTAL is the mass of the gas sample and N is the total number of molecules in the sample, then the above equation can be written as v=
kT
(mTOTAL / N )
=
N kT
mTOTAL
(1)
For an ideal gas, PV = NkT, so that Equation (1) becomes, v=
PV
mTOTAL
(1.67)(3.5105 Pa)(2.5 m3 ) = = 8.0 102 m/s 2.3 kg
______________________________________________________________________________ 44. REASONING The speed vtruck of the truck is equal to the distance x it travels divided by the time ttruck it takes to travel that distance:
vtruck = x ttruck
(1)
Since sound also moves at a constant speed, the distance x it travels in reaching the truck is the product of the speed v of sound and the time tsound for it to travel that distance, or x = v tsound. The time for the sound to reach the truck is one-half the round-trip time tRT, so
tsound = 1 tRT . Thus, the distance traveled by the sound can be written as 2
x=v
( 1 tRT ) 2
x v
Substituting this expression for x into Equation (1) gives
vtruck = =
( 1 tRT ) 2
ttruck
ttruck
(2)
Since the air is assumed to be an ideal gas, the speed of sound is related to the Kelvin temperature T and the average mass m of an air molecule by Equation 16.5:
v=
kT
m
834 WAVES AND SOUND
where is the ratio of the specific heat capacity of air at constant pressure to that at constant volume, and k is Boltzmann's constant. Substituting this expression for v into Equation (2) gives kT 1 t m 2 RT vtruck = ttruck
(
)
SOLUTION The temperature of the air must be its Kelvin temperature, which is related to the Celsius temperature Tc by T = Tc + 273.15 (Equation 12.1): T = 56 C + 273.15 = 329 K. The average mass m (in kg) of an air molecule is the same as that determined in Example 4, namely, m = 4.80 10-26 kg. Thus, the speed of the truck is
4.80 10 kg m = 5.46 m/s = ttruck 4.00 s ______________________________________________________________________________ vtruck =
kT
(
1t 2 RT
)
7 5
(1.38 10-23 J/K )( 329 K )
-26
1 ( 0.120 s ) 2
45. REASONING AND SOLUTION The speed of sound in an ideal gas is
v=
kT
m
so that the mass of a gas molecule is m=
kT
v
2
=
(1.67) (1.38 10-23 J/K )( 3.00 102 K )
( 363 m/s )
2
= 5.25 10-26 kg = 5.25 10-23 g
We now need to determine what fraction of the gas is argon and what fraction is neon. First find the mass of each molecule.
mar = mne = 39.9 g/mol = 6.63 10-23 g 23 6.022 10 /mol 20.2 g/mol = 3.35 10-23 g 23 6.022 10 /mol
Let q be the fraction of gas that is in the form of argon and p be the fraction that is neon. We know that q + p = 1. Also, qmar + pmne = m. Substituting p = 1 q in this equation gives qmar + (1 q)mne = m and we can now solve for the fraction
Chapter 16 Problems
835
q=
m - mne mar - mne
Suppressing the units and algebraically canceling the factor of 1023, we find q= and p = 1 q = 1 0.57 = 0.43 = 43% neon ______________________________________________________________________________ 46. REASONING AND SOLUTION a. The distance traveled is x = vt = (343 m/s)(0.580 s) = 199 m. The one-way distance is half this value or 99.5 m . b. Since the speed is proportional to the square-root of the absolute temperature, i.e., v2 T = 2 v1 T1 or v2 = v1 T2 298 K = (343 m/s) = 346 m/s T1 293 K 5.25 - 3.35 = 0.57 = 57% argon 6.63 - 3.35
The distance computed by the theodolite at the higher temperature would be (1/2) vt2, where v = 343 m/s and t2 = (199 m)/(346 m/s). Then (1/2)(343 m/s)(199 m)/(346 m/s) = 98.6 m The percent error is then % error = [(99.5 m 98.6 m)/(99.5 m)] 100% = 0.9 % ______________________________________________________________________________ 47. SSM WWW REASONING We must determine the time t for the warning to travel the vertical distance h = 10.0 m from the prankster to the ears of the man when he is just under the window. The desired distance above the man's ears is the distance that the balloon would travel in this time and can be found with the aid of the equations of kinematics.
SOLUTION Since sound travels with constant speed vs , the distance h and the time t are related by h = vst . Therefore, the time t required for the warning to reach the ground is
836 WAVES AND SOUND
t=
h 10.0 m = = 0.0292 s vs 343 m/s
We now proceed to find the distance that the balloon travels in this time. To this end, we must find the balloon's final speed vy, after falling from rest for 10.0 m. Since the balloon is
2 2 dropped from rest, we use Equation 3.6b ( v y = v0 y + 2a y y ) with v0y = 0 m/s:
2 v y = v0 y + 2a y y =
( 0 m/s )2 +2(9.80 m/s2 )(10.0 m) = 14.0 m/s
Using this result, we can find the balloon's speed 0.0292 seconds before it hits the man by solving Equation 3.3b ( v y = v0 y + a y t ) for v0 y : v0 y = v y - a y t = (14.0 m/s) - (9.80 m/s 2 )(0.0292 s) = 13.7 m/s Finally, we can find the desired distance y above the man's head from Equation 3.5b:
y = v0 y t + a y t 2 = (13.7 m/s)(0.0292 s) + (9.80 m/s 2 )(0.0292 s)2 = 0.404 m
1 2 1 2
______________________________________________________________________________ 48. REASONING AND SOLUTION We have I = P/A. Therefore, P = IA = (3.2 106 W/m2)(2.1 103 m2) = 6.7 10 9 W (16.8)
______________________________________________________________________________ 49. SSM WWW REASONING AND SOLUTION Since the sound radiates uniformly in all directions, at a distance r from the source, the energy of the sound wave is distributed over the area of a sphere of radius r. Therefore, according to Equation 16.9 [ I = P /(4 r 2 ) ] with r = 3.8 m, the power radiated from the source is
P = 4 Ir 2 = 4 (3.6 10 2 W/m 2 )(3.8 m)2 = 6.5 W
______________________________________________________________________________ 50. REASONING AND SOLUTION Since the sound spreads out uniformly in all directions, the intensity is uniform over any sphere centered on the source. From text Equation 16.9,
I= P 4 r 2
Then,
Chapter 16 Problems
837
I1 /(4 P r12 ) r22 = = I 2 P /(4 r22 ) r12 Solving for r2, we obtain I1 2.0 10-6 W/m 2 = (120 m ) = 190 m I2 0.80 10-6 W/m 2 ______________________________________________________________________________ r2 = r1 51. REASONING AND SOLUTION The intensity of the sound falls off according to the square of the distance according to Equation 16.9. Therefore, the intensity at the new location will be 22 m -4 2 5 2 I 78 = ( 3.0 10 W/m ) = 2.4 10 W/m 78 m ______________________________________________________________________________ 52. REASONING AND SOLUTION The intensity of the "direct" sound is given by text Equation 16.9, P I DIRECT = 4 r 2 The total intensity at the point in question is ITOTAL = IDIRECT + IREFLECTED 1.1 10-3 W + 4.410-6 W/m 2 = 1.4 10 5 W/m 2 = 2 4 (3.0 m) ______________________________________________________________________________ 53. SSM REASONING AND SOLUTION According to Equation 16.8, the power radiated by the speaker is P = IA = I r 2 , where r is the radius of the circular opening. Thus, the radiated power is P = (17.5 W/m 2 )( )(0.0950 m)2 = 0.496 W As a percentage of the electrical power, this is 0.496 W 100 % = 1.98 % 25.0 W ______________________________________________________________________________ 54. REASONING According to Equation 16.8, the intensity I of a sound wave is equal to the sound power P divided by the area A through which the power passes; I = P/A. The area is
2
838 WAVES AND SOUND
that of a circle, so A = r2, where r is the radius. The power, on the other hand, is the energy E per unit time, P = E/t, according to Equation 6.10b. Thus, we have
E P I= = t2 A r
All the variables in this equation are known except for the time.
SOLUTION Solving the expression above for the time yields
t= 4800 J E = 2 I r 5.9 103 W/m 2 1.8 10-2 m = 8.0 102 s
(
) (
)
2
______________________________________________________________________________ 55. REASONING
P , according to A Equation 16.8. The area is given directly, but the power is not. Therefore, we need to recast this expression in terms of the data given in the problem. Power is the change in energy per unit time, according to Equation 6.10b. In this case the energy is the heat Q that causes the Q temperature of the lasagna to increase. Thus, the power is P = , where t denotes the time. t As a result, Equation 16.8 for the intensity becomes
Intensity I is power P divided by the area A, or I =
I=
P Q = A tA
(1)
According to Equation 12.4, the heat that must be supplied to increase the temperature of a substance of mass m by an amount DT is Q = cmDT, where c is the specific heat capacity. Substituting this expression into Equation (1) gives
I= Q cmT = tA tA
(2)
SOLUTION Equation (2) reveals that the intensity of the microwaves is
I= cmT 3200 J/ ( kg C ) ( 0.35 kg )( 72 C ) = = 7.6 103 W/m 2 -2 2 tA ( 480 s ) 2.2 10 m
(
)
______________________________________________________________________________ 56. REASONING Equation 16.9 gives the intensity I (the power per unit area) as
x
123 m
Source 1
Source 2
Chapter 16 Problems
839
I = P/(4 r2) for uniformly radiating sources. The drawing at the right shows the two sound sources and the point where the intensity of each sound is the same; this point is located a distance x from the origin. We will apply Equation 16.9 to each source and set the two intensities equal. The resulting quadratic expression will allow us to find the two values for x.
SOLUTION Using P1 and P2 to denote the power emitted by source 1 and source 2 and applying Equation 16.9, we have
P 1 4 x 2
Intensity of sound 1
=
4 (123 m x )
Intensity of sound 2
P2
2
or
x = =4 P2 123 m x
P 1
2
where we have used the fact that P1 = 4P2. Taking both the positive and negative square roots of the equation on the right gives
Positive root x = +2 123 m x x = 2 (123 m x ) x = 82.0 m Negative root x = 2 123 m x x = 2 (123 m x ) x = 246 m
