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- Tipler 6e Ch15 ISM.pdf

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1539 Chapter 15 Traveling Waves Conceptual Problems 1 • [SSM] A rope hangs vertically from the ceiling. A pulse is sent up the rope. Does the pulse travel faster, slower, or at a constant speed as it moves toward the ceiling? Explain your answer. Determine the Concept The speed of a transverse wave on a rope is given by μ T Fv = where F T is the tension in the rope and μ is its linear density. The waves on the rope move faster as they move toward the ceiling because the tension increases due to the weight of the rope below the pulse. 2 • A pulse on a horizontal taut string travels to the right. If the rope’s mass per unit length decreases to the right, what happens to the speed of the pulse as it travels to the right? (a) It slows down. (b) It speeds up. (c) Its speed is constant. (d) You cannot tell from the information given. Determine the Concept The speed v of a pulse on the string varies with the tension F T in the string and its mass per unit length μ according to μ T Fv = . Because the rope’s mass per unit length decreases to the right, the speed of the pulse increases. ()b is correct. 3 • As a sinusoidal wave travels past a point on a taut string, the arrival time between successive crests is measured to be 0.20 s. Which of the following is true? (a) The wavelength of the wave is 5.0 m. (b) The frequency of the wave is 5.0 Hz. (c) The velocity of propagation of the wave is 5.0 m/s. (d) The wavelength of the wave is 0.20 m. (e) There is not enough information to justify any of these statements. Determine the Concept The distance between successive crests is one wavelength and the time between successive crests is the period of the wave motion. Thus, T = 0.20 s and f = 1/T = 5.0 Hz. )(b is correct. 4 • Two harmonic waves on identical strings differ only in amplitude. Wave A has an amplitude that is twice that of wave B’s. How do the energies of these waves compare? (a) E A = E B . (b) E A = 2E B . (c) E A = 4E B . (d) There is not enough information to compare their energies. Picture the Problem The average energy transmitted by a wave on a string is proportional to the square of its amplitude and is given by ( ) xAE ΔΔ 22 2 1 av μω= where A is the amplitude of the wave, μ is the linear density (mass per unit length) Chapter 15 1540 of the string, ω is the angular frequency of the wave, and Δx is the length of the string. Because the waves on the strings differ only in amplitude, the energy of the wave on string A is given by: 2 AA cAE = Express the energy of the wave on string B: 2 BB cAE = Divide the first of these equations by the second and simplify to obtain: 2 B A 2 B 2 A B A ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ == A A cA cA E E Because A A = 2A B : 4 2 2 B B B A = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ = A A E E ⇒ BA 4EE = ( )c is correct. 5 • [SSM] To keep all of the lengths of the treble strings (unwrapped steel wires) in a piano all about the same order of magnitude, wires of different linear mass densities are employed. Explain how this allows a piano manufacturer to use wires with lengths that are the same order of magnitude. Determine the Concept The resonant (standing wave) frequencies on a string are inversely proportional to the square root of the linear density of the string ( )λμ T Tf = . Thus extremely high frequencies (which might otherwise require very long strings) can be accommodated on relatively short strings if the strings are linearly denser that the high frequency strings. High frequencies are not a problem as they utilize short strings anyway. 6 • Musical instruments produce sounds of widely varying frequencies. Which sounds waves have the longer wavelengths? (a) The lower frequencies. (b) The higher frequencies. (c) All frequencies have the same wavelength. (d) There is not enough information to compare the wavelengths of the different frequency sounds. Determine the Concept Once the sound has been produced by a vibrating string, membrane, or air column, the speed with which it propagates is the product of its frequency and wavelength ().λfv = For a given medium and temperature, v is constant. Hence the wavelength and frequency for a given sound are inversely proportional and ()a is correct. Traveling Waves 1541 7 • In Problem 7, which sound waves have the higher speeds? (a) The lower frequency sounds. (b) The higher frequency sounds. (c) All frequencies have the same wave speed. (d) There is not enough information to compare their speeds. Determine the Concept Once the sound has been produced by a vibrating string, membrane, or air column, the wave speed with which it propagates depends on the properties of the medium in which it is propagating and is independent of the frequency and wavelength of the sound. ( )c is correct. 8 • Sound travels at 343 m/s in air and 1500 m/s in water. A sound of 256 Hz is made under water, but you hear the sound while walking along the side of the pool. In the air, the frequency is (a) the same, but the wavelength of the sound is shorter, (b) higher, but the wavelength of the sound stays the same, (c) lower, but the wavelength of the sound is longer, (d) lower, and the wavelength of the sound is shorter, (e) the same, and the wavelength of the sound stays the same. Determine the Concept In any medium, the wavelength, frequency, and speed of a sound wave are related through λ = v/f. Because the frequency of a wave is determined by its source and is independent of the nature of the medium, if v is greater in water than in air, λ will be greater in water than in air. )(a is correct. 9 • While out on patrol, the battleship Rodger Young hits a mine, begins to burn, and ultimately explodes. Sailor Abel jumps into the water and begins swimming away from the doomed ship, while Sailor Baker gets into a life raft. Comparing their experiences later, Abel tells Baker, ″I was swimming underwater, and heard a big explosion from the ship. When I surfaced, I heard a second explosion. What do you think it could be?″ Baker says, ″I think it was your imagination—I only heard one explosion. ″ Explain why Baker only heard one explosion, while Abel heard two. Determine the Concept There was only one explosion. Sound travels faster in water than air. Abel heard the sound wave in the water first, then, surfacing, heard the sound wave traveling through the air, which took longer to reach him. 10 • True or false: A 60-dB sound has twice the intensity of a 30-dB sound. Picture the Problem The intensity level β, measured in decibels, is given by ()() 0 logdB10 II=β where I 0 = 10 −12 W/m 2 is defined to be the threshold of hearing. Chapter 15 1542 Express the intensity of the 60-dB sound: () 0 60 logdB10dB60 I I = ⇒ 0 6 60 10 II = Express the intensity of the 30-dB sound: () 0 30 logdB10dB30 I I = ⇒ 0 3 30 10 II = Because I 60 = 10 3 I 30 , the statement is false. 11 • [SSM] At a given location, two harmonic sound waves have the same amplitude, but the frequency of sound A is twice the frequency of sound B. How do their average energy densities compare? (a) The average energy density of A is twice the average energy density of B. (b) The average energy density of A is four times the average energy density of B. (c) The average energy density of A is 16 times the average energy density of B. (d) You cannot compare the average energy densities from the data given. Determine the Concept The average energy density of a sound wave is given by 2 0 2 2 1 av sρωη = where ρ is the average density of the medium, s 0 is the displacement amplitude of the molecules making up the medium, and ω is the angular frequency of the sound waves. Express the average energy density of sound A: 2 A,0 2 AA2 1 A av, sωρη = The average energy density of sound B is given by: 2 B,0 2 BB2 1 B av, sωρη = Dividing the first of these equation by the second yields: 2 B,0 2 BB2 1 2 A,0 2 AA2 1 B av, A av, s s ωρ ωρ η η = Because the sound waves are identical except for their frequencies: 2 B A 2 B A 2 B 2 A B av, A av, 2 2 ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ == f f f f π π ω ω η η Because f A = 2f B : 4 2 2 B B B av, A av, = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ = f f η η ⇒ B av,A av, 4ηη = ( )b is correct. 12 • At a given location, two harmonic sound waves have the same frequency but the amplitude of sound A is twice the amplitude of sound B. How do their average energy densities compare? (a) The average energy density of A is Traveling Waves 1543 twice the average energy density of B. (b) The average energy density of A is four times the average energy density of B (c) The average energy density of A is 16 times the average energy density of B (d) You cannot compare the average energy densities from the data given. Determine the Concept The average energy density of a sound wave is given by 2 0 2 2 1 av sρωη = where ρ is the average density of the medium, s 0 is the displacement amplitude of the molecules making up the medium, and ω is the angular frequency of the sound waves. Express the average energy density of sound A: 2 A,0 2 AA2 1 A av, sωρη = The average energy density of sound B is given by: 2 B,0 2 BB2 1 B av, sωρη = Dividing the first of these equation by the second yields: 2 B,0 2 BB2 1 2 A,0 2 AA2 1 B av, A av, s s ωρ ωρ η η = Because the sound waves are identical except for their displacement amplitudes: 2 B,0 A,0 2 B,0 2 A,0 B av, A av, ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ == s s s s η η Because B,0A,0 2ss = 4 2 2 B,0 B,0 B av, A av, = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ = s s η η ⇒ B av,A av, 4ηη = ( )b is correct. 13 • What is the ratio of the intensity of normal conversation to the sound intensity of a soft whisper (at a distance of 5.0 m)? (a) 10 3 , (b) 2, (c) 10 −3 , (d) 1/2. Hint: See Table 15-1. Determine the Concept The problem is asking us for the ratio of the intensities corresponding to normal conversation and a soft whisper. Table 15-1 includes the quantities we need in order to find this ratio. Use Table 15-1 to find the ratio of I/I 0 , where I 0 is the threshold of hearing, for normal conversation: 6 onconversati normal 0 10= ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ I I Use Table 15-1 to find the ratio of I/I 0 for a soft whisper: 3 whisper soft 0 10= ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ I I Chapter 15 1544 Dividing the first of these equation by the second yields: 3 3 6 whisper soft 0 onconversati normal 0 10 10 10 == ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ I I I I ( )a is correct. 14 • What is the ratio of the intensity level of normal conversation to the sound intensity level of a soft whisper (at a distance of 5.0 m)? (a) 10 3 , (b) 2, (c) 10 −3 , (d) 1/2. Hint: See Table 15-1. Determine the Concept The problem is asking us for the ratio of the sound intensity levels (decibel levels) corresponding to normal conversation and a soft whisper. Table 15-1 includes the quantities we need in order to find this ratio. From Table 15-1, the sound intensity level corresponding to normal conversation is: dB 60 onconversati normal =β From Table 15-1, the sound intensity level corresponding to a soft whisper is: dB 30 whisper soft =β Dividing the first of these equation by the second yields: 2 dB 30 dB 60 whisper soft onconversati normal == β β ( )b is correct. 15 • To increase the sound intensity level by 20 dB requires the sound intensity to increase by what factor? (a) 10, (b) 100, (c) 1000, (d) 2. Determine the Concept The sound intensity level is given by () ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ = 0 log dB 10 I I β where I is the intensity of a given sound and I 0 is the intensity of the threshold of hearing. The sound intensity level for a sound whose intensity is I 1 is given by: () ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ = 0 1 1 log dB 10 I I β Traveling Waves 1545 The sound intensity level for a sound whose intensity is I 2 is given by: () ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ = 0 2 2 log dB 10 I I β Subtract the second equation from the first and simplify to obtain: () ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ =− 2 1 21 log dB 10 I I ββ Solving for I 1 /I 2 yields: dB 10 2 1 21 10 ββ − = I I For a 20-dB increase in the sound intensity level: 2 dB 10 dB 20 2 1 1010 == I I ⇒ ( )b is correct. 16 • You are using a hand-held sound level meter to measure the intensity level of the roars produced by a lion prowling in the high grass. To decrease the measured sound intensity level by 20 dB requires the lion move away from you until its distance from you has increased by what factor? (a) 10, (b) 100, (c) 1000, (d) You cannot tell the required distance from the data given. Determine the Concept The sound intensity level is given by () ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ = 0 log dB 10 I I β where I is the intensity of a given sound and I 0 is the intensity of the threshold of hearing. The intensity of the a given sound varies with distance from its source according to 2 av 4 r P I π = . The sound intensity level for a sound whose intensity is I 1 is given by: () ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ = 0 1 1 log dB 10 I I β The sound intensity level for a sound whose intensity is I 2 is given by: () ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ = 0 2 2 log dB 10 I I β Subtract the second equation from the first and simplify to obtain: () ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ =− 2 1 21 log dB 10 I I ββ Solving for I 1 /I 2 yields: dB 10 2 1 21 10 ββ − = I I Chapter 15 1546 Letting r 1 be the lion’s initial distance from you and r 2 be the lion’s final distance from you, substitute for I 1 and I 2 and simplify to obtain: dB 10 2 1 2 2 2 2 av 2 1 av 21 10 4 4 ββ π π − == r r r P r P ⇒ dB 10 1 2 21 10 ββ − = r r For a 20-dB decrease in the sound intensity level: 1010 dB 10 dB 20 1 2 == r r ⇒ ( )b is correct. 