_____________________________________________________________________________________________
57. SSM REASONING Since the sound is emitted from the rocket uniformly in all directions, the energy carried by the sound wave spreads out uniformly over concentric spheres of increasing radii r1, r2, r3, . . . as the wave propagates. Let r1 represent the radius when the measured intensity at the ground is I (position 1 for the rocket) and r2 represent the radius when the measured intensity at the ground is
1I 3
(position 2 of the rocket). The
time required for the energy to spread out over the sphere of radius r1 is t1 = r1 / vs , where vs is the speed of sound. Similarly, the time required for the energy to spread out over a sphere of radius r2 is t2 = r2 / vs . As the sound wave emitted at position 1 spreads out uniformly, the rocket continues to accelerate upward to position 2 with acceleration ay for a time t12. Therefore, the time that elapses between the two intensity measurements is t = t2 - t1 + t12 From Equation 3.3b, the time t12 is (1)
840 WAVES AND SOUND
t12 =
( v2 - v1 )
ay
(2)
where v1 and v2 are the speeds of the rocket at positions 1 and 2, respectively. Combining Equations (1) and (2), we obtain
t = t2 - t1 +
( v2 - v1 )
ay
(3)
2 The respective values of v1 and v2 can be found from Equation 3.6b ( v 2 = v0 y + 2a y y ) with y
v0 y = 0 m/s, and y = r . Once values for v1 and v2 are known, Equation (3) can be solved directly.
SOLUTION From the data given in the problem statement, r1 = 562 m . We can find r2 using the following reasoning: from the definition of intensity, I1 = P /(4 r12 ) and
I 2 = P /(4 r22 ) . Since I1 = 3I2 , we have P /(4 r12 ) = 3P /(4 r22 ) The times t1 and t2 are t1 = and t2 = r2 973 m = = 2.84 s vs 343 m/s r1 562 m = = 1.64 s vs 343 m/s or r2 = r1 3 = 973 m
Then, taking up as the positive direction, we have from Equation 3.6b, v1 = 2a y r1 = 2(58.0 m/s 2 )(562 m) = 255 m/s and v2 = 2a y r2 = 2(58.0 m/s 2 )(973 m) = 336 m/s Substituting these values into Equation (3), we have 2.6 s 58.0 m/s 2 ______________________________________________________________________________ t = 2.84 s - 1.64 s +
( 336 m/s - 255 m/s ) =
Chapter 16 Problems
841
58. REASONING This is a situation in which the intensities Iman and Iwoman (in watts per square
meter) detected by the man and the woman are compared using the intensity level , expressed in decibels. This comparison is based on Equation 16.10, which we rewrite as follows: I = (10 dB ) log man I woman
SOLUTION Using Equation 16.10, we have
= 7.8 dB = (10 dB ) log
I man I woman
or
I log man I woman
7.8 dB = 0.78 = 10 dB
Solving for the intensity ratio gives I man = 100.78 = 6.0
I woman ______________________________________________________________________________ 59. REASONING AND SOLUTION I The intensity level in dB is = (10 dB)log , I0 Equation 16.10, where b = 14 dB. Therefore,
I = 1010 dB = 101.4 = 25 I0 ______________________________________________________________________________
60. REASONING The sound intensity level outside the room is
outside = (10 dB ) log
I outside I0
(16.10)
where I0 is the threshold of hearing. Solving for the intensity Ioutside gives
outside
I outside = I 01010 dB These two relations will allow us to find the sound intensity outside the room.
SOLUTION From the problem statement we know that outside = inside + 44.0 dB. We can evaluate inside by applying Equation 16.10 to the inside of the room:
842 WAVES AND SOUND
inside
I inside 1.20 10-10 W/m 2 = (10 dB ) log = 10 dB ) log 1.00 10-12 W/m 2 = 20.8 dB I ( 0
Thus, the sound intensity level outside the room is outside = 20.8 d + 44.0 dB = 64.8 dB. The sound intensity outside the room is
outside
I outside = I 010
10 dB
= 1.00 10
(
-12
W/m 10
2
)
64.8 dB 10 dB
= 3.02 10-6 W/m 2
______________________________________________________________________________ 61. SSM REASONING AND SOLUTION The intensity level b in decibels (dB) is related to the sound intensity I according to Equation 16.10:
= (10 dB ) log
I I0
where the quantity I0 is the reference intensity. Therefore, we have
2 - 1 = (10 dB ) log
I2 I I 2 / I0 I - (10 dB ) log 1 = (10 dB ) log = (10 dB ) log 2 I I I /I 1 I0 0 1 0
Solving for the ratio I2 / I1 , we find I 30.0 dB = (10 dB ) log 2 I 1 or I2 = 10 3.0 = 1000 I1
Thus, we conclude that the sound intensity increases by a factor of 1000 . ______________________________________________________________________________ 62. REASONING The intensity level is related to the sound intensity I according to Equation 16.10: I = (10 dB ) log I 0 where I0 is the reference level sound intensity. Solving for I gives
10 dB
I = I 010
Chapter 16 Problems
843
SOLUTION Taking the ratio of the sound intensity Irock at the rock concert to the intensity Ijazz at the jazz fest gives
rock jazz
10 dB
I rock I jazz
=
I 0 10 I 0 10
10 dB
=
10
115 dB 10 dB 95 dB 10 dB
= 1.0 102
10
____________________________________________________________________________________________
63. REASONING AND SOLUTION a.
= (10 dB) log
PA PB
250 W = 7.4 dB = (10 dB) log 45 W
b. No , A will not be twice as loud as B, since it requires an increase of 10 dB to double the loudness. ______________________________________________________________________________ 64. SSM REASONING We must first find the intensities that correspond to the given sound intensity levels (in decibels). The total intensity is the sum of the two intensities. Once the total intensity is known, Equation 16.10 can be used to find the total sound intensity level in decibels.
SOLUTION Since, according to Equation 16.10, = (10 dB ) log (I / I 0 ) , where I0 is the
reference intensity corresponding to the threshold of hearing I 0 = 1.00 10-12 W/m 2 , it follows that I = I 0 10 / (10 dB) . Therefore, if 1 = 75.0 dB and 2 = 72.0 dB at the point in question, the corresponding intensities are
I1 = I 0 10 1 / (10 dB) = (1.00 10 12 W/m 2 ) 10(75.0 dB) / (10 dB) = 3.16 10-5 W/m 2 I 2 = I 0 10 2 / (10 dB ) = (1.00 10 12 W/m 2 ) 10(72.0 dB) / (10 dB) = 1.58 10-5 W/m 2
(
)
Therefore, the total intensity Itotal at the point in question is I total = I1 + I 2 = 3.16 10-5 W/m 2 + 1.58 10-5 W/m 2 = 4.74 10-5 W/m 2 and the corresponding intensity level btotal is
(
) (
)
844 WAVES AND SOUND
I total 4.74 10-5 W/m 2 = (10 dB ) log 1.00 10-12 W/m 2 = 76.8 dB I0 ______________________________________________________________________________
total = (10 dB ) log
65. REASONING AND SOLUTION Let 2 and I2 denote the intensity level and the intensity,
respectively, when two rifles are shot. Let 1 and I1 denote the intensity level and intensity, respectively, when one rifle is shot. Then
2 = (10 dB ) log
I2 2I = (10 dB ) log 1 I I0 0
where I1 is the intensity of a single rifle. Therefore, 2 I1 = 10 2 / (10 dB) I0 1.00 10-12 W/m 2 8.00 I = 5.00 10-5 W/m 2 I1 = 0 10 2 / (10 dB) = 10 2 2 I1 5.00 10-5 W/m 2 1 = (10 dB ) log = (10 dB ) log = 77.0 dB -12 W/m 2 1.00 10 I0 ______________________________________________________________________________ 66. REASONING The energy incident on the eardrum is equal to the sound power P (which is assumed to be constant) times the time interval t (see Equation 6.10b): Energy = Pt (1)
According to Equation 16.8, the sound power is equal to the intensity I of the wave times the area A of the eardrum through which the power passes, or P = I A . Substituting this relation into Equation (1) gives Energy = Pt = ( I A ) t (2) We can obtain an expression for the sound intensity by employing the definition of the sound intensity level (Equation 16.10)
= (10 dB ) log
I I0
where I0 is the threshold of hearing. Solving this expression for I gives
Chapter 16 Problems
845
I = I 010
10 dB
Substituting this relation for I into Equation (2), we have that
I 1010 dB At Energy = ( I A ) t = 0
SOLUTION Noting that I0 = 1.00 10-12 W/m2 and that t = 9.0 h = 3.24 104 s, the energy incident on the eardrum is
I 1010 dB At Energy = 0
90.0 dB 2) -12 = (1.00 10 W/m 10 10 dB ( 2.0 10-4 m 2 )( 3.24 104 s ) = 6.5 10-3 J ______________________________________________________________________________
67. REASONING The sound intensity level heard by the gardener increases by 10.0 dB because distance between him and the radio decreases. The intensity of the sound is greater when the radio is closer than when it is further away. Since the unit is emitting sound uniformly (neglecting any reflections), the intensity is inversely proportional to the square of the distance from the radio, according to Equation 16.9. Combining this equation with Equation 16.10, which relates the intensity level in decibels to the intensity I, we can use the 10.0-dB change in the intensity level to find the final vertical position of the radio. Once that position is known, we can then use kinematics to determine the fall time.
SOLUTION Let the sound intensity levels at the initial and final positions of the radio be i
and f , respectively. Using Equation 16.10 for each, we have If Ii - (10 dB ) log I 0 I0
f - i = (10 dB ) log
I f / I0 = (10 dB ) log I /I i 0
= (10 dB ) log
If I i
(1)
Since the radiation is uniform, Equation 16.9 can be used to substitute for the intensities Ii and If , so that Equation (1) becomes
I f - i = (10 dB ) log f I i P / 4 h 2 h f = (10 dB ) log = (10 dB ) log i 2 hf P / 4 hi
(
(
)
)
h = ( 20 dB ) log i hf
2
(2)
Since f i = 10.0 dB, Equation (2) gives
846 WAVES AND SOUND
f - i = 10.0 dB = ( 20 dB ) log
h i hf
or
hi (10.0 dB ) / ( 20 dB) = 10 = 100.500 = 3.16 hf
We can now determine the final position of the radio as follows: hi hf = 5.1 m = 3.16 hf or hf = 5.1 m = 1.61 m 3.16
It follows, then, that the radio falls through a distance of 5.1 m 1.61 m = 3.49 m. Taking upward as the positive direction and noting that the radio falls from rest (v0 = 0 m/s), we can
1 solve Equation 2.8 y = v 0 t + at 2 from the equations of kinematics for the fall time t: 2
2 ( 3.49 m ) 2y = = 0.84 s a 9.80 m/s 2 ______________________________________________________________________________
t=
68. REASONING AND SOLUTION The intensity level at each point is given by
I= P 4 r 2
2
Therefore, I1 r2 = I 2 r1
Since the two intensity levels differ by 2.00 dB, the intensity ratio is I1 = 100.200 = 1.58 I2 Thus,
r2 = 1.58 r 1
2
We also know that r2 r1 = 1.00 m. We can then solve the two equations simultaneously by substituting, i.e., r2 = r1 1.58 gives r1 1.58 r1 = 1.00 m so that
Chapter 16 Problems
847
r1 = (1.00 m)/[ 1.58 1] = 3.9 m and r2 = 1.00 m + r1 = 4.9 m ______________________________________________________________________________ 69. SSM REASONING AND SOLUTION The sound intensity level b in decibels (dB) is related to the sound intensity I according to Equation 16.10, = (10 dB ) log ( I / I 0 ) , where the quantity I0 is the reference intensity. According to the problem statement, when the sound intensity level triples, the sound intensity also triples; therefore, 3I 3 = (10 dB ) log I 0 Then, 3I I 3I / I 0 3 = (10 dB ) log (10 dB ) log = (10 dB ) log I I I/I 0 0 0 Thus, 2 = (10 dB ) log 3 and
= ( 5 dB ) log 3 = 2.39 dB
______________________________________________________________________________ 70. REASONING The observed frequency changes because of the Doppler effect. As you drive toward the parked car (a stationary source of sound), the Doppler effect is that given by Equation 16.13. As you drive away from the parked car, Equation 16.14 applies.
SOLUTION Equations 16.13 and 16.14 give the observed frequency fo in each case:
f o, toward = fs (1 + vo /v )
Driving toward parked car
and
f o, away = fs (1 vo /v )
Driving away from parked car
Subtracting the equation on the right from the one on the left gives the change in the observed frequency: f o, toward fo, away = 2 fs vo /v Solving for the observer's speed (which is your speed), we obtain
vo = v f o, toward fo, away
17 m/s 2 fs 2 ( 960 Hz ) ______________________________________________________________________________
(
) = (343 m/s )(95 Hz ) =
71. REASONING You hear a frequency fo that is 1.0% lower than the frequency fs emitted by the source. This means that the frequency you observe is 99.0% of the emitted frequency, so
848 WAVES AND SOUND
that fo = 0.990 fs. You are an observer who is moving away from a stationary source of sound. Therefore, the Doppler-shifted frequency that you observe is specified by Equation 16.14, which can be solved for the bicycle speed vo.
SOLUTION Equation 16.14, in which v denotes the speed of sound, states that
v f o = fs 1 - o v Solving for vo and using the fact that fo = 0.990 fs reveal that 0.990 fs f vo = v 1 - o = ( 343 m/s ) 1 - = 3.4 m/s fs fs ______________________________________________________________________________ 72. REASONING Both the observer (you) and the source (the eagle) are moving toward each other. According to Equation 16.15, the frequency fo heard by the observer is related to the frequency fs of the sound emitted by the source by
vo 1+ v f o = fs 1 - vs v
where vo and vs are, respectively, the speeds of the observer and source.
SOLUTION Substituting in the given data, we find that the frequency heard by the observer is
vo 39 m/s 1+ v 1 + 330 m/s = 4.0 103 Hz f o = fs = ( 3400 Hz ) 18 m/s vs 1- 1- 330 m/s v ______________________________________________________________________________ 73. SSM REASONING Since you detect a frequency that is smaller than that emitted by the car when the car is stationary, the car must be moving away from you. Therefore, according to Equation 16.12, the frequency fo heard by a stationary observer from a source moving away from the observer is given by
Chapter 16 Problems
849
1 f o = fs v 1 + s v
where fs is the frequency emitted from the source when it is stationary with respect to the observer, v is the speed of sound, and vs is the speed of the moving source. This expression can be solved for vs .
SOLUTION We proceed to solve for vs and substitute the data given in the problem statement. Rearrangement gives vs f = s 1 v fo
Solving for vs and noting that f o / fs = 0.86 yields
f 1 vs = v s 1 = (343 m/s) 1 = 56 m/s f 0.86 o ______________________________________________________________________________
74. REASONING This problem deals with the Doppler effect in a situation where the source of the sound is moving and the observer is stationary. Thus, the observed frequency is given by Equation 16.11 when the car is approaching the observer and Equation 16.12 when the car is moving away from the observer. These equations relate the frequency fo heard by the observer to the frequency fs emitted by the source, the speed vs of the source, and the speed v of sound. They can be used directly to calculate the desired ratio of the observed frequencies. We note that no information is given about the frequency emitted by the source. We will see, however, that none is needed, since fs will be eliminated algebraically from the solution.
SOLUTION Equations 16.11 and 16.12 are
1 f oApproach = fs 1- v / v s
(16.11)
1 f oRecede = fs 1+ v / v s
(16.12)
The ratio is
f oApproach f oRecede 1 9.00 m/s fs 1+ 1 - vs / v = 1 + vs / v = 343 m/s = 1.054 = 1 1 - vs / v 1 - 9.00 m/s fs 343 m/s 1+ v / v s
850 WAVES AND SOUND
As mentioned in the REASONING, the unknown source frequency fs has been eliminated algebraically from this calculation.
75. SSM REASONING AND SOLUTION In this situation, both the source and the observer are moving through the air. Therefore, the frequency of sound heard by the crew is given by Equation 16.15 using the plus signs in both the numerator and the denominator.
vo 13.0 m/s 1+ v 1 + 343 m/s f o = fs = 1350 Hz = (1550 Hz) 67.0 m/s 1 + vs 1+ 343 m/s v ______________________________________________________________________________
76. REASONING AND SOLUTION The speed of the Bungee jumper, vs, after she has fallen a distance y is given by 2 2 vs = v0 + 2ay Since she falls from rest, v0 = 0 m/s, and
vs = 2ay = 2 -9.80 m/s 2 ( -11.0 m ) = 14.7 m/s
(
)
Then, from Equation 16.11
1 1 f o = fs = (589 Hz) = 615 Hz . 1- v / v 1 - (14.7 m/s ) / ( 343 m/s ) s ______________________________________________________________________________
77. REASONING The Doppler shift that occurs here arises because both the source and the observer of the sound are moving. Therefore, the expression for the Doppler-shifted observed frequency fo is given by Equation 16.15 as
1 vo / v f o = fs 1 v / v s
where fs is the frequency emitted by the source, vo is the speed of the observer, vs is the speed of the source, and v is the speed of sound. The observer is moving toward the source, so we use the plus sign in the numerator. The source is moving toward the observer, so we use the minus sign in the denominator. Thus, Equation 16.15 becomes
Chapter 16 Problems
851
1 + vo / v f o = fs 1- v / v s
Recognizing that both trucks move at the same speed, we can substitute vo = vs = vTruck and solve for vTruck.
SOLUTION Using Equation 16.15 as described in the REASONING and substituting vo = vs = vTruck, we have
1 + vTruck / v f o = fs 1- v Truck / v
or
fo
f v v - o Truck = 1 + Truck fs fs v v
Rearranging, with a view toward solving for vTruck/v, gives
fo fs -1 = vTruck v f v + o Truck f v s
or
vTruck fo fo -1 1 + = v fs fs
Finally, we obtain
fo -1 fs 1.14 - 1 0.14 = = = f o 1 + 1.14 2.14 1+ fs
vTruck v
or
0.14 vTruck = ( 343 m/s ) = 22 m/s 2.14
78. REASONING Since the car is accelerating, its velocity is changing. The acceleration as of the car (the "source") is given by Equation 2.4 as
as = vs, final - vs, initial t
(1)
where vs, final and vs, initial are the final and initial velocities of the source and t is the elapsed time. To find vs, final and vs, initial we note that the frequency fo of the sound heard by the observer depends on the speed vs of the source. When the source is moving away from a stationary observer, Equation 16.12 gives this relation as
1 f o = fs v 1+ s v
852 WAVES AND SOUND
where fs is the frequency of the sound emitted by the source and v is the speed of sound. Solving for vs gives
f vs = v s - 1 fo
(2)
We assume that the car is accelerating along the +x axis, so its velocity is always positive. Since the speed of the car is the magnitude of the velocity, the speed has the same numerical value as the velocity. According to Equation (2), the final and initial velocities of the car are:
[Final velocity]
fs - 1 vs, final = v f o, final fs - 1 vs, initial = v f o, initial
(3)
[Initial velocity]
(4)
Substituting Equations (3) and (4) into Equation (1) gives
fs fs - 1 - v - 1 v f o, final f o, initial v f 1 1 = s = - t t f o, final f o, initial
as =
vs, final - vs, initial t
SOLUTION The acceleration of the car is
v f 1 1 1 1 ( 343 m/s ) (1.00 103 Hz ) 2 - - as = s = 1.5 m/s = 14.0 s t f o, final f o, initial 912.0 Hz 966.0 Hz
______________________________________________________________________________ 79. SSM REASONING a. Since the two submarines are approaching each other head on, the frequency fo detected by the observer (sub B) is related to the frequency fs emitted by the source (sub A) by
vo 1 + v f o = fs 1 - vs v
(16.15)
Chapter 16 Problems
853
where vo and vs are the speed of the observer and source, respectively, and v is the speed of the underwater sound b. The sound reflected from submarine B has the same frequency that it detects, namely, fo. Now sub B becomes the source of sound and sub A is the observer. We can still use Equation 16.15 to find the frequency detected by sub A.
SOLUTION a. The frequency fo detected by sub B is
vo 1 + v f o = fs 1 - vs v
8 m/s 1 + 1522 m/s = 1570 Hz = (1550 Hz ) 1 - 12 m/s 1522 m/s
b. The sound reflected from submarine B has the same frequency that it detects, namely, 1570 Hz. Now sub B is the source of sound whose frequency is fs = 1570 Hz. The speed of sub B is vs = 8 m/s. The frequency detected by sub A (whose speed is vo = 12 m/s) is 12 m/s vo 1 + 1522 m/s 1 + v = 1590 Hz f o = fs = (1570 Hz ) 1 - 8 m/s 1 - vs 1522 m/s v ______________________________________________________________________________ 80. REASONING AND SOLUTION The maximum observed frequency is fomax, and the minimum observed frequency is fomin. We are given that fomax fomin = 2.1 Hz, where