17 • One end of a very light (but strong) thread is attached to an end of a thicker and denser cord. The other end of the thread is fastened to a sturdy post and you pull the other end of the cord so the thread and cord are taut. A pulse is sent down the thicker, denser cord. True or false: (a) The pulse that is reflected back from the thread-cord attachment point is inverted compared to the initial incoming pulse. (b) The pulse that continues past the thread-cord attachment point is not inverted compared to the initial incoming pulse. (c) The pulse that continues past the thread-cord attachment point has an amplitude that is smaller than the pulse that is reflected. (a) False. Because the reflection medium (cord) in which the pulse travels initially has a greater linear density than the transmission medium (thread), there is no phase shift in the reflected pulse. (b) True. Because the thread-cord attachment point of the media is more like a loose end than a fixed end, the pulse transmitted into the light thread is in phase with the incoming pulse in the thicker, denser cord. (c) True. Because the string is attached to a thicker, denser cord, the reflected pulse behaves almost as though it was reflected from a fixed end and is, therefore, inverted. The pulse in the light thread is in phase with the incoming pulse in the thicker, denser cord. Because energy is conserved and there is some energy transmitted and some reflected, the transmitted wave cannot have an amplitude as large as the amplitude of the pulse in the thicker, denser cord. 18 • Light traveling in air strikes a glass surface at an incident angle of 45°. True or false: (a) The angle between the reflected light ray and the incident ray is 90°. (b) The angle between the reflected light ray and the refracted light ray is less than 90°. Determine the Concept The ray diagram shows the relationship between the incident, reflected, and refracted rays. Traveling Waves 1547 air glass °45 θ 1 rayIncident ray Reflected ray Refracted glassair vv > (a) True. Because the angle of incidence is equal to the angle of reflection, the sum of these angles in this example is 90°. Note: this is a special case and is not generally true. (b) False. Because light travels slower in glass than in air, the refracted ray makes an angle of less than 45° with the normal. Hence the angle it makes with the reflected ray is greater than 90°. 19 • [SSM] Sound waves in air encounter a 1.0-m wide door into a classroom. Due to the effects of refraction, the sound of which frequency is least likely to be heard by all the students in the room, assuming the room is full? (a) 600 Hz, (b) 300 Hz, (c) 100 Hz, (d) All the sounds are equally likely to be heard in the room. (e) Diffraction depends on wavelength not frequency, so you cannot tell from the data given. Determine the Concept If the wavelength is large relative to the door, the diffraction effects are large and the waves spread out as they pass through the door. Because we’re interested in sounds that are least likely to be heard everywhere in the room, we want the wavelength to be short and the frequency to be high. Hence ()a is correct. 20 • Microwave radiation in modern microwave ovens has a wavelength on the order of centimeters. Would you expect significant diffraction if this radiation was aimed at a 1.0-m wide door? Explain. Determine the Concept No. Because the wavelength of the radiation is small relative to the door, the diffraction effects are small and the waves do not spread out significantly as they pass through the door. 21 •• [SSM] Stars often occur in pairs revolving around their common center of mass. If one of the stars is a black hole, it is invisible. Explain how the Chapter 15 1548 existence of such a black hole might be inferred by measuring the Doppler frequency shift of the light observed from the other, visible star. Determine the Concept The light from the visible star will be shifted about its mean frequency periodically due to the relative approach toward and recession away from Earth as the star revolves around the common center of mass. 22 •• Figure 15-30 shows a wave pulse at time t = 0 moving to the right. (a) At this particular time, which segments of the string are moving up? (b) Which segments are moving down? (c) Is there any segment of the string at the pulse that is instantaneously at rest? Answer these questions by sketching the pulse at a slightly later time and a slightly earlier time to see how the segments of the string are moving. Determine the Concept The graph shown below shows the pulse at an earlier time (t < 0) and later time (t > 0). (a) One can see that at t = 0, the portion of the string between 1 cm and 2 cm is moving down, (b) the portion between 2 cm and 3 cm is moving up, and (c) the string at x = 2 cm is instantaneously at rest. 012345678910 x , cm y t = 0 t > 0 t < 0 23 •• Make a sketch of the velocity of each string segment versus position for the pulse shown in Figure 15-30. Determine the Concept The velocity of the string at t = 0 is shown. Note that the velocity is negative for 1 cm < x < 2 cm and is positive for 2 cm < x < 3 cm. Traveling Waves 1549 012345678910 x , cm v y 24 •• An object of mass m hangs on a very light rope that is connected to the ceiling. You pluck the rope just above the object, and a wave pulse travels up to the ceiling and back. Compare the round-trip time for such a wave pulse to the round-trip time of a wave pulse on the same rope if an object of mass 9m is hung on the rope instead. (Assume the rope does not stretch, that is, the mass-to-ceiling distance is the same in each case.) Determine the Concept If the mass providing the tension in the rope is increased by a factor of n, then the tension in the rope increases by the same factor and the speed of the pulse on the rope increases by a factor of n and the round trip time for the pulse is decreased by a factor of n1. Let null represent the distance the pulse travels. Then the time for the round trip is given by: v t null = (1) Express the speed of the wave when the object of mass m provides the tension in the rope: μ m m F v T, = Express the speed of the wave when the object of mass 9m provides the tension in the rope: μ m m F v T,9 9 = Substituting for m v 9 in equation (1) yields: m m m F F t T,9 T,9 9 μ μ null null == (2) Similarly, the travel time when the object whose mass is m is providing the tension is given by: m m m F F t T, T, μ μ null null == (3) Chapter 15 1550 Divide equation (2) by equation (3) and simplify to obtain: m m m m F F t t T,9 T, 9 = Because mm FF T,T,9 9= : 3 1 T, T, 9 9 == m m m m F F t t ⇒ mm tt 3 1 9 = 25 •• The explosion of a depth charge beneath the surface of the water is recorded by a helicopter hovering above its surface, as shown in Figure 15-31. Along which path, A, B, or C, will the sound wave take the least time to reach the helicopter? Explain why you chose the path you did. Determine the Concept Path C. Because the wave speed is highest in the water, and more of path C is underwater than A or B, the sound wave will spend the least time on path C. 26 •• Does a speed of Mach 2 at an altitude of 60,000 feet mean the same as a speed of Mach 2 near ground level? Explain clearly. Determine the Concept No. The term Mach 2 means that the speed is twice the speed of sound at a given altitude. Because the speed of sound is lower at high altitudes than at ground level (due to lower density and colder temperature), Mach 2 means less than twice the speed of sound at sea level. Estimation and Approximation 27 •• Many years ago, Olympic 100 m dashes were started by the sound from a starter’s pistol, with the starter positioned several meters down the track, just on the inside of the track. (Today, the pistol that is used is often only a trigger, which is used to electronically activate speakers behind each sprinter’s starting blocks. This method avoids the problem of one runner hearing sound before the other runners.) Estimate the time advantage the inside lane (relative to the runner at the outside lane of 8 runners) would have if all runners started when they heard the sound from the starter’s pistol. Picture the Problem Due to the differences in the distances from the starter to the inside- and outside-lane sprinters, the sound of the starting pistol took more time to reach the outside-lane sprinter than it did to reach the inside-lane sprinter. Express the difference in time for the two sprinters in terms of the distances the sound must travel and the speed of sound: vv ttt lane insidelane outside lane insidelane ouside Δ nullnull −= −= Traveling Waves 1551 Assuming that the separation of the inside- and outside-lane sprinters is w: 22 lane insidelane outside w+= nullnull Substitute for lane outside null to obtain: v w t lane inside 22 lane inside Δ nullnull −+ = Assuming that the separation of the inside- and outside-lane sprinters is 7.0 m and that the starter is 5.0 m from the inside-lane sprinter yields: ()() ms 11 m/s 343 m 0.5m 0.7m 0.5 Δ 22 = −+ =t Remarks: Because a 100-m sprint is often decided by one or two hundredths of a second, this difference could be significant. This difference led to the invention of the electronic starter’s sound. 28 •• Estimate the speed of the bullet as it passes through the helium balloon in Figure 15-32. Hint: A protractor would be beneficial. Picture the Problem You can use a protractor to measure the angle of the shock cone and then estimate the speed of the bullet using .sin uv=θ The speed of sound in helium at room temperature (293 K) is 977 m/s. Relate the speed of the bullet u to the speed of sound v in helium and the angle of the shock cone θ : u v =θsin ⇒ θsin v u = Measuring θ yields: °≈ 70θ Substitute numerical values and evaluate u: km/s0.1 70sin m/s977 = ° =u 29 •• The new student townhouses at a local college are in the form of a semicircle half-enclosing the track field. To estimate the speed of sound in air, an ambitious physics student stood at the center of the semicircle and clapped his hands rhythmically at a frequency at which he couldn’t hear the echo of the clap, because it reached him at the same time as his next clap. This frequency was about 2.5 claps/s. Once he established this frequency, he paced off the distance to the townhouses, which was 30 double strides. Assuming that the length of each stride is equal to half his height (5 ft 11 in), estimate the speed of sound in air using this data. How far off is this from the commonly accepted value of 343 m/s? Chapter 15 1552 Picture the Problem Let d be the distance to the townhouses. We can relate the speed of sound to the distance to the townhouses to the frequency of the clapping for which no echo is heard. 5 ft 11 in is equal to 1.8 m. Relate the speed of sound to the distance it travels to the townhouses and back to the elapsed time: t d v Δ = 2 Express d in terms of the number of strides and distance covered per stride: ( )( ) m54m/stride8.1strides30 ==d Relate the elapsed time Δt to the frequency f of the clapping: s40.0 claps/s5.2 11 Δ === f t Substitute numerical values and evaluate v: ( ) km/s27.0m/s 270 s40.0 m542 ===v The percent difference between this result and 343 m/s is: %21 m/s343 m/s270m/s343 diff % = − = Speed of Waves 30 • (a) The bulk modulus of water is 2.00 × 10 9 N/m 2 . Use this value to find the speed of sound in water. (b) The speed of sound in mercury is 1410 m/s. What is the bulk modulus of mercury (ρ = 13.6 × 10 3 kg/m 3 )? Picture the Problem The speed of sound in a fluid is given by ρBv = where B is the bulk modulus of the fluid and ρ is its density. (a) Express the speed of sound in water in terms of its bulk modulus: ρ B v = Substitute numerical values and evaluate v: km/s1.41 km/s 414.1 kg/m101.00 N/m102.00 33 29 = = × × =v (b) Solving ρBv = for B yields: 2 vB ρ= Substitute numerical values and evaluate B: ( )( ) 210 233 N/m102.70 m/s1410kg/m1013.6 ×= ×=B Traveling Waves 1553 31 • Calculate the speed of sound waves in hydrogen gas (M = 2.00 g/mol and γ = 1.40) at T = 300 K. Picture the Problem The speed of sound in a gas is given by MRTv γ= where R is the gas constant, T is the absolute temperature, M is the molecular mass of the gas, and γ is a constant that is characteristic of the particular molecular structure of the gas. Because hydrogen gas is diatomic, γ = 1.4. Express the dependence of the speed of sound in hydrogen gas on the absolute temperature: M RT v γ = Substitute numerical values and evaluate v: ( )( ) km/s32.1 kg/mol1000.2 K300KJ/mol314.84.1 3 = × ⋅ = − v 32 • A 7.00-m-long guitar string has a mass of 100 g and is under a tension of 900 N. What is the speed of a transverse wave pulse on this string? Picture the Problem The speed of a transverse wave pulse on a string is given by μ T Fv = where F T is the tension in the string, m is its mass, L is its length, and μ is its mass per unit length. Express the dependence of the speed of the pulse on the tension in the string: μ T F v = where μ is the mass per unit length of the string. Substitute numerical values and evaluate v: m/s251 m7.00 kg0.100 N900 ==v 33 •• [SSM] (a) Compute the derivative of the speed of a wave on a string with respect to the tension dv/dF T , and show that the differentials dv and dF T obey TT2 1 FdFvdv = . (b) A wave moves with a speed of 300 m/s on a string that is under a tension of 500 N. Using to the differential approximation, estimate how much the tension must be changed to increase the speed to 312 m/s. (c) Calculate ΔF T exactly and compare it to the differential approximation result in Part (b). Assume that the string does not stretch with the increase in tension. Chapter 15 1554 Picture the Problem (a) The speed of a transverse wave on a string is given by μ T Fv = where T F is the tension in the wire and μ is its linear density. We can differentiate this expression with respect to F T and then separate the variables to show that the differentials satisfy . TT2 1 FdFvdv = (b) We’ll approximate the differential quantities to determine by how much the tension must be changed to increase the speed of the wave to 312 