v f omax = fs 1 + o v
(16.13)
and
v f omin = fs 1 - o v
(16.14)
We have
v v v f omax - f omin = fs 1 + o - fs 1 - o = 2 fs o v v v
We can now solve for the maximum speed vo of the microphone:
f omax - f omin vo = v 2 fs 2.1 Hz = (343 m/s) = 0.82 m/s 2(440 Hz)
854 WAVES AND SOUND
Using vmax = vo = Aw, Equation 10.6, where A is the amplitude of the simple harmonic 2 2 = = 3.1 rad/s , we have motion and w is the angular frequency, = 2.0 s T
0.82 m/s = 0.26 m 3.1 rad/s ______________________________________________________________________________
A= = vo
81. REASONING Knowing that the threshold of hearing corresponds to an intensity of I0 = 1.00 1012 W/m2, we can solve Equation 16.10 directly for the desired intensity.
SOLUTION Using Equation 16.10, we find
= 115 dB = (10 dB ) log
Solving for I gives
I I0
or
I = 10(115 dB) / (10 dB) = 1011.5 I0
I = I 01011.5 = 1.00 10 12 W/m 2 1011.5 = 0.316 W/m 2
(
)
______________________________________________________________________________ 82. REASONING The speed v, frequency f, and wavelength of the sound are related according to v = f (Equation 16.1). This expression can be solved for the wavelength in terms of the speed and the frequency. The speed of sound in seawater is 1522 m/s, as given in Table 16.1. While the frequency is not given directly, the period T is known and is related to the frequency according to f = 1/T (Equation 10.5).
SOLUTION Substituting the frequency from Equation 10.5 into Equation 16.1 gives
1 v = f = T
Solving this result for the wavelength yields
= vT = (1522 m/s ) 71 10-3 s = 110 m
______________________________________________________________________________ 83. SSM
REASONING AND SOLUTION The speed of sound in a liquid is given by
(
)
Equation 16.6, v = Bad / , where Bad is the adiabatic bulk modulus and r is the density of
2 the liquid. Solving for Bad, we obtain Bad = v . Values for the speed of sound in fresh
Chapter 16 Problems
855
water and in ethyl alcohol are given in Table 16.1. The ratio of the adiabatic bulk modulus of fresh water to that of ethyl alcohol at 20C is, therefore,
2 ( Bad )water vwater water (1482 m/s) 2 (998 kg/m3 ) = 2 = = ( Bad )ethyl alcohol vethyl alcohol ethyl alcohol (1162 m/s)2 (789 kg/m3 )
2.06
______________________________________________________________________________ 84. REASONING The observer of the sound (the bird-watcher) is stationary, while the source (the bird) is moving toward the observer. Therefore, the Doppler-shifted observed frequency is given by Equation 16.11. This expression can be solved to give the ratio of the bird's speed to the speed of sound, from which the desired percentage follows directly.
SOLUTION According to Equation 16.11, the observed frequency fo is related to the frequency fs of the source, and the ratio of the speed of the source vs to the speed of sound v by 1 fo fs v 1 f o = fs or or = = 1- s 1- v / v fs 1 - vs / v fo v s Solving for vs/v gives vs f 1250 Hz = 1- s = 1- = 0.031 v fo 1290 Hz
This ratio corresponds to 3.1% .
85. SSM
REASONING
The tension F in the violin string can be found by solving
Equation 16.2 for F to obtain F = mv 2 / L , where v is the speed of waves on the string and can be found from Equation 16.1 as v = f .
SOLUTION Combining Equations 16.2 and 16.1 and using the given data, we obtain
2 mv 2 2 F= = (m / L) f 2 2 = 7.8 10-4 kg/m ( 440 Hz ) 65 10-2 m = 64 N L ______________________________________________________________________________
(
)
(
)
86. REASONING AND SOLUTION a. Since 15 boxcars pass by in 12.0 s, the boxcars pass by with a frequency of
f =
15 = 1.25 Hz 12.0 s
b. Since the length of a boxcar corresponds to the wavelength of a wave, we have, from Equation 16.1, that
856 WAVES AND SOUND
v = f = (1.25 Hz )(14.0 m ) = 17.5 m/s ______________________________________________________________________________
87. SSM REASONING AND SOLUTION The sound intensity level b in decibels (dB) is related to the sound intensity I according to Equation 16.10, = (10 dB ) log ( I / I 0 ) , where
I0 is the reference intensity. If b1 and b2 represent two different sound level intensities, then, I I I /I I 2 - 1 = (10 dB ) log 2 - (10 dB ) log 1 = (10 dB ) log 2 0 = (10 dB ) log 2 I I I I /I 1 0 0 1 0 Therefore, I2 = 10( 2 1 ) / (10 dB) I1
When the difference in sound intensity levels is 2 1 = 1.0 dB , the ratio of the sound intensities is I2 = 10(1 dB) / (10 dB) = 1.3 I1 ______________________________________________________________________________ 88. REASONING AND SOLUTION The energy carried by the sound into the ear is Energy = IAt = (3.2 105 W/m2)(2.1 103 m2)(3600 s) = 2.4 10 4 J ______________________________________________________________________________ 89. SSM REASONING As the transverse wave propagates, the colored dot moves up and down in simple harmonic motion with a frequency of 5.0 Hz. The amplitude (1.3 cm) is the magnitude of the maximum displacement of the dot from its equilibrium position.
SOLUTION The period T of the simple harmonic motion of the dot is T = 1/ f = 1/(5.0 Hz) = 0.20 s . In one period the dot travels through one complete cycle of its motion, and covers a vertical distance of 4 (1.3 cm) = 5.2 cm . Therefore, in 3.0 s the dot will have traveled a total vertical distance of
3.0 s (5.2 cm) = 78 cm 0.20 s ______________________________________________________________________________
90. REASONING The speed v of a wave is equal to its frequency f times its wavelength (Equation 16.1). The wavelength is the horizontal length of one cycle of the wave. From the left graph in the text it can be seen that this distance is 0.040 m. The frequency is the
Chapter 16 Problems
857
reciprocal of the period, according to Equation 10.5, and the period is the time required for one complete cycle of the wave to pass. From the right graph in the text, it can be seen that the period is 0.020 s, so the frequency is 1/(0.020 s).
SOLUTION Since the wavelength is = 0.040 m and the period is T = 0.020 s, the speed of the wave is 1 1 v = f = = (0.040 m) = 0.20 m/s T 0.20 s
_____________________________________________________________________________________________
91. SSM REASONING AND SOLUTION: METHOD 1 Since the police car is matching the speed of the speeder, there is no relative motion between the police car and the listener (the speeder). Therefore, the frequency that the speeder hears when the siren is turned on is the same as if the police car and the listener are stationary. Thus, the speeder hears a frequency of 860 Hz .
REASONING AND SOLUTION: METHOD 2 In this situation, both the source (the siren) and the observer (the speeder) are moving through the air. Therefore, the frequency of the siren heard by the speeder is given by Equation 16.15 using the minus signs in both the numerator and the denominator. vo 38 m/s 1 v 1 343 m/s f o = fs = 860 Hz = ( 860 Hz ) 38 m/s 1 vs 1 343 m/s v ______________________________________________________________________________
92. REASONING AND SOLUTION The speed of sound in an ideal gas is given by text Equation 16.5 kT v= m where m is the mass of a single gas particle (atom or molecule). Solving for T gives
T= mv 2 k
(1)
The mass of a single helium atom is 4.003 g/mol 1 kg -27 kg = 6.65010 6.0221023 mol-1 1000 g The speed of sound in oxygen at 0 C is 316 m/s. Since helium is a monatomic gas, = 1.67. Then, substituting into Equation (1) gives
858 WAVES AND SOUND
( 6.65 10-27 kg ) (316 m/s )2 = 28.8 K T= 1.67 (1.38 10-23 J/K )
______________________________________________________________________________ 93. SSM REASONING According to Equation 16.10, the sound intensity level in decibels (dB) is related to the sound intensity I according to = (10 dB ) log ( I / I 0 ) , where the quantity I0 is the reference intensity. Since the sound is emitted uniformly in all directions, the intensity, or power per unit area, is given by I = P /(4 r 2 ) . Thus, the sound intensity at position 1 can be written as I1 = P /(4 r12 ) , while the sound intensity at position 2 can be written as I 2 = P /(4 r22 ) . We can obtain the sound intensity levels from Equation 16.10 for these two positions by using these expressions for the intensities.
SOLUTION Using Equation 16.10 and the expressions for the intensities at the two positions, we can write the difference in the sound intensity levels 21 between the two positions as follows: I I 21 = 2 - 1 = (10 dB ) log 2 - (10 dB ) log 1 I I 0 0
I /I I = (10 dB ) log 2 0 = (10 dB ) log 2 I I /I 1 1 0 P /(4 r22 ) r2 r = (10 dB ) log 12 = (10 dB ) log 1 21 = (10 dB ) log 2 P /(4 r1 ) r2 r2 r = ( 20 dB ) log 1 r 2
2
r = ( 20 dB ) log 1 = ( 20 dB ) log (1/ 2 ) = 6.0 dB 2r 1
The negative sign indicates that the sound intensity level decreases. ______________________________________________________________________________ 94. REASONING AND SOLUTION The emitted power is 2.0 105 J/s = 2.0 105 W. The intensity is, therefore,
I= P P 2.0 105 W = = = 2.2 W/m 2 2 2 A 4 r 4 ( 85 m )
(16.9)
Thus, the sound intensity level is
Chapter 16 Problems
859
I 2.2 W/m 2 = (10 dB ) log = (10 dB ) log (16.10) = 120 dB -12 W/m 2 1.00 10 I0 ______________________________________________________________________________
95. REASONING Suppose we knew the sound intensity ratio knew the sound intensity ratio ratio
IC IB IC IA
IA . In addition, suppose we IB
. Then, all that would be necessary to obtain the desired
would be to multiply the two known ratios:
I A IC I I B A IC = I B
This is exactly the procedure we will follow, except that first we will obtain the intensity ratios from the given intensity levels (expressed in decibels). The intensity level in decibels is given by Equation 16.10.