m/s. (c) We can use μ T Fv = to obtain an exact expression for ΔF T , (a) Evaluate dv/dF T : TT T T 2 11 2 1 F v F F dF d dF dv ⋅== ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ = μμ Separate the variables to obtain: T T 2 1 F dF v dv = (b) Solve the equation derived in Part (a) for dF T : v dv FdF TT 2= Approximate dF T with ΔF T and dv with Δv to obtain: v v FF Δ 2Δ TT = Substitute numerical values and evaluate ΔF T : () N40 m/s300 m/s 300m/s312 N5002Δ T = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ − =F (c) The exact value for (ΔF) exact is given by: ( ) T,1T,2exact Δ FFF −= (1) Express the wave speeds for the two tensions: μ T,1 1 F v = and μ T,2 2 F v = Dividing the second equation by the first and simplifying yields: T,1 T,2 T,1 T,2 1 2 v F F F F v == μ μ ⇒ 2 1 2 F,1T,2 ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ = v v FF Traveling Waves 1555 Substituting for F T,2 in equation (1) yields: () ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎣ ⎡ − ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ = − ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ = 1 Δ 2 1 2 F,1 T,1 2 1 2 F,1exactT v v F F v v FF Substitute numerical values and evaluate (ΔF T ) exact : () ( ) N 8.40 1 m/s 300 m/s 312 N 500Δ 2 exactT = ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎣ ⎡ − ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ =F The percent error between the exact and approximate values for ΔF T is: ( ) () %2 N 8.40 N 40.0N 8.40 Δ ΔΔ exactT TexactT ≈ − = − F FF 34 •• (a) Compute the derivative of the speed of sound in air with respect to the absolute temperature, and show that the differentials dv and dT obey TdTvdv 2 1 = . (b) Use this result to estimate the percentage change in the speed of sound when the temperature changes from 0 to 27°C. (c) If the speed of sound is 331 m/s at 0°C, estimate its value at 27°C using the differential approximation. (d) How does this approximation compare with the result of an exact calculation? Picture the Problem The speed of sound in a gas is given by MRTv γ= where R is the gas constant, T is the absolute temperature, M is the molecular mass of the gas, and γ is a constant that is characteristic of the particular molecular structure of the gas. We can differentiate this expression with respect to T and then separate the variables to show that the differentials satisfy . 2 1 TdTvdv = We’ll approximate the differential quantities to estimate the percentage change in the speed of sound when the temperature increases from 0 to 27°C. Lacking information regarding the nature of the gas, we can express the ratio of the speeds of sound at 300 K and 273 K to obtain an expression that involves just the temperatures. (a) Evaluate dv/dT: T v M R RT M M RT dT d dT dv 2 1 2 1 = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ = ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ = γ γ γ Chapter 15 1556 Separate the variables to obtain: T dT v dv 2 1 = (b) Approximate dT with ΔT and dv with Δv and substitute numerical values to obtain: %0.5 K 273 K 273K 300 2 1Δ = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ − = v v (c) Using a differential approximation, approximate the speed of sound at 300 K: ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ Δ += Δ +≈ v v v v v vvv 1 K273 K273K273K300 Substitute numerical values and evaluate v 300 K : ( )( ) m/s347 0495.01m/s331 K300 = +=v (d) Use MRTv γ= to express the speed of sound at 300 K: ( ) M R v K300 K300 γ = Use MRTv γ= to express the speed of sound at 273 K: ( ) M R v K273 K732 γ = Divide the first of these equations by the second and solve for and evaluate v 300 K : ( ) () 273 300 K273 K300 K273 K300 == M R M R v v γ γ and () m/s347 273 300 m/s331 K300 ==v Note that these two results agree to three significant figures. 35 ••• Derive a convenient formula for the speed of sound in air at temperature t in Celsius degrees. Begin by writing the temperature as T = T 0 + ΔT, where T 0 = 273 K corresponds to 0°C and ΔT = t, the Celsius temperature. The speed of sound is a function of T, v(T). To a first-order approximation, you can write () ()()TdTdvTvTv T Δ 0 0 +≈ , where dv / dT( ) T 0 is the derivative evaluated at T = T 0 . Compute this derivative, and show that the result leads to v = 331 m / s()1+ t 2T 0 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = 331+ 0.606t() m/s. Traveling Waves 1557 Picture the Problem We can use the approximate expression for v(T) given in the problem statement and the result of Problem 34 to show that ()m/s 606.0331 tv += . The speed of sound as a function of T is given by: () () T dT dv TvTv T Δ 0 0 ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ +≈ (1) From Problem 44, dv/dT is given by: ( ) T Tv dt dv 2 1 = Evaluating 0 TT dT dv = ⎥ ⎦ ⎤ yields: ( ) 0 0 2 1 0 T Tv dT dv TT = ⎥ ⎦ ⎤ = (2) Substitute equation (2) in equation (1) and simplify to obtain: () () ( ) () ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ += ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ +≈ 0 0 0 0 0 2 Δ 1 Δ 2 1 T T Tv T T Tv TvTv For T 0 = 273° K, () m/s 331 0 =Tv , and tT =Δ : () () ()m/s 606.0331 K 2732 1m/s 331 t t v += ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ += The Wave Equation 36 • Show explicitly that the following functions satisfy the wave equation 2 2 22 2 1 t y vx y ∂ ∂ = ∂ ∂ : (a) y(x,t) = k(x + vt) 3 , (b) y(x, t) = Ae ik(x – vt) , where A and k are constants and 1−=i , and (c) y(x,t) = ln k(x – vt). Picture the Problem To show that each of the functions satisfies the wave equation, we’ll need to find their first and second derivatives with respect to x and t and then substitute these derivatives in the wave equation. (a) Find the first two spatial derivatives of () ( ): 3 vtxktx,y += () 2 3 vtxk x y += ∂ ∂ and ()vtxk x y += ∂ ∂ 6 2 2 (1) Chapter 15 1558 Find the first two temporal derivatives of () ( ): 3 vtxktx,y += () 2 3 vtxkv t y += ∂ ∂ and ()vtxkv t y += ∂ ∂ 2 2 2 6 (2) Express the ratio of equation (1) to equation (2): () () 22 2 2 2 2 1 6 6 vvtxkv vtxk t y x y = + + = ∂ ∂ ∂ ∂ confirming that ( )( ) 3 , vtxktxy += satisfies the general wave equation. (b) Find the first two spatial derivatives of () ()vtxik Aetx,y − = : ()vtxik ikAe x y − = ∂ ∂ , ()vtxik Aeki x y − = ∂ ∂ 22 2 2 or ()vtxik Aek x y − −= ∂ ∂ 2 2 2 (3) Find the first two temporal derivatives of () ()vtxik Aetx,y − = : ()vtxik ikvAe t y − −= ∂ ∂ , ()vtxik Aevki t y − = ∂ ∂ 222 2 2 or ()vtxik Aevk t y − −= ∂ ∂ 22 2 2 (4) Express the ratio of equation (3) to equation (4): () () 222 2 2 2 2 2 1 vAevk Aek t y x y vtxik vtxik = − − = ∂ ∂ ∂ ∂ − − confirming that ( ) ()vtxik Aetxy − =, satisfies the general wave equation. (c) Find the first two spatial derivatives of () ( )vtxktx,y −= ln : vtx k x y − = ∂ ∂ and () 2 2 2 2 vtx k x y − −= ∂ ∂ (5) Traveling Waves 1559 Find the first two temporal derivatives of () ( )vtxktx,y −= ln : vtx vk t y − −= ∂ ∂ and () 2 22 2 2 vtx kv t y − −= ∂ ∂ (6) Express the ratio of equation (5) to equation (6): () () 2 2 22 2 2 2 2 2 2 1 v vtx kv vtx k t y x y = − − − − = ∂ ∂ ∂ ∂ confirming that ( )()vtxktx,y −= ln satisfies the general wave equation. 37 • Show that the function y = A sin kx cos ωt satisfies the wave equation. Picture the Problem The wave equation is 2 2 22 2 1 t y vx y ∂ ∂ = ∂ ∂ . To show that tkxAy ωcossin= satisfies this equation, we’ll need to find the first and second derivatives of y with respect to x and t and then substitute these derivatives in the wave equation. Find the first two spatial derivatives of tkxAy ωcossin= : tkxAk x y ωcoscos= ∂ ∂ and tkxAk x y ωcossin 2 2 2 −= ∂ ∂ (1) Find the first two temporal derivatives of tkxAy ωcossin= : tkxA t y ωω sinsin−= ∂ ∂ and tkxA t y ωω cossin 2 2 2 −= ∂ ∂ (2) Chapter 15 1560 Express the ratio of equation (1) to equation (2): 2 2 2 2 2 2 2 2 2 1 cossin cossin v k tkxA tkxAk t y x y = = − − = ∂ ∂ ∂ ∂ ωωω ω confirming that tkxAy ωcossin= satisfies the general wave equation. Harmonic Waves on a String 38 • One end of a string 6.0 m long is moved up and down with simple harmonic motion at a frequency of 60 Hz. If the wave crests travel the length of the string in 0.50 s, find the wavelength of the waves on the string. Picture the Problem We can find the wavelength of the waves on the string using the relationship λfv = . Express the wavelength of the waves: vf =λ ⇒ f v =λ The speed of the wave is equal to the distance divided by the time: m/s12 s0.50 m6.0 Δ Δ === t x v Substitute numerical values to obtain: cm20 s60 m/s12 1 == − λ 39 • [SSM] A harmonic wave on a string with a mass per unit length of 0.050 kg/m and a tension of 80 N, has an amplitude of 5.0 cm. Each point on the string moves with simple harmonic motion at a frequency of 10 Hz. What is the power carried by the wave propagating along the string? Picture the Problem The average power propagated along the string by a harmonic wave is v,AP 22 2 1 av μω= where v is the speed of the wave, and μ, ω, and A are the linear density of the string, the angular frequency of the wave, and the amplitude of the wave, respectively. Express and evaluate the power propagated along the string: vAP 22 2 1 av μω= Traveling Waves 1561 The speed of the wave on the string is given by: μ T F v = Substitute for v to obtain: μ μω T22 2 1 av F AP = Substitute numerical values and evaluate P av : ()()()() W9.9 kg/m0.05 N80 m0.050s10kg/m0.0504 2 2 12 2 1 av == − πP 40 • A 2.00-m-long rope has a mass of 0.100 kg. The tension is 60.0 N. An oscillator at one end sends a harmonic wave with an amplitude of 1.00 cm down the rope. The other end of the rope is terminated so that all of the energy of the wave is absorbed and none is reflected. What is the frequency of the oscillator if the power transmitted is 100 W? Picture the Problem The average power propagated along the rope by a harmonic wave is v,AP 22 2 1 av μω= where v is the velocity of the wave, and μ, ω, and A are the linear density of the string, the angular frequency of the wave, and the amplitude of the wave, respectively. Rewrite the power equation in terms of the frequency of the wave: vAfvAP 22222 2 1 av 2 μπμω == Solve for the frequency and simplify to obtain: v P AvA P f μπμπ av 22 av 2 2 1 2 == The wave speed is given by: μ T F v = Substitute for v and simplify to obtain: T av T av 2 2 12 2 1 F P A F P A f μ π μ μ π == Chapter 15 1562 Substitute numerical values and evaluate f: () ( ) () Hz 171 N 0.60 m 2.00 kg 0.100 W1002 m 0100.02 1 = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ = π f 41 •• The wave function for a harmonic wave on a string is y(x, t) = (1.00 mm) sin(62.8 m –1 x + 314 s –1 t). (a) In what direction does this wave travel, and what is the wave’s speed? (b) Find the wavelength, frequency, and period of this wave. (c) What is the maximum speed of any point on the string? Picture the Problem () ( )tkxAtxy ω−= sin, describes a wave traveling in the +x direction. For a wave traveling in the −x direction, we have ( )()tkxAtxy ω+= sin, . We can determine A, k, and ω by examination of the wave function. The wavelength, frequency, and period of the wave can, in turn, be determined from k and ω. (a) Because the sign between the kx and ωt terms is positive, the wave is traveling in the −x direction. The speed of the wave is: m/s00.5 m8.62 s314 1 1 === − − k v ω (b) The coefficient of x is k and: λ π2 =k ⇒ k π λ 2 = Substitute numerical values and evaluate λ: cm0.10 m8.62 2 1 == − π λ The coefficient of t is ω and: Hz0.50 2 s314 2 1 === − ππ ω f The period of the wave motion is the reciprocal of its frequency: s0200.0 s0.50 11 1 === − f T (c) Express and evaluate the maximum speed of any string segment: ( )( ) m/s0.314 rad/s314mm00.1 max = == ωAv Traveling Waves 1563 42 •• A harmonic wave on a string with a frequency of 80 Hz and an amplitude of 0.025 m travels in the +x direction with a speed of 12 m/s. (a) Write a suitable wave function for this wave. (b) Find the maximum speed of a point on the string. (c) Find the maximum acceleration of a point on the string. Picture the Problem () ( )tkxAx,ty ω−= sin describes a wave traveling in the +x direction. We can find ω and k from the data included in the problem statement and substitute in the general equation. The maximum speed of a point on the string can be found from ωAv = max and the maximum acceleration from . 2 max ωAa = (a) Express the general form of the equation of a harmonic wave traveling to the right: ( ) ( )tkxAx,ty ω−= sin Evaluate ω: ( ) 12 11 s100.5 s503s8022 − −− ×= === ππω f Determine k: 1 1 m42 m/s12 s503 − − === v k ω Substitute to obtain: ()( ) ( ) ( )[ ]txx,ty 121 s100.5m42sinm025.0 −− ×−= (b) The maximum speed of a point on the string is: ( )( ) m/s13 s503m0.025 1 max = == − ωAv (c) Express the maximum acceleration of a point on the string: 2 max ωAa = Substitute numerical values and evaluate a max : ( )( ) 2 2 1 max km/s3.6s503m0.025 == − a 43 •• A 200-Hz harmonic wave with an amplitude 1.2 cm moves along a 40-m-long string that has a mass of 0.120 kg and a tension of 50 N. (a) What is the average total energy of the waves on a 20-m-long segment of the string? (b) What is the power transmitted past a given point on the string? Picture the Problem The average power propagated along the string is . 22 2 1 av vAP μω= The average energy on a segment of length L is the average power multiplied by the time for the wave to travel the distance L Chapter 15 1564 (a) Express the power transmitted past a given point on the string. The average energy passing a point in time t is the average power multiplied by the time: vAP 22 2 1 av μω= and tPE avav = The time for the wave to travel the length L is L/v: v L t = Combine the previous steps and solve for the average energy on the string: () 22222 2 1 22 2 1 22 2 1 av 22 ALfAfL AL v L AvE μππμ ωμωμ == == Substitute numerical values and evaluate E av : ()()( ) J 8.6m0.012m 20s200 m40 kg0.120 2 2 2 12 av = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ = − πE (b) Express P av : 22 2 1 av AvP ωμ= To calculate P av we need an expression for the wave speed v: μ T F v = Substitute for v to obtain: () μ μπ μ πμ T222 T2 2 2 1 av 2 2 F Af F AfP = = Substitute numerical values and evaluate P av : ()() W44 m 20 kg 060.0 N 50 m 012.0s 200 m 20 kg 060.0 2 2 2 12 av = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ = − πP 44 •• On a real string, some of the energy of the wave dissipates as the wave travels down the string. Such a situation can be described by a wave function whose amplitude A(x) depends on x: y = A(x) sin (kx – ωt) where A(x) = (A 0 e –bx ). What is the power transported by the wave as a function of x, where x > 0? Traveling Waves 1565 Picture the Problem The power propagated along the string by a harmonic wave is vAP 22 2 1 μω= where v is the velocity of the wave, and μ, ω, and A are the linear density of the string, the angular frequency of the wave, and the amplitude of the wave, respectively. Express the power associated with the wave at the origin: vAP 22 2 1 μω= (1) Express the amplitude of the wave at x: ( ) bx eAxA − = 0 Substitute in equation (1) to obtain: ( ) ( ) bx bx veA veAxP 22 0 2 2 1 2 0 2 2 1 − − = = μω μω 45 •• [SSM] Power is to be transmitted along a taut string by means of transverse harmonic waves. The wave speed is 10 m/s and the linear mass density of the string is 0.010 kg/m. The power source oscillates with an amplitude of 0.50 mm. (a) What average power is transmitted along the string if the frequency is 400 Hz? (b) The power transmitted can be increased by increasing the tension in the string, the frequency of the source, or the amplitude of the waves. By how much would each of these quantities have to increase to cause an increase in power by a factor of 100 if it is the only quantity changed? Picture the Problem The average power propagated along a string by a harmonic wave is vAP 22 2 1 av μω= where v is the speed of the wave, and μ, ω, and A are the linear density of the string, the angular frequency of the wave, and the amplitude of the wave, respectively. (a) Express the average power transmitted along the string: vAfvAP 22222 2 1 av 2 μπμω == Substitute numerical values and evaluate P av : ( )( ) ()() mW79 m/s10m100.50 s400kg/m0.0102 2 3 2 12 av = ×× = − − πP (b) Because 2 av fP ∝ , increasing f by a factor of 10 would increase P av by a factor of 100. Chapter 15 1566 Because 2 av AP ∝ , increasing A by a factor of 10 would increase P av by a factor of 100. Because vP ∝ av and Fv ∝ , increasing F by a factor of 100 would increase v by a factor of 100 and P av by a factor of 100. 46 ••• Two very long strings are tied together at the point x = 0. In the region x < 0, the wave speed is v 1 , while in the region x > 0, the speed is v 2 . A sinusoidal wave is incident on the knot from the left (x < 0); part of the wave is reflected and part is transmitted. For x < 0, the displacement of the wave is describable by y(x,t) = A sin (k 1 x – ωt) + B sin(k 1 x + ωt), while for x > 0, y(x,t) = C sin (k 2 x – ωt), where ω/k 1 = v 1 and ω/k 2 = v 2 . (a) If we assume that both the wave function y and its first spatial derivative ∂y/∂x must be continuous at x = 0, show that C/A = 2v 2 /( v 1 + v 2 ), and that B/A = (v 1 – v 2 )/( v 1 + v 2 ). (b) Show that B 2 + (v 1 /v 2 )C 2 = A 2 . Picture the Problem We can use the assumption that both the wave function and its first spatial derivative are continuous at x = 0 to establish equations relating A, B, C, k 1 , and k 2 . Then, we can solve these simultaneous equations to obtain expressions for B and C in terms of A, v 1 , and v 2 . (a) Let y 1 (x, t) represent the wave function in the region x < 0, and y 2 (x, t) represent the wave function in the region x > 0. Express the continuity of the two wave functions at x = 0: ( ) ( )tyty ,0,0 21 = and ( )[ ] ()[ ] ()[]tkC tkBtkA ω ωω −= ++− 0sin 0sin0sin 2 11 or ( )()tCtBtA ωωω −=+− sinsinsin Because the sine function is odd; that is, () θθ sinsin −=− : tCtBtA ωωω sinsinsin −=+− and CBA =− (1) Differentiate the wave functions with respect to x to obtain: () ()txkBk txkAk x y ω ω ++ −= ∂ ∂ 11 11 1 cos cos and ()txkCk x y ω−= ∂ ∂ 22 2 cos Traveling Waves 1567 Express the continuity of the slopes of the two wave functions at x = 0: ∂y 1 ∂x x =0 = ∂y 2 ∂x x =0 and ( )[ ] ()[] ()[]tkCk tkBktkAk ω ωω −= ++− 0cos 0cos0cos 22 1111 or ( ) ()tCk tBktAk ω ωω −= +− cos coscos 2 11 Because the cosine function is even; that is, () θθ coscos =− : tCktBktAk ωωω coscoscos 211 =+ and CkBkAk 211 =+ (2) Multiply equation (1) by k 1 and add it to equation (2) to obtain: ( )CkkAk 211 2 += Solving for C yields: A kk A kk k C 1221 1 1 22 + = + = Solve for C/A and substitute ω/v 1 for k 1 and ω/v 2 for k 2 to obtain: 21 2 2112 2 1 2 1 2 vv v vvkkA C + = + = + = Substitute in equation (1) to obtain: A vv v BA ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ + =− 12 2 2 Solving for B/A yields: 21 21 vv vv A B + − = Chapter 15 1568 (b) We wish to show that B 2 + (v 1 /v 2 )C 2 = A 2 Use the results of (a) to obtain the expressions B = −[(1 − α)/(1 + α)] A and C = 2A/(1 + α), where α = v 1 /v 2 . Substitute these expressions into B 2 + (v 1 /v 2 )C 2 = A 2 and check to see if the resulting equation is an identity: () () () () () () 11 1 1 1 1 1 21 1 1 421 1 1 41 1 1 2 1 1 1 2 1 1 2 2 2 2 2 2 2 2 22 22 2 2 2 22 2 1 2 = = + + = + ++ = + ++− = + +− = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ + + ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ + − = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ + + ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ + − =+ α α α αα α ααα α αα α α α α α α α α AAA AC v v B The equation is an identity: Therefore, 22 2 1 2 AC v v B =+ Remarks: Our result in (a) can be checked by considering the limit of B/A as v 2 /v 1 → 0. This limit gives B/A = +1, telling us that the transmitted wave has zero amplitude and the incident and reflected waves superpose to give a standing wave with a node at x = 0. Harmonic Sound Waves 47 • A sound wave in air produces a pressure variation given by () ()txtxp 343 2 cos75.0, −= π , where p is in pascals, x is in meters, and t is in seconds. Find (a) the pressure amplitude, (b) the wavelength, (c) the frequency, and (d) the wave speed. Picture the Problem The pressure variation is of the form () ( )vtxkpx,tp −= cos 0 where k is the wave number, p 0 is the pressure amplitude, and v is the wave speed. We can find λ from k and f from ω and k. (a) By inspection of the equation: Pa75.0 0 =p Traveling Waves 1569 (b) Because : 2 2 π λ π ==k m00.4=λ (c) Solve k f k v πω 2 == for f to obtain: π2 kv f = Substitute numerical values and evaluate f: () Hz8.85 2 m/s343 2 == π π f (d) By inspection of the equation: m/s343=v 48 • (a) Middle C on the musical scale has a frequency of 262 Hz. What is the wavelength of this note in air? (b) The frequency of the C an octave above middle C is twice that of middle C. What is the wavelength of this note in air? Picture the Problem The frequency, wavelength, and speed of the sound waves are related by v = fλ. (a) The wavelength of middle C is given by: m30.1 s262 m/s340 1 === − f v λ (b) Evaluate λ for a frequency twice that of middle C: () m649.0 s2622 m/s340 1 === − f v λ 49 • [SSM] (a) What is the displacement amplitude for a sound wave with a frequency of 100 Hz and a pressure amplitude of 1.00 × 10 –4 atm? (b) The displacement amplitude of a sound wave of frequency 300 Hz is 1.00 ×10 –7 m. Assuming the density of air is 1.29 kg/m 3 , what is the pressure amplitude of this wave? Picture the Problem The pressure amplitude depends on the density of the medium ρ, the angular frequency of the sound wave μ, the speed of the wave v, and the displacement amplitude s 0 according to . 00 vsp ρω= (a) Solve 00 vsp ρω= for s 0 : v p s ρω 0 0 = Chapter 15 1570 Substitute numerical values and evaluate s 0 : ( )( ) ())() m 4.36m 1064.3 m/s 343s 100kg/m 1.292 Pa/atm 1001325.1atm 1000.1 5 13 54 0 μ π =×= ×× = − − − s (b) Use 00 vsp ρω= to find p 0 : ()( )( )( ) mPa4.83m101.00m/s343s300kg/m29.12 713 0 =×= −− πp 50 • (a) What is the displacement amplitude of a sound wave that has a frequency of 500 Hz at the pain-threshold pressure amplitude of 29.0 Pa? (b) Assuming the density of air is 1.29 kg/m 3 , what is the displacement amplitude of a sound wave that has the same pressure amplitude as the wave in Part (a), but has a frequency of 1.00 kHz? Picture the Problem The pressure amplitude depends on the density of the medium ρ, the angular frequency of the sound wave μ, the wave speed v, and the displacement amplitude s 0 according to 00 vsp ρω= . (a) Solve 00 vsp ρω= for s 0 : v p s ρω 0 0 = Substitute numerical values and evaluate s 0 : ()()() m9.20 m/s343Hz500kg/m1.292 Pa0.29 3 0 μ= = π s (b) Proceed as in (a) with f = 1.00 kHz: ()()() m4.10 m/s343kHz 00.1kg/m1.292 Pa0.29 3 0 μ= = π s 51 • A typical loud sound wave that has a frequency of 1.00 kHz has a pressure amplitude of about 1.00 × 10 –4 atm. (a) At t = 0, the pressure is a maximum at some point x 1 . What is the displacement at that point at t = 0? (b) Assuming the density of air is 1.29 kg/m 3 , what is the maximum value of the displacement at any time and place? Picture the Problem The pressure or density wave is 90° out of phase with the displacement wave. When the displacement is zero, the pressure and density changes are either a maximum or a minimum. When the displacement is a Traveling Waves 1571 maximum or minimum, the pressure and density changes are zero. We can use 00 vsp ρω= to find the maximum value of the displacement at any time and place. (a) If the pressure is a maximum at x 1 when t = 0, the displacement s is zero. (b) Solve 00 vsp ρω= for s 0 : v p s ρω 0 0 = Substitute numerical values and evaluate s 0 : ()( ) ()()() m64.3 m/s343kHz 1.00kg/m1.292 Pa/atm101.01325atm101.00 3 54 0 μ= ×× = − π s 52 • An octave represents a change in frequency by a factor of two. Over how many octaves can a typical person hear? Picture the Problem A human can hear sounds between roughly 20 Hz and 20 kHz; a factor of 1000. An octave represents a change in frequency by a factor of 2. We can evaluate 2 N = 1000 to find the number of octaves heard by a person who can hear this range of frequencies. Relate the number of octaves to the difference between 20 kHz and 20 Hz: 10002 = N Take the logarithm of both sides of the equation to obtain: 3 10log2log = N ⇒ 32log =N Solving for N yields: 1097.9 2log 3 ≈==N 53 •• In the oceans, whales communicate by sound transmission through the water. A whale emits a sound of 50.0 Hz to tell a wayward calf to catch up to the pod. The speed of sound in water is about 1500 m/s. (a) How long does it take the sound to reach the calf if he is 1.20 km away? (b) What is the wavelength of this sound in the water? (c) If the whales are close to the surface, some of the sound energy might refract out into the air. What would be the frequency and wavelength of the sound in the air? Chapter 15 1572 Picture the Problem (a) We can use the definition of average speed to find the time required for the sound to travel to the calf. (b) We can use the relationship between wavelength, frequency, and speed to find the wavelength of the sound in water. (c) The frequency of the sound does not change as it travels from water to air, but its wavelength changes because of the difference in the speed of sound in water and in air. (a) Relate the time it takes the sound to reach the calf to the distance from the whale to the calf and the speed of sound in water: s 80.0 m/s1500 km 20.1 Δ === v d t (b) The wavelength of this sound in water is the ratio of its speed in water to its frequency: m 30 Hz 50.0 m/s 1500 water water === f v λ (c) Because the frequency does not change as it travels from water to air: Hz 0.50=f The wavelength of this sound in air is the ratio of its speed in air to its frequency: m .866 Hz 50.0 m/s 433 air air === f v λ Waves in Three Dimensions: Intensity 54 • A spherical sinusoidal source radiates sound uniformly in all directions. At a distance of 10.0 m, the sound intensity level is 1.00 × 10 –4 W/m 2 . (a) At what distance from the source is the intensity 1.00 × 10 –6 W/m 2 ? (b) What power is radiated by this source? Picture the Problem The intensity of the sound from the spherical sinusoidal source varies inversely with the square of the distance from the source. The power radiated by the source is the product of the intensity of the radiation and the surface area over which it is distributed. (a) Relate the intensity at 10.0 m to the distance from the source: 2 av 4 r P I π = or () 2 av 24 m 0.104 W/m1000.1 π P =× − Traveling Waves 1573 Letting r′ represent the distance at which the intensity is 10 −6 W/m 2 , express the intensity as above: 2 av 26 4 W/m1000.1 r' P π =× − Divide the first of these equations by the second to obtain: () 2 av 2 av 26 24 4 m0.104 W/m1000.1 W/m1000.1 r' P P π π = × × − − Solving for r′ yields: ( )() m100m0.101000.1 2 2 =×=r' (b) Solve 2 av 4 r P I π = for P av : IrP 2 av 4π= Substitute numerical values and evaluate P av : ( ) ( ) mW126 W/m1000.1m0.104 242 av = ×= − πP 55 • [SSM] A loudspeaker at a rock concert generates a sound that has an intensity level equal to 1.00 × 10 –2 W/m 2 at 20.0 m and has a frequency of 1.00 kHz. Assume that the speaker spreads its energy uniformly in three dimensions. (a) What is the total acoustic power output of the speaker? (b) At what distance will the sound intensity be at the pain threshold of 1.00 W/m 2 ? (c) What is the sound intensity at 30.0 m? Picture the Problem Because the power radiated by the loudspeaker is the product of the intensity of the sound and the area over which it is distributed, we can use this relationship to find the average power, the intensity of the radiation, or the distance to the speaker for a given intensity or average power. (a) Use IrP 2 av 4π= to find the total acoustic power output of the speaker: ( ) ( ) W3.50 W27.50 W/m1000.1m0.204 222 av == ×= − πP (b) Relate the intensity of the sound at 20 m to the distance from the speaker: () 2 av22 m 0.204 W/m1000.1 π P =× − Relate the threshold-of-pain intensity to the distance from the speaker: 2 av2 4 W/m00.1 r P π = Chapter 15 1574 Divide the first