SOLUTION Applying Equation 16.10 to persons A and B gives
A/B = 1.5 dB = (10 dB ) log
I log A I B 1.5 dB = 10 dB = 0.15
IA IB IA = 100.15 IB
or
In a similar fashion for persons C and A we obtain
C/A = 2.7 dB = (10 dB ) log
I 2.7 dB log C = I 10 dB = 0.27 A
IC IA IC IA
or
= 100.27
Multiplying the two intensity ratios reveals that
I = A IB IB IC IC I A 0.15 100.27 = 100.42 = 2.6 = 10
(
)(
)
860 WAVES AND SOUND
96. REASONING AND SOLUTION First find the speed of the record at a distance of 0.100 m from the center: v = rw = (0.100 m)(3.49 rad/s) = 0.349 m/s (8.9) The wavelength is, then,
l = v/f = (0.349 m/s)/(5.00 103 Hz) = 6.98 10 5 m
(16.1)
______________________________________________________________________________ 97. SSM WWW REASONING Using the procedures developed in Chapter 4 for using Newton's second law to analyze the motion of bodies and neglecting the weight of the wire relative to the tension in the wire lead to the following equations of motion for the two blocks: Fx = F - m1 g (sin 30.0) = 0 (1)
Fy = F - m2 g = 0
(2)
where F is the tension in the wire. In Equation (1) we have taken the direction of the +x axis for block 1 to be parallel to and up the incline. In Equation (2) we have taken the direction of the +y axis to be upward for block 2. This set of equations consists of two equations in three unknowns, m1, m2, and F. Thus, a third equation is needed in order to solve for any of the unknowns. A useful third equation can be obtained by solving Equation 16.2 for F:
F = ( m / L ) v2
(3)
Combining Equation (3) with Equations (1) and (2) leads to (m / L)v 2 - m1g sin 30.0 = 0 (m / L)v 2 - m2 g = 0 Equations (4) and (5) can be solved directly for the masses m1 and m2.
SOLUTION Substituting values into Equation (4), we obtain
(4) (5)
m1 =
( m / L )v 2 (0.0250 kg/m)(75.0 m/s)2 = = 28.7 kg g sin 30.0 (9.80 m/s 2 ) sin 30.0
Similarly, substituting values into Equation (5), we obtain
Chapter 16 Problems
861
(m / L)v 2 (0.0250 kg/m)(75.0 m/s) 2 = = 14.3 kg g (9.80 m/s 2 ) ______________________________________________________________________________
m2 =
98. REASONING A particle of the string is moving in simple harmonic motion. The maximum speed of the particle is given by Equation 10.8 as vmax = A, where A is the amplitude of the wave and is the angular frequency. The angular frequency is related to the frequency f by Equation 10.6, = 2 f, so the maximum speed can be written as vmax = 2 f A. The speed v of a wave on a string is related to the frequency f and wavelength by Equation 16.1, v = f . The ratio of the maximum particle speed to the speed of the wave is
vmax 2 f A 2 A = = v f
The equation can be used to find the wavelength of the wave.
SOLUTION Solving the equation above for the wavelength, we have
2 ( 4.5 cm ) 2 A = = 9.1 cm 3.1 vmax v ______________________________________________________________________________
=
99. REASONING AND SOLUTION a. According to Equation 16.2, the speed of the wave is
v= 315 N F = = 2.20 10 2 m/s -3 m/ L 6.50 10 kg/m
b. According to Equations 10.6 and 10.8, the maximum speed of the point on the wire is
v max = (2 f )A = 2 ( 585 Hz ) 2.50 10 3 m = 9.19 m/s ______________________________________________________________________________
(
)
100. REASONING If I1 is the sound intensity produced by a single person, then N I1 is the sound intensity generated by N people. The sound intensity level generated by N people is given by Equation 16.10 as NI N = (10 dB ) log 1 I 0
862 WAVES AND SOUND
where I0 is the threshold of hearing. Solving this equation for N yields
N I N = 0 1010 dB I 1
(1)
We also know that the sound intensity level for one person is
I 1 = (10 dB ) log 1 I 0
1
or
I1 = I 0 10
10 dB
(2)
Equations (1) and (2) are all that we need in order to find the number of people at the football game.
SOLUTION Substituting the expression for I1 from Equation (2) into Equation (1) gives the desired result.
N
N=
I 0 10
10 dB
1
10 dB
=
10
109 dB 10 dB
10 I 0 10 ______________________________________________________________________________ 101. REASONING AND SOLUTION The sound emitted by the plane at A reaches the person after a time t. This time, t, required for the sound wave at A to reach the person is the same as the time required for the plane to fly from A to B. The figure at the right shows the relevant geometry. During the time t, the plane travels the distance x, while the sound wave travels the distance d. The sound wave travels with constant speed. The plane has a speed of v0 at A and a speed v at B and travels with constant acceleration. Thus,
d = vsound t
60.0 dB 10 dB
= 79 400
x
A
B
d
Person
and
x=
1 2
( v + v0 ) t
sin =
x = d
1 2
From the drawing, we have
( v + v0 ) t =
vsound t
v + v0
2vsound
Solving for v gives
Chapter 16 Problems
863
v = 2vsound sin v0 = 2 ( 343 m/s ) sin 36.0 164 m/s = 239 m/s
______________________________________________________________________________ 102. REASONING AND SOLUTION The speed of sound in fresh water is
vfresh = Bad = 2.20 109 Pa = 1.48 103 m/s 3 3 1.00 10 kg/m
The speed of sound in salt water is
vsalt = Bad 2.37109 Pa = 1.52103 m/s 3 1025 kg/m
=
The ratio of these two speeds is
vsalt vfresh = 1.52103 m/s = 1.03 1.48103 m/s
Thus, the sonar unit erroneously calculates that the objects are 1.03 times the distance measured in salt water: d actual = (1.03)(10.0 m) = 10.3 m ______________________________________________________________________________ 103. CONCEPT QUESTIONS a. The wavelength is the horizontal distance between two successive crests. The horizontal distance between successive crests of wave B is two times greater than that of wave A. Therefore, B has the greater wavelength. b. The frequency f of a wave is related to its speed v and wavelength by Equation 16.1, f = v/ . Since the speed is the same for both waves, the wave with the smaller wavelength has the larger frequency. Therefore wave A, having the smaller wavelength, has the larger frequency.
A, where A is the amplitude of the wave and is the angular frequency, = 2 f. Wave A has both a larger amplitude and frequency, so the maximum particle speed is greater for A.
SOLUTION a. From the drawing, we determine the wavelength of each wave to be
c. The maximum speed of a particle attached to a wave is given by Equation 10.8 as vmax =
A = 2.0 m
and B = 4.0 m .
864 WAVES AND SOUND
b. The frequency of each wave is given by Equation 16.1 as:
fA = v
A
=
12 m/s = 6.0 Hz 2.0 m
and
fB =
v
B
=
12 m/s = 3.0 Hz 4.0 m
c. The maximum speed for a particle moving in simple harmonic motion is given by Equation 10.8 as vmax = A. The amplitude of each wave can be obtained from the drawing: AA = 0.50 m and AB = 0.25 m.
Wave A Wave B
vmax = AA A = AA 2 f A = ( 0.50 m ) 2 ( 6.0 Hz ) = 19 m/s vmax = AB B = AB 2 f B = ( 0.25 m ) 2 ( 3.0 Hz ) = 4.7 m/s
______________________________________________________________________________ 104. CONCEPT QUESTIONS a. Generally, the speed of sound in a liquid like water is greater than in a gas. And, in fact, according to Table 16.1, the speed of sound in water is greater than the speed of sound in air. b. Since the speed of sound in water is greater than in air, an underwater ultrasonic pulse returns to the ruler in a shorter time than a pulse in air. The ruler has been designed for use in air, not in water, so this quicker return time fools the ruler into believing that the object is much closer than it actually is. Therefore, the reading on the ruler is less than the actual distance.
SOLUTION Let x be the actual distance from the ruler to the object. The time it takes for the ultrasonic pulse to reach the object and return to the ruler, a distance of 2x, is equal to the distance divided by the speed of sound in water vwater: t = 2x/vwater. The speed of sound
in water is given by Equation 16.6 as vwater = Bad / , where Bad is the adiabatic bulk modulus and is the density of water. Thus, the time it takes for the pulse to return is
t=
2x 2x = = vwater Bad
2 ( 25.0 m ) 2.37 10 Pa 1025 kg/m3
9
= 3.29 10-2 s
The ruler measures this value for the time and computes the distance to the object by using the speed of sound in air, 343 m/s. The distance xruler displayed by the ruler is equal to the speed of sound in air multiplied by the time 1 t it takes for the pulse to go from the ruler to 2 the object: xruler = vair ( 1 t ) = ( 343 m/s ) ( 1 ) 3.29 10-2 s = 5.64 m 2 2
(
)
Chapter 16 Problems
865
Thus, the ruler displays a distance of xruler = 5.64 m. As expected, the reading on the ruler's display is less than the actual distance. ______________________________________________________________________________ 105. CONCEPT QUESTION You don't want to run the car any faster than you have to, so you'll try and break the sound barrier when the speed of sound in air has its smallest value. The speed of sound in an ideal gas depends on its temperature T through Equation 16.5 as kT , where = 1.40 for air, k is Boltzmann's constant, and m is the mass of a v= m molecule in the air. Thus, the speed of sound depends on the air temperature, with a lower temperature giving rise to a smaller speed of sound. Thus, you should attempt to break the sound barrier in the early morning when the temperature is lower.
SOLUTION The speed of sound in air is v =
, where the temperature T must be m expressed on the Kelvin scale (T = Tc + 273, Equation 12.1). Taking the ratio of the speed of sound at 43 C to that at 0 C, we have
kT
v43 C v0 C
k ( 43 + 273)
=
316 K m = 273 K k ( 0 + 273) m
The speed of sound at 43 C is
v43 C = v0 C 316 K 316 K = ( 331 m/s ) = 356 m/s 273 K 273 K
Therefore, the speed of your car must be vcar = 356 m/s . ______________________________________________________________________________ 106. CONCEPT QUESTIONS a. The source emits sound uniformly in all directions, so the sound intensity I at any distance r is given by Equation 16.9 as I = P / 4 r 2 , where P is the sound power emitted