of these equations by the second and solve for r: ( )() m00.2m0.201000.1 22 =×= − r (c) Use 2 av 4 r P I π = to find the intensity at 30.0 m: () () 23 2 W/m1045.4 m0.304 W3.50 m0.30 − ×= = π I 56 •• When a pin of mass 0.100 g is dropped from a height of 1.00 m, 0.050 percent of its energy is converted into a sound pulse that has a duration of 0.100 s. (a) Estimate how far away the dropped pin can be heard if the minimum audible intensity is 1.00 × 10 –11 W/m 2 . (b) Your result in Part (a) is much too large in practice because of background noise. If you assume that the intensity must be at least 1.00 × 10 –8 W/m 2 for the sound to be heard, estimate how far away the dropped pin can be heard. (In both parts, assume that the intensity is P/4πr 2 .) Picture the Problem We can use conservation of energy to find the acoustical energy resulting from the dropping of the pin. The power developed can then be found from the given time during which the energy was transformed from mechanical to acoustical form. We can find the range at which the dropped pin can be heard from I = P/4π r 2 . (a) Assuming that I = P/4π r 2 , express the distance at which one can hear the dropped pin: I P r π4 = Use conservation of energy to determine the sound energy generated when the pin falls: ()() ()() J10905.4 m00.1m/s81.9 kg10100.000050.0 7 2 3 − − ×= × ×= = mghE ε The power of the sound pulse is given by: W104.905 s0.100 J104.905 Δ 6 7 − − ×= × == t E P Substitute numerical values and evaluate r: () km20.0 W/m1000.14 W10905.4 211 6 = × × = − − π r (b) Repeat the last step in (a) with I = 1.00 × 10 –8 W/m 2 : () m2.6 W/m1000.14 W10905.4 28 6 = × × = − − π r Traveling Waves 1575 *Intensity Level 57 • [SSM] What is the intensity level in decibels of a sound wave of intensity (a) 1.00 × 10 –10 W/m 2 and (b) 1.00 × 10 –2 W/m 2 ? Picture the Problem The intensity level β of a sound wave, measured in decibels, is given by ()( ) 0 logdB10 II=β where I 0 = 10 −12 W/m 2 is defined to be the threshold of hearing. (a) Using its definition, calculate the intensity level of a sound wave whose intensity is 1.00 × 10 –10 W/m 2 : () dB0.2010log10 W/m10 W/m101.00 logdB10 2 212 201 == ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ × = − − β (b) Proceed as in (a) with I = 1.00 × 10 –2 W/m 2 : () dB10010log10 W/m10 W/m101.00 logdB10 10 212 22 == ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ × = − − β 58 • What is the intensity of a sound wave if, at a particular location, the intensity level is (a) β = 10 dB and (b) β = 3 dB? Picture the Problem (a) and (b) The intensity level β of a sound wave, measured in decibels, is given by ( ) ( ) 0 logdB10 II=β where I 0 = 10 −12 W/m 2 is defined to be the threshold of hearing. (a) Solve ()() 0 logdB10 II=β for I to obtain: ( ) 0 dB10 10 II β = Evaluate I for β = 10 dB: ( ) ( ) () 211212 00 dB10dB10 W/m10W/m1010 1010 −− == == III (b) Proceed as in (a) with β = 3 dB: ( ) ( ) () 212212 00 dB10dB3 W/m102W/m102 210 −− ×== == III 59 • At a certain distance, the sound intensity level of a dog’s bark is 50 dB. At that same distance, the sound intensity level of a rock concert is 10,000 times that of the dog’s bark. What is the sound intensity level of the rock concert? Chapter 15 1576 Picture the Problem The intensity level β of a sound wave, measured in decibels, is given by ()( ) 0 logdB10 II=β where I 0 = 10 −12 W/m 2 is defined to be the threshold of hearing. Express the sound level of the rock concert: () ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ = 0 concert concert logdB10 I I β (1) Express the sound level of the dog’s bark: () ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ = 0 dog logdB10dB50 I I Solve for the intensity of the dog’s bark: ( ) 27 2125 0 5 dog W/m10 W/m101010 − − = == II Express the intensity of the rock concert in terms of the intensity of the dog’s bark: ( ) 23 274 dog 4 concert W/m10 W/m101010 − − = == II Substitute in equation (1) and evaluate β concert : () () dB9010logdB10 W/m10 W/m10 logdB10 9 212 23 concert == ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ = − − β 60 • What fraction of the acoustic power of a noise would have to be eliminated to lower its sound intensity level from 90 to 70 dB? Picture the Problem We can express the intensity levels at both 90 dB and 70 dB in terms of the intensities of the sound at those levels. By subtracting the two expressions, we can solve for the ratio of the intensities at the two levels and then find the fractional change in the intensity that corresponds to a decrease in intensity level from 90 dB to 70 dB. Express the intensity level at 90 dB: () ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ = 0 90 logdB10dB90 I I Express the intensity level at 70 dB: () ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ = 0 70 logdB10dB70 I I Traveling Waves 1577 Express Δβ = β 90 − β 70 : () () () ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ − ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ = =Δ 70 90 0 70 0 90 logdB10 logdB10logdB10 dB20 I I I I I I β Solving for I 90 yields: 7090 100II = Express the fractional change in the intensity from 90 dB to 70 dB: %99 100 100 70 7070 90 7090 = − = − I II I II Because IP ∝ , the fractional change in power is %99 . 61 •• A spherical source radiates sound uniformly in all directions. At a distance of 10 m, the sound intensity level is 80 dB. (a) At what distance from the source is the intensity level 60 dB? (b) What power is radiated by this source? Picture the Problem The intensity at a distance r from a spherical source varies with distance from the source according to .4 2 av rPI π= We can use this relationship to relate the intensities corresponding to an 80-dB intensity level (I 80 ) and the intensity corresponding to a 60-dB intensity level (I 60 ) to their distances from the source. We can relate the intensities to the intensity levels through ()() 0 logdB10 II=β . (a) Express the intensity of the sound where the intensity level is 80 dB: 2 80 av 80 4 r P I π = Express the intensity of the sound where the intensity level is 60 dB: 2 60 av 60 4 r P I π = Divide the first of these equations by the second to obtain: () 2 2 60 2 60 av 2 av 60 80 m100 4 m104 r r P P I I == π π Solving for r 60 yields: () 60 80 60 m10 I I r = Chapter 15 1578 Find the intensity of the 80-dB sound level radiation: () ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ = 0 80 logdB10dB80 I I and 24 0 8 80 W/m1010 − == II Find the intensity of the 60-dB sound level radiation: () ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ = 0 60 logdB10dB60 I I and 26 0 6 60 W/m1010 − == II Substitute numerical values for I 80 and I 60 and evaluate r 60 : () km10.0 W/m10 W/m10 m10 26 24 60 == − − r (b) Using the intensity corresponding to an intensity level of 80 dB, express and evaluate the power radiated by this source: ( ) ()[ ] W13.0 m104πW/m10 2 24 80 = == − AIP 62 •• Harry and Sally are sitting on opposite sides of a circus tent when an elephant trumpets a loud blast. If Harry experiences a sound intensity level of 65 dB and Sally experiences only 55 dB, what is the ratio of the distance between Sally and the elephant to the distance between Harry and the elephant? Picture the Problem The intensity of the sound heard by Harry and Sally depends inversely on the square of the distance between the elephant and each of them. We can use the definition of sound intensity (decibel) level ( ()() 0 logdB 10 II=β ) and the definition of intensity ( API av = ) to find the ratio of these distances. Express the sound intensity level at Harry’s location: () () 0 2 H av 0 H H 4 logdB 10 logdB 10 Ir P I I π β = = Similarly, the sound intensity level at Sally’s location is: () 0 2 S av S 4 logdB 10 Ir P π β = The difference in decibel level’s at the two locations is given by: SH Δ βββ −= Traveling Waves 1579 Substituting for β H , β S , and Δβ yields: () () dB 10dB 55dB 65 4 logdB 10 4 logdB 10Δ 0 2 S av 0 2 H av =−=−= Ir P Ir P ππ β Simplifying this expression yields: 1log 2 H 2 S = r r ⇒ 1 2 H 2 S 10= r r Solving for the ratio HS rr yields: 2.310 H S == r r 63 •• Three noise sources produce intensity levels of 70, 73, and 80 dB when acting separately. When the sources act together, the resultant intensity is the sum of the individual intensities. (a) Find the sound intensity level in decibels when the three sources act at the same time. (b) Discuss the effectiveness of eliminating the two least intense sources in reducing the intensity level of the noise. Picture the Problem We can find the intensities of the three sources from their intensity levels and, because their intensities are additive, find the intensity level when all three sources are acting. (a) Express the sound intensity level when the three sources act at the same time: () () 0 807370 0 sources3 sources3 logdB10 logdB10 I III I I ++ = =β Find the intensities of each of the three sources: () 0 7 70 0 70 10logdB10dB70 II I I =⇒= () 0 3.7 73 0 73 10logdB10dB73 II I I =⇒= and () 0 8 80 0 80 10logdB10dB80 II I I =⇒= Substituting in the expression for β 3 sources yields: () ()() dB81 101010logdB10 101010 logdB10 83.77 0 0 8 0 3.7 0 7 sources3 = ++= ++ = I III β Chapter 15 1580 (b) Find the intensity level with the two least intense sources eliminated: () dB80 10 logdB10 0 0 8 80 == I I β Eliminating the two least intense sources does not reduce the intensity level significantly. 64 •• Show that if two people are different distances away from a sound source, the difference Δβ between the intensity levels reaching the people, in decibels, will always be the same, no matter the power radiated by the source. Picture the Problem Let the two people be identified by the numerals 1 and 2 and use the definition of the intensity level, in decibels, to express Δβ as a function of the distances r 1 and r 2 of the two people from the source. Express the difference in the sound intensity level heard by the two people: 12 Δ βββ −= The sound level intensity (decibel level) heard by the second person is given by: () () () 0 2 2 av 02 av 0 2 2 4 logdB 10 logdB 10logdB 10 Ir P IA P I I π β = == In like manner: () 0 2 1 av 1 4 logdB 10 Ir P π β = Substitute in the expression for Δβ to obtain: () () 0 2 1 av 0 2 2 av 4 logdB 10 4 logdB 10Δ Ir P Ir P ππ β −= Simplifying yields: () () ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ = ⎟ ⎟ ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎜ ⎜ ⎝ ⎛ = 2 2 2 1 0 2 1 av 0 2 2 av logdB 10 4 4 logdB 10Δ r r Ir P Ir P π π β This result shows that the difference in sound level intensities depends only on the distances to the source and not on the source’s power output. Traveling Waves 1581 65 ••• Everyone at a party is talking equally loudly. One person is talking to you and the sound intensity level at your location is 72 dB. Assuming that all 38 people at the party are at the same distance from you as the person who you are talking to, find the sound intensity level at your location. Picture the Problem The sound intensity level can be found from the intensity of the sound due to the 38 people. When 38 people are talking, the intensities add. Express the sound level when all 38 people are talking: () () () () dB88 dB7238logdB10 logdB1038logdB10 38 logdB10 0 1 0 1 38 = += ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ += ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ = I I I I β An equivalent but longer solution: Express the sound intensity level when all 38 people are talking: () ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ = 0 1 38 38 logdB10 I I β Express the sound intensity level when only one person is talking: () ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ == 0 1 1 logdB10dB72 I I β Solving for I 1 yields: ( ) 25 2122.7 0 2.7 1 W/m101.58 W/m101010 − − ×= == II Express the sound intensity when all 38 people are talking: 138 38II = The sound intensity level is: () ( ) dB88 W/m10 W/m101.5838 logdB10 212 25 38 = × = − − β 66 ••• When a violinist pulls the bow across a string, the force with which the bow is pulled is fairly small, about 0.60 N. Suppose the bow travels across the A string, which vibrates at 440 Hz, at 0.50 m/s. A listener 35 m from the performer hears a sound of 60-dB intensity. Assuming that the sound radiates uniformly in Chapter 15 1582 all directions, with what efficiency is the mechanical energy of bowing converted to sound energy? Picture the Problem Let η represent the efficiency with which mechanical energy is converted to sound energy. Because we’re given information regarding the rate at which mechanical energy is delivered to the string and the rate at which sound energy arrives at the location of the listener, we’ll take the efficiency to be the ratio of the sound power delivered to the listener divided by the power delivered to the string. We can calculate the power input directly from the given data. We’ll calculate the intensity of the sound at 35 m from its intensity level at that distance and use this result to find the power output. Express the efficiency of the conversion of mechanical energy to sound energy: in out P P =η Find the power delivered by the bow to the string: ( )( ) W0.30m/s0.50N0.60 in === FvP Using ()() 0 logdB10 II=β , find the intensity of the sound at 35 m: () ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ = 0 m35 logdB10dB60 I I and 26 0 6 m35 W/m1000.110 − ×== II The rate at which sound energy is emitted is given by: ( )() W0.0154 m35W/m101.004π 2 26 out = ×== − IAP Substitute numerical values and evaluate η: %1.5 W0.30 W0.0154 ==η 67 ••• [SSM] The noise intensity level at some location in an empty examination hall is 40 dB. When 100 students are writing an exam, the noise level at that location increases to 60 dB. Assuming that the noise produced by each student contributes an equal amount of acoustic power, find the noise intensity level at that location when 50 students have left. Picture the Problem Because the sound intensities are additive, we’ll find the noise intensity level due to one student by subtracting the background noise intensity from the intensity due to the students and dividing by 100. Then, we’ll use this result to calculate the intensity level due to 50 students. Traveling Waves 1583 Express the intensity level due to 50 students: () ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ = 0 1 50 50 logdB10 I I β Find the sound intensity when 100 students are writing the exam: () ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ = 0 100 logdB10dB60 I I and 26 0 6 100 W/m1010 − == II Find the sound intensity due to the background noise: () ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ = 0 background logdB10dB40 I I and 28 0 4 background W/m1010 − == II Express the sound intensity due to the 100 students: 26 2826 background100 W/m10 W/m10W/m10 − −− ≈ −=− II Find the sound intensity due to 1 student: 28background100 W/m10 100 − = − II Substitute numerical values and evaluate the noise intensity level due to 50 students: () ( ) dB57 W/m10 W/m101.0050 logdB10 212 28 50 = × = − − β String Waves Experiencing Speed Changes 68 • A 3.00-m-long piece of string with a mass of 25.0 g is tied to 4.00 m of heavy twine with a mass of 75.0 g and the combination is put under a tension of 100 N. If a transverse pulse is sent down the less dense string, determine the reflection and transmission coefficients at the junction point. Picture the Problem Let the subscript ″s″ refer to the string and the subscript ″t″ to the heavy twine. We can use the definitions of the reflection and transmission coefficients and the expression for the speed of waves on a string (Equation 15-3) to find the speeds of the pulse on the string and the heavy twine. Chapter 15 1584 Use their definitions to express the reflection and transmission coefficients: t s t s st st 1 1 v v v v vv vv r + − = + − = (1) and t s st t 1 22 v v vv v + = + =τ (2) Use Equation 15-3 to express v t and v s : t T t μ F v = and s T s μ F v = Dividing the expression for v s by the expression for v t and simplifying yields: s t t T s T t s μ μ μ μ == F F v v Substitute for t s v v in equation (1) to obtain: s t s t 1 1 μ μ μ μ + − =r (3) Express the ratio of μ t to μ s : ts st s s t t s t null null null null m m m m == μ μ Substituting in equation (3) yields: ts st ts st 1 1 null null null null m m m m r + − = Substitute numerical values and evaluate r: ( )( ) ()() () ()() 20.0 m 00.4kg 100.25 m 00.3kg 100.75 1 m 00.4kg 100.25 m 00.3kg 100.75 1 3 3 3 3 −= × × + × × − = − − − − r Traveling Waves 1585 Substitute for t s v v in equation (2) to obtain: ts st s t 1 2 1 2 null null m m + = + = μ μ τ Substitute numerical values and evaluateτ: ()() () 80.0 m 00.4kg 100.25 m 00.3kg 100.75 1 2 3 3 = × × + = − − τ Remarks: Because 1=+τr , we could have used this relationship to find τ. 69 • [SSM] Consider a taut string with a mass per unit length μ 1 , carrying transverse wave pulses that are incident upon a point where the string connects to a second string with a mass per unit length μ 2 . (a) Show that if μ 2 = μ 1 , then the reflection coefficient r equals zero and the transmission coefficient τ equals +1. (b) Show that if μ 2 >> μ 1 , then r ≈ –1 and τ ≈ 0; and (c) if μ 2 << μ 1 then r ≈ +1 and τ ≈ +2. Picture the Problem We can use the definitions of the reflection and transmission coefficients and the expression for the speed of waves on a string (Equation 15-3) to r and t in terms of the linear densities of the strings. (a) Use their definitions to express the reflection and transmission coefficients: 2 1 2 1 12 12 1 1 v v v v vv vv r + − = + − = (1) and 2 1 12 2 1 22 v v vv v + = + =τ (2) Use Equation 15-3 to express v 2 and v 1 : 2 T 2 μ F v = and 1 T 1 μ F v = Dividing the expression for v 1 by the expression for v 2 and simplifying yields: 1 2 2 T 1 T 2 1 μ μ μ μ == F F v v Chapter 15 1586 Substitute for 2 1 v v in equation (1) to obtain: 1 2 1 2 1 1 μ μ μ μ + − =r (3) Substitute for 2 1 v v in equation (2) to obtain: 1 2 1 2 μ μ τ + = (4) If μ 2 = μ 1 : 0 11 11 = + − =r and 1 11 2 1 2 1 2 = + = + = μ μ τ (b) From equations (1) and (2), if μ 2 >> μ 1 then v 1 >> v 2 : 1 1 1 12 12 −= − ≈ + − = v v vv vv r and 0 1 22 2 1 12 2 ≈ + = + = v v vv v τ (c) If μ 2 << μ 1 , then v 2 >> v 1 : 1 1 1 2 1 2 1 12 12 ≈ + − = + − = v v v v vv vv r and 2 1 22 2 1 12 2 ≈ + = + = v v vv v τ 70 •• Verify the validity of ( ) 2 21 2 1 τvvr += (Equation 15-36) by substituting the expressions for r and τ into it. Picture the Problem Making the indicated substitutions will lead us to the identity 1 = 1. Traveling Waves 1587 The reflection and transmission coefficients are given by: 12 12 vv vv r + − = and 12 2 2 vv v + =τ Substituting into the equation given in the problem statement yields: 2 12 2 2 1 2 12 12 2 1 ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ + + ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ + − = vv v v v vv vv Simplify this expression algebraically to obtain: () () () () () () 1 242 4 1 2 12 2 12 2 12 2 112 2 2 2 12 12 2 112 2 2 2 12 2 12 2 2 12 = + + = + ++ = + ++− = + ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ +− = vv vv vv vvvv vv vvvvvv vv v v vvv 71 ••• Consider a taut string that has a mass per unit length μ 1 carrying transverse wave pulses of the form y = f(x – v 1 t) that are incident upon a point P where the string connects to a second string with mass per unit length μ 2 . Derive () 2 21 2 1 τvvr += by equating the power incident on point P to the power reflected at P plus the power transmitted at P. Picture the Problem Choose the direction of propagation of the incident pulse as the positive x direction and let x = 0 at point P. Energy is conserved as the incident pulse is partially reflected and partially transmitted at point P. From the conservation of energy we have: trin PPP =+ (1) From Equation 15-20, the power transmitted in the direction of increasing x is given by: x y x y FP ∂ ∂ ∂ ∂ −= T Substituting in equation (1) yields: ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ∂ ∂ ∂ ∂ −=⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ∂ ∂ ∂ ∂ −+⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ∂ ∂ ∂ ∂ − x y x y F x y x y F x y x y F tt T rr T inin T or, upon simplification, x y x y x y x y x y x y ∂ ∂ ∂ ∂ = ∂ ∂ ∂ ∂ + ∂ ∂ ∂ ∂ ttrrinin (2) Chapter 15 1588 Because the incident pulse is given by ()tvxfy 1in −= , the reflected and transmitted pulses are given by: ( )tvxrfy 1r −−= and [] ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ −= tvx v v fy 2 2 1 t τ Evaluating x y ∂ ∂ and t y ∂ ∂ using the chain rule yields: xd df x y ∂ ∂ = ∂ ∂ η η and td df t y ∂ ∂ = ∂ ∂ η η where η is the argument of the wave function. For the transmitted pulse: [] 2 1 2 2 1t v v d df tvx v v xd df x y η τ η τ = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ − ∂ ∂ = ∂ ∂ and [] () η τ η τ η τ d df v v d df tvx v v td df t y 1 12 2 1t −= −= ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ − ∂ ∂ = ∂ ∂ For the reflected pulse: () ηη d df rtvx xd df r x y −=−− ∂ ∂ = ∂ ∂ 1 r and ()() η ηη d df rv v d df rtvx td df r t y 1 11 r −= −=−− ∂ ∂ = ∂ ∂ For the incident pulse: () ηη d df tvx xd df x y =− ∂ ∂ = ∂ ∂ 1 in and ()() η ηη d df v v d df tvx td df t y 1 11 in −= −=− ∂ ∂ = ∂ ∂ Substitute in equation (2) to obtain: ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ − ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ − ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ −+ ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ − η τ η τ ηηηη d df v v v d df d df rv d df r d df v d df 1 2 1 11 Traveling Waves 1589 Simplifying and rearranging terms yields: 2 2 1 2 1 τ v v r += The Doppler Effect 72 • A sound source is moving at 80 m/s toward a stationary listener that is standing in still air. (a) Find the wavelength of the sound in the region between the source and the listener. (b) Find the frequency heard by the listener. Picture the Problem We can use Equation 15-38 ( s s f uv± =λ ) to find the wavelength of the sound between the source and the listener and Equation15-41a ( s s r r f uv uv f ± ± = ) to find the frequency heard by the listener. (a) Apply Equation 15-38 to find λ: m32.1 s200 m/s80m/s343 1 s s s s = − = − = ± = − f uv f uv λ (b) Apply Equation 15-41a to obtain f r : () Hz261 s200 m/s80m/s343 m/s343 0 1 s s s r r = − = − ± = ± ± = − s f uv v f uv uv f 73 • Consider the situation described in Problem 72 from the reference frame of the source. In this frame, the listener and the air are moving toward the source at 80 m/s and the source is at rest. (a) At what speed, relative to the source, is the sound traveling in the region between the source and the listener? (b) Find the wavelength of the sound in the region between the source and the listener. (c) Find the frequency heard by the listener. Picture the Problem (a) In the reference frame of the source, the speed of sound from the source to the listener is reduced by the speed of the air. (b) We can find the wavelength of the sound in the region between the source and the listener from v = fλ. (c) Because the sound waves in the region between the source and the listener will be compressed by the motion of the listener, the frequency of the sound heard by the listener will be higher than the frequency emitted by the source and can be calculated using s s r r f uv uv f ± ± = (Equation 15-41a). Chapter 15 1590 (a) The speed of sound in the reference frame of the source is: m/s263 m/s80m/s433 wind = −=−= uvv' (b) Noting that the frequency is unchanged, express the wavelength of the sound: m32.1 s200 m/s263 1 === − f v' λ (c) Apply Equation 15-41a to obtain: () Hz261 s200 m/s263 m/s80m/s263 0 1 s r s s r r = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ + = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ± + = ± ± = − f v' uv' f uv' uv' f 74 • A sound source is moving away from the stationary listener at 80 m/s. (a) Find the wavelength of the sound waves in the region between the source and the listener. (b) Find the frequency heard by the listener. Picture the Problem We can use ( ) ss fuv±=λ ( Equation 15-38) to find the wavelength of the sound in the region between the source and the listener and s s r r f uv uv f ± ± = (Equation 15-41a) to find the frequency heard by the listener. Because the sound waves in the region between the source and the listener will be spread out by the motion of the listener, the frequency of the sound heard by the listener will be lower than the frequency emitted by the source. (a) Because the source is moving away from the listener, use the positive sign in the numerator of Equation 15-38 to find the wavelength of the sound between the source and the listener: m12.2 s200 m/s80m/s343 1 s s s s = + = + = ± = − f uv f uv λ (b) Because the listener is at rest and the source is receding, u r = 0 and the denominator of Equation 15-41a is the sum of the two speeds: () Hz162 s200 m/s80m/s343 m/s343 0 1 s s s s r r = + = + ± = ± ± = − f uv v f uv uv f Traveling Waves 1591 75 • The listener is moving 80 m/s from the stationary source that is in rest relative to the air. Find the frequency heard by the listener. Picture the Problem Because the listener is moving away from the source, we know that the frequency he/she will hear will be less than the frequency emitted by the source. We can use s s r r f uv uv f ± ± = (Equation 15-41a), with u s = 0 and the minus sign in the numerator, to determine its value. Relate the frequency heard by the listener to that of the source: () Hz153 s200 m/s343 m/s80m/s343 0 1 s r s s r r = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ − = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ± − = ± ± = − f v uv f uv uv f 76 •• You have made the trek to observe a Space Shuttle landing. Near the end of its descent, the ship is traveling at Mach 2.50 at an altitude of 5000 m. (a) What is the angle that the shock wave makes with the line of flight of the shuttle? (b) How far horizontally are you from the shuttle by the time you hear its shock wave, assuming the shuttle maintains both a constant heading and a constant 5000-m altitude after flying directly over your head? Picture the Problem The diagram shows the position of the shuttle at time t after it was directly over your head (located at point P) Let u represent the speed of the shuttle and v the speed of sound. We can use trigonometry to determine the angle of the shock wave as well as the location of the shuttle x when you hear the shock wave. (a) Referring to the diagram, express θ in terms of v, u, and t: 5.2 11 sin === vuut vt θ Solving for θ yields: °=°= ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ = − 6.2358.23 50.2 1 sin 1 θ Chapter 15 1592 (b) Using the diagram, relate θ to the altitude h of the shuttle and the distance x: x h =θtan ⇒ θtan h x = Substitute numerical values and evaluate x: km5.11 tan23.58 m5000 = ° =x 77 •• The SuperKamiokande neutrino detector in Japan is a water tank the size of a 14-story building. When neutrinos collide with electrons in water, most of their energy is transferred to the electrons. As a consequence, the electrons then fly off at speeds that approach c. The neutrino is counted by detecting the shock wave, called Cerenkov radiation, that is produced when the high-speed electrons travel through the water at speeds greater than the speed of light in water. If the maximum angle of the Cerenkov shock-wave cone is 48.75°, what is the speed of light in water? Picture the Problem The angle θ of the Cerenkov shock wave is related to the speed of light in water v and the speed of light in a vacuum c according to .sin cv=θ Relate the speed of light in water v to the angle of the Cerenkov cone: c v =θsin ⇒ θsincv = Substitute numerical values and evaluate v: ( ) m/s10254.2 75.48sinm/s10998.2 8 8 ×= °×=v 78 •• You are in charge of calibrating the radar guns for a local police department. One such device emits microwaves at a frequency of 2.00 GHz. During the trials, you have it arranged so that these waves are reflected from a car moving directly away from the stationary emitter. In this situation, you detect a frequency difference (between the received microwaves and the ones sent out) of 293 Hz. Find the speed of the car. Picture the Problem Because the car is moving away from the stationary emitter at a speed u r , the frequency f r it receives will be less than the frequency emitted by the emitter. The microwaves reflected from the car, moving away from a stationary detector, will be of a still lower frequency f r ′. We can use the Doppler shift equations to derive an expression for the speed of the car in terms of difference of these frequencies. Express the frequency f r received by the moving car in terms of f s , u r , and c: s r s s r r 0 f c uc f uc uc f ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ± − = ± ± = (1) Traveling Waves 1593 Express the frequency f r ′ received by the stationary source in terms of f r , u r , and c: r r r s r r 0 f c uc f uc uc 'f ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ± − = ± ± = (2) Substitute equation (1) in equation (2) to eliminate f r : . provided 2 1 1 rs r s 2 r s 2 r r cuf c u f c u f c uc 'f <<⎟ ⎠ ⎞ ⎜ ⎝ ⎛ −≈ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ −=⎟ ⎠ ⎞ ⎜ ⎝ ⎛ − = The frequency difference detected at the source is given by: s r s r srs 2 2 1Δ f c u f c u f'fff = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ −−=−= Solving for u r yields: f f c u Δ 2 s r = Substitute numerical values and evaluate u r : () () km/h1.79 h s3600 m10 km1 s m 0.22 Hz293 GHz2.002 m/s10998.2 3 8 r = ××= × =u 79 •• [SSM] The Doppler effect is routinely used to measure the speed of winds in storm systems. As the manager of a weather monitoring station in the Midwest, you are using a Doppler radar system that has a frequency of 625 MHz to bounce a radar pulse off of the raindrops in a swirling thunderstorm system 50 km away. You measure the reflected radar pulse to be up-shifted in frequency by 325 Hz. Assuming the wind is headed directly toward you, how fast are the winds in the storm system moving? Hint: The radar system can only measure the component of the wind velocity along its ″line of sight.