(
)
by the source. Since patches 1 and 2 are at the same distance from the source of sound, the sound intensity at each location is the same, so I1 = I2. Patch 3 is farther from the sound source, so the intensity I3 is smaller for points on that patch. Therefore, patches 1 and 2 have equal intensities, each of which is greater than the intensity at patch 3. b. According to Equation 16.8, the sound intensity I is defined as the sound power P that passes perpendicularly through a surface divided by the area A of that surface, I = P/A. The area of the surface is, then, A = P/I. Since the same sound power passes through patches 1
866 WAVES AND SOUND
and 2, and the intensity at each one is the same, their areas must also be the same, A1 = A2. The same sound power passes through patch 3, but the intensity at that surface is smaller than that at patches 1 and 2. Thus, the area A3 of patch 3 is larger than that of surface 1 or 2. In summary, A3 is the largest area, followed by A1 and A2, which are equal.
SOLUTION a. The sound intensity at the inner spherical surface is given by Equation 16.9 as
IA = 2.3 W P 2 = = 0.51 W/m 2 2 4 rA 4 ( 0.60 m )
This intensity is the same at all points on the inner surface, since all points are equidistant from the sound source. Therefore, the sound intensity at patches 1 and 2 are equal; I1 = I2 =
0.51 W/m
2
.
The sound intensity at the outer spherical surface is
IB = P 2.3 W 2 = = 0.29 W/m 2 2 4 rB 4 ( 0.80 m )
This intensity is the same at all points on outer surface. Therefore, the sound intensity at patch 3 is I3 = 0.29 W/m 2 . b. The area of a surface is given by Equation 16.8 as the sound power passing perpendicularly through that area divided by the sound intensity, A = P/I. The areas of the three surfaces are: -3 P 1.8 10 W 2 -3 A1 = = = 3.5 10 m Surface 1 2 I1 0.51 W/m
Surface 2
A2 = A3 = P 1.8 10 W 2 -3 = = 3.5 10 m 2 I2 0.51 W/m
-3 -3
P 1.8 10 W 2 -3 = = 6.2 10 m Surface 3 2 I 3 0.29 W/m ______________________________________________________________________________
107. CONCEPT QUESTIONS a. A threshold of hearing of 8.00 dB means that this individual can hear a sound whose intensity is less than I0 = 1.00 10-12 W/m2, which is the intensity of the reference level. b. The person with a threshold of hearing of +12.0 dB can only hear sounds that have intensities greater than I0 = 1.00 10-12 W/m2. Thus, the person whose threshold of
Chapter 16 Problems
867
hearing is -8.00 dB has the better hearing, because he can hear sounds with intensities less than 1.00 10-12 W/m2.
SOLUTION The relation between the sound intensity level and the sound intensity I is given by Equation 16.10:
I = (10 dB ) log or I = I 0 1010 dB I 0
The threshold of hearing intensities for the two people are
1 2
I1 = I 0 10
10 dB
and I 2 = I 0 10
10 dB
Taking the ratio I1/ I2 gives
1
10 dB
I 10 I1 10 = 0 = = 1.00 10-2 2 +12.0 dB I2 10 10 dB I 0 1010 dB ______________________________________________________________________________
-8.00 dB 10 dB
108. CONCEPT QUESTIONS Velocity of Sound Source (Toward the Observer) (a) (b) (c) 0 m/s Velocity of Observer (Toward the Source) 0 m/s 0 m/s
Wavelength
Frequency Heard by Observer
Remains the same Decreases Decreases
Remains the same Increases Increases
a. Since the sound source and the observer are stationary, there is no Doppler effect. The wavelength remains the same and the frequency of the sound heard by the observer remains the same as that emitted by the sound source. b. When the sound source moves toward a stationary observer, the wavelength decreases (see Figure 16.29b). This decrease arises because the condensations "bunch-up" as the source moves toward the observer. The frequency heard by the observer increases, because,
868 WAVES AND SOUND
according to Equation 16.1, the frequency is inversely proportional to the wavelength; a smaller wavelength gives rise to a greater frequency. c. The wavelength decreases for the same reason given in part (b). The increase in frequency is due to two effects; the decrease in wavelength, and the fact that the observer intercepts more wave cycles per second as she moves toward the sound source.
SOLUTION a. The frequency of the sound is the same as that emitted by the siren; f o = fs = 2450 Hz . The wavelength is given by Equation 16.1 as
=
343 m /s v = = 0.140 m fs 2450 Hz
b. According to the discussion in Section 16.9 (see the subsection "Moving source") the wavelength of the sound is given by = - vs T , where vs is the speed of the source and T is the period of the sound. However, T = 1/fs so that
= - vs T = -
vs fs
= 0.140 m -
26.8 m /s = 0.129 m 2450 Hz
The frequency fo heard by the observer is equal to the speed of sound v divided by the shortened wavelength : v 343 m /s = = 2660 Hz fo = 0.129 m c. The wavelength is the same as that in part (b), so = 0.129 m . The frequency heard by the observer can be obtained from Equation 16.15, where we use the fact that the observer is moving toward the sound source:
14.0 m/s v 1+ o 1 + 343 m/s v = ( 2450 Hz ) = 2770 Hz fo = fs vs 1 - 26.8 m/s 1- 343 m/s v ______________________________________________________________________________
109. CONCEPT QUESTIONS a. The tension in the rope is greater near the top than near the bottom. This is because the rope has weight. The part of the rope near the top must support more of that weight than the part of the rope near the bottom. In fact, the very top end must support all the weight of the rope beneath it. In contrast, the very bottom end supports no weight at all, since nothing hangs beneath it.
Chapter 16 Problems
869
b. The speed of the wave is greater near the top of the rope. This follows directly from Equation 16.2, which indicates that the speed of the wave is proportional to the square root of the tension. Since the tension near the top of the rope is greater than the tension near the bottom, the speed is greater near the top. c. The weight is the mass of the section of rope times the acceleration g due to gravity. Since the rope is uniform, the mass of the section is simply the total mass m of the rope times the fraction y/L, which is the length of the section divided by the total length of the y rope. Thus, the weight of the section is m g . L SOLUTION a. According to Equation 16.2, the speed v of the wave is
v= F m/ L
where F is the tension, m is the total mass of the rope, and L is the length of the rope. At a point y meters above the bottom end, the rope is supporting the weight of the section y beneath that point, which is m g , as discussed in Concept Question (c). The rope L supports the weight by virtue of the tension in the rope. Since the rope does not accelerate y upward or downward, the tension must be equal to m g , according to Newton's L second law of motion. Substituting this tension for F in Equation 16.2 reveals that the speed at a point y meters above the bottom end is
y m g L = v= m/ L
yg
b. Using the expression just derived, we find the following speeds
[y = 0.50 m]
v= v=
yg = yg =
( 0.50 m ) ( 9.80 m/s 2 ) = ( 2.0 m ) ( 9.80 m/s 2 ) =
2.2 m/s
[y = 2.0 m]
4.4 m/s
As expected, the speed is greater at the spot higher up the rope.
110. CONCEPT QUESTIONS
870 WAVES AND SOUND
a. Following its release from rest, the platform begins to move down the incline, picking up speed as it goes. Thus, the platform's velocity points down the incline, and its magnitude increases with time. The reason for the increasing velocity is gravity. The acceleration due to gravity points vertically downward and has a component along the length of the incline. b. The changing velocity is related to the acceleration of the platform according to Equation 2.4, which gives the acceleration as the change in the velocity divided by the time interval during which the change occurs. c. The frequency detected by the microphone at the instant the platform is released from rest is the same as the frequency broadcast by the speaker. However, as the platform begins to move away from the speaker, the microphone detects fewer wave cycles per second than the speaker broadcasts. In other words, the microphone detects a frequency that is smaller than that broadcast by the speaker. This is an example of the Doppler effect and occurs because the platform is moving in the same direction as the sound is traveling. As the platform picks up speed, the microphone detects an ever decreasing frequency. d. The speaker is the source of the sound, and the microphone is the "observer." Since the source is stationary and the observer is moving away from the source, the Doppler-shifted observed frequency is given by Equation 16.14.
SOLUTION The acceleration a is directed down the incline and is the change in the velocity divided by the time interval during which the change occurs. The change in the velocity is the velocity vo,t at a later time t minus the velocity vo,t at an earlier time t0.
0
Thus, according to Equation 2.4, the acceleration is
a= vo, t - vo, t t - t0
0
Equation 16.14 gives the frequency fo detected by the microphone in terms of the frequency fs emitted by the speaker, the speed vo, and the speed v of sound:
v f o = fs 1 - o v
Solving for the speed of the mike gives
f vo = v 1 - o fs
Using this result, we can determine the speed of the microphone-platform at the two given times:
Chapter 16 Problems
871
[t = 1.5 s]
9939 Hz vo,1.5 s = ( 343 m/s ) 1 - = 2.1 m/s 1.000 104 Hz 9857 Hz vo, 3.5 s = ( 343 m/s ) 1 - = 4.9 m/s 1.000 104 Hz
[t = 3.5 s]
Using these two values for the velocity, we can now obtain the acceleration using Equation 2.4 vo, 3.5 s - vo,1.5 s 4.9 m/s - 2.1 m/s = = 1.4 m/s 2 a= t - t0 3.5 s - 1.5 s ______________________________________________________________________________