″ Picture the Problem Because the wind is moving toward the weather station (radar device), the frequency f r the raindrops receive will be greater than the frequency emitted by the radar device. The radar waves reflected from the raindrops, moving toward the stationary detector at the weather station, will be of a still higher frequency f r ′. We can use the Doppler shift equations to derive an expression for the radial speed u of the wind in terms of difference of these frequencies. Chapter 15 1594 Use Equation 15-41a to express the frequency f r received by the raindrops in terms of f s , u r , and c: s r s s r r 0 f c uc f uc uc f ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ± + = ± ± = (1) Express the frequency f r ′ received by the stationary source at the weather station in terms of f r , u r , and c: r r r s r r 0 f c uc f uc uc 'f ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ± + = ± ± = (2) Substitute equation (1) in equation (2) to eliminate f r : . provided 2 1 1 rs r s 2 r s 2 r r cuf c u f c u f c uc 'f <<⎟ ⎠ ⎞ ⎜ ⎝ ⎛ +≈ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ +=⎟ ⎠ ⎞ ⎜ ⎝ ⎛ + = The frequency difference detected at the source is: s r ss r sr 2 2 1Δ f c u ff c u f'ff = −⎟ ⎠ ⎞ ⎜ ⎝ ⎛ +=−= Solving for u r yields: f f c u Δ 2 s r = Substitute numerical values and evaluate u r : () () mi/h174 m/s4470.0 mi/h1 m/s 95.77 Hz253 MHz2562 m/s10998.2 8 r = ×= × =u 80 •• A stationary destroyer is equipped with sonar that sends out 40-MHz- pulses of sound. The destroyer receives pulses back from a submarine directly below with a time delay of 80 ms at a frequency of 39.958 MHz. If the speed of sound in seawater is 1.54 km/s, (a) what is the depth of the submarine? (b) What is its vertical speed? Picture the Problem Let the depth of the submarine be represented by D and its vertical speed by u. The submarine acts as both a receiver and source. We can apply the definition of average speed to determine the depth of the submarine and use the Doppler shift equations to derive an expression for the vertical speed of the submarine in terms of the frequency difference. The frequency received is lower than the frequency of the source, so the sub is descending. Traveling Waves 1595 (a) Using the definition of average speed, relate the depth of the submarine to the time delay between the transmitted and reflected pulses: tvD Δ=2 ⇒ tvD Δ= 2 1 Substitute numerical values and evaluate D: ( )( ) m62ms80km/s1.54 2 1 ==D (b) Use s s r r f uv uv f ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ ± ± = to express the frequency f sub received by the submarine: 0sub f v uv f ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ − = (1) where u is the vertical speed of the submarine. Use s s r r f uv uv f ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ ± ± = to express the frequency f r ′ received by the destroyer: subr f uv v 'f ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ + = (2) Substitute equation (1) in equation (2) to eliminate f sub : 0r ' f uv uv f ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ + − = ⇒ v ff ff u ' ' r0 r0 + − = Substitute numerical values and evaluate u: ()m/s81.0km/s 54.1 MHz 958.39MHz 000.40 MHz 958.39MHz 000.40 =⎟ ⎠ ⎞ ⎜ ⎝ ⎛ + − =u and so, because u is positive, the sub is descending. 81 •• A police radar unit transmits microwaves of frequency 3.00 × 10 10 Hz, and their speed in air is 3.00 × 10 8 m/s. Suppose a car is receding from the stationary police car at a speed of 140 km/h. (a) What is the frequency difference between the transmitted signal and the signal received from the receding car? (b) Suppose the police car is, instead, moving at a speed of 60 km/h in the same direction as the other vehicle. What is the difference in frequency between the emitted and the reflected signals? Picture the Problem The radar wave strikes the speeding car at frequency f r .This frequency is less than f s because the car is moving away from the source. The frequency shift is given by Equation 15-42 (the low-speed, relative to light, approximation). The car then acts as a moving source emitting waves of frequency f r . The police car detects waves of frequency f r ′ < f r because the source (the speeding car) is moving away from the police car. The total frequency shift Chapter 15 1596 is the sum of the two frequency shifts. (a) Express the frequency difference Δf as the sum of the frequency difference sr1 Δ fff −= and the frequency difference rr2 Δ f'ff −= : 21 ΔΔΔ fff += (1) Using Equation 15-42, substitute for the frequency differences in equation (1): () rsrs Δ ff c u f c u f c u f +−=−−= (2) where rs uuu ±= is the speed of the source relative to the receiver. Apply Equation 15-42 to 1 Δf to obtain: c u f ff f f −= − = s sr s 1 Δ where we’ve used the minus sign because we know the frequency difference is a downshift. Solving for f r yields: sr 1 f c u f ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ −= Substitute for f r in equation (2) and simplify to obtain: s ss 2 1Δ f c u c u f c u f c u f ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ −−= ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ −+−= Because u/c is negligible compared to 2: c u ff s 2Δ −≈ Substitute numerical values and evaluate Δf: () kHz 78.7 m/s 1000.3 s3600 h1 h km 140 Hz103.002Δ 8 10 −= × ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ × ×−≈f Traveling Waves 1597 (b) Use the result derived in (a) with u equal to the difference in the speeds of the car and the police cruiser to obtain: () kHz .44 m/s 1000.3 s3600 h1 h km 60 h km 140 Hz103.002Δ 8 10 −= × ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ − ×−≈f 82 •• In modern medicine, the Doppler effect is routinely used to measure the rate and direction of blood flow in arteries and veins. High frequency ″ultrasound″ (sound at frequencies above the human hearing range) is typically employed. Suppose you are in charge of measuring the blood flow in a vein (located in the lower leg of an older patient) that returns blood upward to the heart. Her varicose veins indicate that perhaps the one-way valves in the vein are not working properly and that the blood is ″pooling″ in the veins and perhaps even that the blood flow is backward toward her feet. Employing sound with a frequency of 50.0 kHz, you point the sound source from above her thigh region down towards her feet and measure the sound reflected from that vein area to be lower than 50.0 kHz. (a) Was your diagnosis of the valve condition correct? If so, explain. (b) Estimate the instrument’s frequency difference capability to enable you to measure speeds down to 6.00 cm/s. Take the speed of sound in flesh to be the same as that in water, 1500 m/s. Picture the Problem (a) Whether your diagnosis was correct depends on whether the signal reflected from the blood is upshifted or downshifted. (b) Applying the Doppler shift equations with the source stationary and the receiver moving initially and then a second time with the source moving and the receiver stationary will allow us to estimate the instrument’s frequency difference capability. (a) Because the received sound frequency is less than the frequency that was sent out, the blood it reflected from must have been moving away from the source, or toward the feet of the patient. The blood is flowing the wrong way and your diagnosis is correct. (b) The blood receives a frequency that is downshifted according to: s r s r s s r r 1 0 f v u f v uv f uv uv f ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ −= ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ± − = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ ± ± = (1) Chapter 15 1598 The frequency given by equation (1) is the frequency emitted by the moving blood back toward the receiver. This results in a second downshift: '' ''' f v u f v u f uv v f uv uv f s s s 1 s s s s s r r 11 0 ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ −≈⎟ ⎠ ⎞ ⎜ ⎝ ⎛ += ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ + ± = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ ± ± = − (2) The source frequency in equation (2) is the received frequency given by equation (1). Note that the receiver and emitter speeds are the same … the speed of the blood. Let this speed be u to obtain: ssr 2111 f v u f v u v u f ' ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ −≈⎟ ⎠ ⎞ ⎜ ⎝ ⎛ −⎟ ⎠ ⎞ ⎜ ⎝ ⎛ −≈ Express the magnitude of the difference in frequencies: s ssrs 2 21Δ f v u f v u ffff ' = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ −−≈−= Substitute numerical values and evaluate Δf: () Hz .004 kHz 0.50 m/s 1500 cm/s .006 2Δ = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ≈f 83 •• [SSM] A sound source of frequency f 0 moves with speed u s relative to still air toward a receiver who is moving with speed u r relative to still air away from the source. (a) Write an expression for the received frequency f′. (b) Use the result that (1 – x) –1 ≈ 1 + x to show that if both u s and u r are small compared to v, then the received frequency is approximately s rel 1' f v u f ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ += , where u rel = u s – u r is the velocity of the source relative to the receiver. Picture the Problem The received and transmitted frequencies are related through s s r r f uv uv f ± ± = (Equation 15-41a), where the variables have the meanings given in the problem statement. Because the source and receiver are moving in the same direction, we use the minus signs in both the numerator and denominator. Traveling Waves 1599 (a) Relate the received frequency f r to the frequency f s of the source: s 1 sr s s r s s r r 11 1 1 f v u v u f v u v u f uv uv f − ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ −⎟ ⎠ ⎞ ⎜ ⎝ ⎛ −= − − = ± ± = (b) Expand () 1 s 1 − − vu binomially and discard the higher-order terms: v u v u s 1 s 11 +≈⎟ ⎠ ⎞ ⎜ ⎝ ⎛ − − Substitute to obtain: s rs s srrs s sr r 1 1 11 f v uu f v u v u v u v u f v u v u f ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ − +≈ ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ −−+= ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ +⎟ ⎠ ⎞ ⎜ ⎝ ⎛ −= because both u s and u r are small compared to v. Because u rel = u s − u r s rel r 1 f v u f ' ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ +≈ 84 ••• To study the Doppler shift on your own, you take an electronic tone generating device that is set to a frequency of middle C (262Hz) to a campus wishing well known as ″The Abyss. ″ When you hold the device at arm’s length (1.00 m), you measure its intensity level to be 80.0 dB. You then drop the tuner down the hole, listening to its sound as it falls. (a) After the tuner has fallen for 5.50 s, what frequency do you hear? (b) Estimate the time at which you can no longer hear the tuner. Picture the Problem (a) As the tuner falls, its speed increases and so the frequency you hear is Doppler shifted. Our concern, however, is with the frequency you hear 5.50 s after you’ve dropped the tuner. The sound that you hear at this time was emitted sometime before the tuner had fallen for 5.50 s. Hence we must answer the question ″At what time was the sound emitted that you heard 5.50 s after dropping the tuner?