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Evan D'Agostino 106003714 November 27, 20073. The market return Market capitalization Value4. The bond's maturity date refers to a future date on which the issuer pays the principal to the investor. Bond maturities usually range from one day up t

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My PhilosophyDorothy WalkerEducation should provide for the needs of all students. Students who possess superior intelligence should have specific programs to address their needs. Students who have challenging difficulties should also have a prog

SUNY Stony Brook - PHI - 104

Alexandria Hurt Mr. Youngren Philosophy My Philosophy I have always been told by those who know me very well that I "over analyze" and question everything. I am the type of person that doesn't just allow things to be accepted just because it is consi

SUNY Stony Brook - BUS - 330

The subway restaurant that would be the best investment would be the small subway restaurant it have the fastest Cumulative cash flow for payback at 1.41 years compared to the medium store at 2.19 years and the large store at 2.97 years to payback. T

SUNY Stony Brook - PHI - 104

Bibliography 1. Craig, William. "The Absurdity of Life Without God." The Meaning of Life. Second Edition (2000): 40-56. 2. Nielsen, Kai. "Death and the Meaning of Life." The Meaning of Life. Second Edition (2000): 153-159. 3. Stamos, David. The Meani

SUNY Stony Brook - BUS - 348

Evan D'Agostino Septemeber 26, 2007 BUS 348Marketing ArticleDell has recently agreed to sell computers to a major retailer in China this is a large step for Dell's global market to keep up with competitors such as Lenovo Group, HewlettPackard who

SUNY Stony Brook - PHI - 104

Name: Questions to accompany Chekhov's A Woman's KingdomPART I: ON THE EVE (pp. 50-61) 1. Summarize the episode described in these pages in NO MORE THAN THREE sentences. Anna akimovna won a court appeal and was awarded 1,500 roubles, she had felt g

SUNY Stony Brook - BUS - 348

Evan D'Agostino October 24, 2007 ID# 106003714 Bus 348Marketing ArticleThe 97th conference of the Association of National Advertisers hosting 1,200 people was held Thursday through yesterday, heard speaker after "speaker address the growing popul

SUNY Stony Brook - BUS - 220

Evan D'Agostino Vijay Arikupurathu October 2, 2007 BUS 2201a. D=800-10P D=800-10($20)= 600 units D=800-10($70)= 100 unitsb. TR=D*Pc. D=800-10(30)= 500 units D=800-10(40)= 400 units D=800-10(50)= 300 units TR=500*30=$15,000 TR=400*40=$16,000 TR=

SUNY Stony Brook - PHI - 104

Samuel Singleton Philosophy 101 Dr. Jerry Miller 19 September 2007 I began writing my paper last week. When I was looking over my paper last night I realized that none of the ideas in my paper was mine. I was just regurgitating the ideas that previou

SUNY Stony Brook - BUS - 348

Marketing Plan December 5, 2007Bello CosmeticsCrme De La Crme Anti-Aging Skin Revitalizer Prevents Wrinkles Reverses Wrinkles Cures Acne Evens skin tone SPF 60 Sun ProtectionSWOT Strengths: Hollywood Industry Cutting edge cosmetic techn

SUNY Stony Brook - PHI - 104

Sean Rivers Philosophy Mr. Felton Wednesday, November 02, 2005Philosophy MidtermSome things change lives. Just as a certain seasoning will alter a food so that no matter what happens to that food you can always tell that that certain seasoning is

SUNY Stony Brook - HUI - 216

The Roman gladiator was a male figure who, just like "Spartacus" and "Maximus" was usually a slave. These men would compete against another man in a large open amphitheatre known as the Coliseum. These fights were a brutal yet popular form of enterta

SUNY Stony Brook - PHI - 104

Alexandria Hurt Mr. Youngren Midterm Part IIMidterm Part 2 Quote: "I was being gratuitously wanton, having no inducement to evil but the evil itself. It was foul, and I loved it. I loved my own undoing. I loved my error-not that for which I erred b

SUNY Stony Brook - BUS - 330

Evan D'Agostino November 5, 2007 BUS3301. The primary concern should be for safety and liquidity rather than maximization of profit it is good during market volatility and it give you lower interest rates.2. For people who wish to enjoy the benef

SUNY Stony Brook - BUS - 330

Evan D'Agostino2. Capital budgeting decisions emphasis remains on cash flow.4. Cost of capital5. Investment decisions in which the acceptance of a project precludes the acceptance of one or more alternative projects. 7. Yes1. 100,000 -50,000

SUNY Stony Brook - BUS - 330

Evan D'Agostino September 26, 2007 Principles of Finance ID#106003714 2. Market value is what the value of an asset or stock is on the open market. Book value = net worth of the firm = shareholder's equity = ASSETS LIABILITIES. "Book value is better

SUNY Stony Brook - PHI - 104

AristotleThe ancient Greek philosopher Aristotle explored the genre of tragedy in his work of literary criticism. His discussion was hugely influenced on the tragedies of the Elizabethan and Jacobean ages. He thought that a tragedy should involve a

SUNY Stony Brook - PHI - 104

Aristotle's LifeAristotle was born in 384 BC in Stagira, Greece. His father was Nicomachus and mother was Phaestis. His father was court physican to king Amyntas of Macedonia. Aristotle's father died when he was only 10 years old. When Aristotle was

SUNY Stony Brook - BUS - 330

1060037143. Money has a time value. This is because people have a preference for current consumption since they gain satisfaction from consuming goods and services. In order to convince someone to save or invest the money, people are given some com

SUNY Stony Brook - BUS - 220

Hart Manufacturing makes three products. Each product requires manufacturing operations in three departments: A, B and C. The labor-hour requirements, by department, are:Department A B CProduct 1 1.50 2.00 0.25Product 2 3.00 1.00 0.25Product

SUNY Stony Brook - BUS - 330

Evan D'Agostino October 17, 2007 ID#106003714 BUS 3301. Break even analysis allows the firm to determine the level of operations it will break even or earn zero profit, also determine relationship between volume, costs, and profits, and helps the m

SUNY Stony Brook - BUS - 220

BUS 220 Intro to Decision Sciences Credit(s): 3 Prerequisite: BUS Maj/Min or MTD/ECO/ISE/or CME major; BUS 110/111/115; BUS 215; MAT 122/125 or higher Note: Wait list capacity - 15 students 53964 LEC 02 TH 06:50-09:50PM JAVITS LECTR 101 Clark,R BUS 4

Penn State - STS - 200

STS 200 Sec.001 Medical Anthropology: Bioethics and Mental Health The area of medical anthropology I want to focus on is bioethics. According to dictionary.com, bioethics is defined as "a field of study concerned with the ethics and philosophical imp

Penn State - SOC - 197A

SOC. 197A Sec.001 Why is Asian American Studies necessary in Collegiate Curricula? "There's an Asian American Studies class at Penn State?" This was the response from a few of my friends after asking what classes I was taking this spring semester. Ev

SUNY Stony Brook - GEO - 107

Evidence for Plate TectonicsFigure 2-19Evidence for Plate Tectonics Magnetic patterns on sea floors Earthquake epicenters and volcanoes map out major plate boundaries Deep ocean trenches Youngest ocean floor rocks are just formed extremely

SUNY Stony Brook - LIN - 101

TodaySyntaxWhat you know when you know a language You know rules for combining morphemes into words (morphology). You know rules for combining words into phrases and phrases into sentences (syntax).The language faculty: when something goes wro

SUNY Stony Brook - GEO - 107

Geology 107: Natural HazardsProfessor: Christiane Stidham ESS Room 338 Office hours: Wed 10:30 11:30 am, Thu 1-2 pm Telephone: 632-8059 Email: christiane.stidham@stonybrook.edu T.A.: Tsvi Pick ESS Room 345A Office hours: Tue 1-2 pm, Thu 9:45 10:45

SUNY Stony Brook - LIN - 101

Today Sound and spelling How sounds are made English consonant soundsGrammar is a system of rules for associating sound and meaning. How do we make speech sounds?How would you say this?choletharSAME LETTERSDIFFERENT SOUNDS THOUGH COUGH

SUNY Stony Brook - LIN - 101

Today: MorphologyReview: When you know a language, you know Vocabulary or the Lexicon (words, pieces of words) Grammar (rules for putting together words and pieces of words) Different varieties of English may have different vocabulary (lexicon)

SUNY Stony Brook - LIN - 101

TodayHow Languages Differ Embedded SentencesFIRST EXAM: Monday, Feb. 25. Exam will be open book and similar to homework problems.English Phrase Structure Rules1) S => NP VP2) NP => (Det) (Adj) N (PP)3) VP => V (NP) (Adverb) (PP)4) PP => P

SUNY Stony Brook - LIN - 101

Today1. More on English word formation2. Morphology of other languages: How different can languages be?Review English speakers have learned a grammar (rules for associating meaning and sound). This grammar is creative-it allows you to form new

SUNY Stony Brook - LIN - 101

TodayToday: Recursion, Wh-Questions, Passives FIRST EXAM: Monday, Feb. 25. Exam will be open book and similar to homework problems.Basic Rules of Sentence Formation1) Every sentence must have a subject and a predicate. The boy walked. The boy thr

SUNY Stony Brook - GEO - 107

Earth History4.57 billion years ago: formation of solar system and EarthMeasured ages of oldest objects in Solar System (Moon rocks, meteorites) Oldest Earth rocks: 4.055 billion year old Canadian gneiss Oldest bits of rocks found on Earth: 4.4 bil

SUNY Stony Brook - GEO - 107

Seismic Waves: Body WavesAnimation: Body Waves emanating from an EarthquakeSeismic Waves: Body WavesAnimation: P wave vs. S waveSeismic Waves: Surface WavesAnimation: Surface Waves in EarthquakesEarth Structure and Seismic Waves S waves

SUNY Stony Brook - GEO - 107

Transform Plate Boundary North Anatolian fault, TurkeyMagnitude 7.4, 7.1 earthquakes in 1999 caused 19,000 deathshttp:/neic.usgs.gov/neis/world/turkey/tec_setting.htmlTransform Plate Boundary: Bam, Iran fault zone 12/26/2003, magnitude 6.6 At

SUNY Stony Brook - LIN - 101

TodayPhrase Structure: NP, VP, PPFIRST EXAM: Monday, Feb. 25. Exam will be open book and similar to homework problems.What do you see?What do you see?Your brain's job: to receive data from the world. to interpret the data: organize it int

SUNY Stony Brook - LIN - 101

Today Semantics & Pragmatics Rules of ConversationFirst: ReviewQuestion IV:The fact that Obama won Iowa surprised Hillary.Main (big) SentenceMain sentence active: Hillary fears that Obama will win the nomination. Main sentence passive: Tha

SUNY Stony Brook - LIN - 101

Linguistics 101Goals of this class To show you how incredibly smart you are.Goals of this class To show you how incredibly smart you are. To make you aware of how much knowledge you have of your language.Goals of this class To show you how