″ (b) We’ll estimate the minimum intensity of the sound from the tuner when it can be heard at the top of the well to be 30.0 dB. Then we can use the intensity level decrease of 50.0 dB to find the depth of the tuner when it can no longer be heard. The time at which you can no longer hear the tuner is the sum of the time it took the tuner to fall to the depth at which you Chapter 15 1600 can no longer hear it and the time it takes the sound to travel back to your ears. (a) Apply Equation 15-41a to express the frequency you’ll hear after 5.50 s: s fall s s r r f vv v f uv uv f + = ± ± = (1) Relate the elapsed time of 5.50 s to the time required for the tuner to fall and the time for the sound to return to you: returnfalltot ΔΔΔ ttt += (2) Use a constant-acceleration equation to relate the time-of-fall to the fall distance: ( ) 2 fall2 1 ΔΔ tgy = ⇒ g y t 2Δ Δ fall = The return time depends on how far the tuner has fallen and on the speed of sound in air: v y t Δ Δ return =⇒ return ΔΔ tvy = Substituting for Δy in the expression for Δt fall yields: g tv t return fall Δ2 Δ = Square both sides of the equation, use equation (2) to substitute for Δt return , and write the result explicitly as a quadratic equation to obtain: () () 0Δ 2 Δ 2 Δ totfall 2 fall =−+ t g v t g v t Substitute numerical values to obtain: () () () ( )( ) 0 m/s81.9 s 50.5m/s 3432 Δ m/s81.9 m/s 3432 Δ 2 fall 2 2 fall =−+ tt or ()( )( ) 0s 6.384Δs 93.69Δ 2 fall 2 fall =−+ tt Use your graphing calculator or the quadratic formula to solve this equation for its positive root: s 124.5Δ fall =t Because fallfall Δtgv = , equation (1) becomes: s fall r Δ f tgv v f + = Traveling Waves 1601 Substitute numerical values and evaluate f r : ()() () Hz 229Hz 262 s 124.5m/s 81.9m/s 343 m/s 343 2 r = + =f (b) The sound will reach your ear when: v y g y ttt ff returnfall 2 ΔΔΔ += += (3) where y f is the depth of the tuner when it can no longer be heard at the top of the well. Express the initial and final intensity levels to the change in intensity level that corresponds to no longer being able to hear the tuner: βββ Δ if += (4) The initial and final intensity levels are given by: () ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ = 0 2 i i i 4 logdB 10 Iy P π β and () ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ = 0 2 f f f 4 logdB 10 Iy P π β Substitute in equation (4) to obtain: () () β ππ Δ 4 logdB 10 4 logdB 10 0 2 i i 0 2 f f + ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ Iy P Iy P or () () ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ − ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ = 0 2 i i 0 2 f f 4 logdB 10 4 logdB 10Δ Iy P Iy P ππ β Assuming a decrease in the decibel level of 50.0 dB and simplifying yields: () () 2 f i 0 2 i i 0 2 f f logdB 10 4 4 logdB 10dB 50 ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ = ⎟ ⎟ ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎜ ⎜ ⎝ ⎛ =− y y Iy P Iy P π π Chapter 15 1602 Simplify further to obtain: 2 f i log5 ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ =− y y or i 5.2 f 10 yy = Substituting for y f in equation (3) yields: v y g y t i 5.2 i 5.2 10102 Δ + × = Substitute numerical values and evaluate Δt: ( ) () s 0.9 m/s 343 m 0.110 m/s 81.9 m 0.1102 Δ 5.2 2 5.2 ≈ + × =t 85 •• You are in a hot-air balloon carried along by a 36-km/h wind and have a sound source with you that emits a sound of 800 Hz as it approaches a tall building. (a) What is the frequency of the sound heard by an observer at the window of this building? (b) What is the frequency of the reflected sound heard by you? Picture the Problem The simplest way to approach this problem is to transform to a reference frame in which the balloon is at rest. In that reference frame, the speed of sound is v = 343 m/s, and u r = 36 km/h = 10 m/s. Then, we can use the equations for a moving receiver and a moving source to find the frequencies heard at the window and on the balloon. (a) Use Equation 15-41a to express the received frequency in terms of the frequency of the source: s r s r s s r s s r r 1 0 1 1 1 1 f v u f v v u f v u v u f uv uv f ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ += ± + = ± ± = ± ± = Substitute numerical values and evaluate f r : () kHz82.0 Hz823Hz800 m/s343 m/s10 1 r = = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ +=f (b) Treating the tall building as a moving source, express the frequency of the reflected sound heard by a person riding in the balloon: Traveling Waves 1603 r s r s r s r r s r r 1 1 1 0 1 1 1 f v u f v u v f v u v u f uv uv 'f ⎟ ⎟ ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎜ ⎜ ⎝ ⎛ − = − ± = ± ± = ± ± = Substitute numerical values and evaluate f r ′: () kHz0.85 Hz823 m/s343 m/s10 1 1 r = ⎟ ⎟ ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎜ ⎜ ⎝ ⎛ − ='f 86 •• A car is approaching a reflecting wall. A stationary observer behind the car hears a sound of frequency 745 Hz from the car horn and a sound of frequency 863 Hz from the wall. (a) How fast is the car traveling? (b) What is the frequency of the car horn? (c) What frequency does the car driver hear reflected from the wall? Picture the Problem We can relate the frequencies f r and f r ′ heard by the stationary observer behind the car to the speed of the car u r and the frequency of the car’s horn f s . Dividing these equations will eliminate the frequency of the car’s horn and allow us to solve for the speed of the car. We can then substitute to find the frequency of the car’s horn. We can find the frequency heard by the driver as a moving receiver by relating this frequency to the frequency reflected from the wall. (a) Use Equation 15-41a to relate the frequency heard by the observer directly from the car’s horn to the speed of the car: s s s s s s r s s r r 1 1 1 0 1 1 1 f v u f v u v f v u v u f uv uv f + = + ± = ± ± = ± ± = (1) Relate the frequency reflected from the wall to the speed of the car: s s s s s s r r 1 1 1 0 1 f v u f v u v f uv uv 'f − = − ± = ± ± = (2) Chapter 15 1604 Divide equation (2) by equation (1) to obtain: v u v u f 'f s s r r 1 1 − + =⇒ v 'f'f f'f u rr rr + − = Substitute numerical values and evaluate u s : () km/h6.90 h s3600 m10 km1 s m 17.25 m/s343 Hz745Hz863 Hz745Hz863 3 s = ××= + − =u (b) Solve equation (1) for f s to obtain: r r s 1 f v u f ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ += Substitute numerical values and evaluate f s : () Hz800 Hz745 m/s343 m/s25.17 1 s = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ +=f (c) The driver is a moving receiver and so we can relate the frequency heard by the driver to the frequency reflected by the wall (the frequency heard by the stationary observer): 'f v u f r r driver 1 ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ += Substitute numerical values and evaluate f driver : () Hz926 H863 m/s343 m/s25.17 1 driver = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ += zf 87 •• The driver of a car traveling at 100 km/h toward a vertical wall briefly sounds the horn. Exactly 1.00 s later she hears the echo and notes that its frequency is 840 Hz. How far from the wall was the car when the driver sounded the horn and what is the frequency of the horn? Picture the Problem Let t = 0 when the driver sounds her horn and let the distance to the wall at that instant be d. The received and transmitted frequencies are related through s s r r f uv uv f ± ± = (Equation 15-41a). Solving this equation for f s will allow us to determine the frequency of the car horn. We can use the total distance the sound travels (car-to-wall plus wall-back to-car … now closer to the Traveling Waves 1605 wall) to determine the distance to the wall when the horn was briefly sounded. Use Equation 15-41a to relate the frequency heard by the driver to her speed and to the frequency of her horn: s s r s s r s s r r 1 1 1 1 f v u v u f v u v u f uv uv f − + = ± ± = ± ± = Solving for f s yields: r r s s 1 1 f v u v u f + − = Substitute numerical values and evaluate f s : () () Hz 714Hz840 m/s343 m/s78.27 1 m/s343 m/s78.27 1 Hz840 m/s343 s3600 h1 h km 100 1 m/s343 s3600 h1 h km 100 1 s = + − = × + × − =f Relate the distance d to the wall at t = 0 to the distance she travels in time Δt = 1 s, her speed u s , and the speed of sound v: ( ) tvtudd ΔΔ s =−+ ⇒ ()tvud Δ s2 1 += Substitute numerical values and evaluate d: ( )( ) m185 s1.00m/s343m/s27.78 2 1 = +=d 88 •• You are on a transatlantic flight traveling due west at 800 km/h. An experimental plane flying at Mach 1.6 and 3.0 km to the north of your plane is also on an east-to-west course. What is the distance between the two planes when you hear the sonic boom from the experimental plane? Picture the Problem You’ll hear the sonic boom when the surface of its cone reaches your plane. In the following diagram, the experimental plane is at C and your plane is at P. The distance h = 3.0 km. The distance between the planes when you hear the sonic boom is d. We can use trigonometry to determine the angle of the shock wave as well as the separation of the planes when you hear the sonic boom. Chapter 15 1606 Using the Pythagorean theorem, relate the separation of the planes d, to the distance h and the angle θ : θ 2222 cosdhd += ⇒ θ 2 cos1 1 − = hd Express θ in terms of v, u, and t: 6.1 11 sin === vuut vt θ so °= ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ = − 7.38 6.1 1 sin 1 θ Substitute numerical values and evaluate d: () km8.4 7.38cos1 1 km0.3 2 = °− =d 89 ••• The Hubble space telescope has been used to determine the existence of planets orbiting around distant stars. The planet orbiting the star will cause the star to ″wobble″ with the same period as the planet’s orbit. Because of wobble, light from the star will be Doppler-shifted up and down periodically. Estimate the maximum and minimum wavelengths of light of nominal wavelength 500 nm emitted by the Sun that is Doppler-shifted by the motion of the Sun due to the planet Jupiter. Picture the Problem The Sun and Jupiter orbit about their effective mass located at their common center of mass. We can apply Newton’s 2 nd law to the Sun to obtain an expression for its orbital speed about the Sun-Jupiter center of mass and then use this speed in the Doppler shift equation to estimate the maximum and minimum wavelengths resulting from the Jupiter-induced motion of the Sun. Letting v be the orbital speed of the sun about the center of mass of the Sun-Jupiter system, express the Doppler shift of the light due to this motion when the Sun is approaching Earth: c v c v c c v c v f ' c f' − + = − + == 1 1 1 1 λλ Traveling Waves 1607 Solving for λ′ and simplifying yields: 2121 1 11 11 1 1 − − ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ + ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ −= ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ +⎟ ⎠ ⎞ ⎜ ⎝ ⎛ −= + − = c v c v c v c v c v c v ' λ λλλ Because v << c, we can expand 21 1 ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ − c v and 21 1 − ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ + c v binomially to obtain: c v c v 2 11 21 −≈⎟ ⎠ ⎞ ⎜ ⎝ ⎛ − and c v c v 2 11 21 −≈⎟ ⎠ ⎞ ⎜ ⎝ ⎛ + − Substitute for 21 1 ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ − c v and 21 1 − ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ + c v to obtain: c v c v c v c v −≈ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ −= + − 1 2 1 1 1 2 When the Sun is receding from Earth: c v c v c v c v +≈ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ += − + 1 2 1 1 1 2 Hence the motion of the Sun will give an observed Doppler shift of: ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ±≈ c v ' 1λλ (1) Apply Newton’s 2 nd law to the Sun: cm 2 S 2 cm effS r v M r MGM =⇒ cm eff r GM v = Measured from the center of the Sun, the distance to the center of mass of the Sun-Jupiter system is: ( ) Js JJ-s Js JJ-sS cm 0 MM Mr MM MrM r + = + + = The effective mass is related to the masses of the Sun and Jupiter according to: Jseff 111 MMM +=⇒ Js Js eff MM MM M + = Chapter 15 1608 Substitute for M eff and r cm to obtain: J-s s Js JJ-s Js Js r GM MM Mr MM MM G v = + + = Using m1078.7 11 J-s ×=r as the mean orbital radius of Jupiter, substitute numerical values and evaluate v: ()( ) m/s10306.1 m1078.7 kg1099.1kg/mN10673.6 4 11 302211 ×= × ×⋅× = − v Substitute numerical values in equation (1) to obtain: () ()() 5 8 4 1036.41nm500 m/s10998.2 m/s10306.1 1nm500' − ×±= ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ × × ±≈λ The maximum and minimum wavelengths are: ( )( ) 5 max 1036.41nm500 − ×+=λ and ( )( ) 5 min 1036.41nm500 − ×−=λ General Problems 90 • At time t = 0, the shape of a wave pulse on a string is given by the function () ( )( ) 2 2 3 m 2.00/m 120.00, xxy += , where x is in meters. (a) Sketch y(x, 0) versus x. (b) Give the wave function y(x,t) at a general time t if the pulse is moving in the +x direction with a speed of 10.0 m/s and if the pulse is moving in the −x direction with a speed of 10.0 m/s. Picture the Problem The equation of a wave traveling in the positive x direction is of the form () )( vtxfx,ty −= and that of a wave traveling in the negative x direction is () )( vtxfx,ty += . Traveling Waves 1609 (a) The pulse at t = 0 shown below was plotted using a spreadsheet program: 0.000 0.005 0.010 0.015 0.020 0.025 0.030 0.035 -6 -4 -2 0 2 4 6 x (m) y ( x ,0) ( m ) (b) If the pulse is moving in the +x direction, the wave function must be of the form: ( )()[ ]txfvtxfx,ty m/s0.10)( −=−= because v = 10.0 m/s Replace x with ()tx m/s0.10− to obtain: () ()( )[] 22 3 m/s0.10m00.2 m120.0 tx x,ty −+ = If the pulse is moving in the −x direction, the wave function must be of the form: ( )()[ ]txfvtxfx,ty m/s0.10)( +=+= because v = 10.0 m/s Replace x with ()tx m/s0.10+ to obtain: () () ( )[] 22 3 m/s0.10m00.2 m120.0 , tx txy ++ = 91 • [SSM] A whistle that has a frequency of 500 Hz moves in a circle of radius 1.00 m at 3.00 rev/s. What are the maximum and minimum frequencies heard by a stationary listener in the plane of the circle and 5.00 m away from its center? Chapter 15 1610 Picture the Problem The diagram depicts the whistle traveling in a circular path of radius r = 1.00 m. The stationary listener will hear the maximum frequency when the whistle is at point 1 and the minimum frequency when it is at point 2. These maximum and minimum frequencies are determined by f 0 and the tangential speed u s = 2π r/T. We can relate the frequencies heard at point P to the speed of the approaching whistle at point 1 and the speed of the receding whistle at point 2. s u m 00.1=r m 00.5 1 2 P s u Relate the frequency heard at point P to the speed of the approaching whistle at point 1: s s max 1 1 f v u f − = Because ωru = s : smax 1 1 f v r f ω − = Substitute numerical values and evaluate f max : () () Hz529Hz500 m/s343 rev rad2 s rev 3.00m00.1 1 1 max = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ × − = π f Relate the frequency heard at point P to the speed of the receding whistle at point 2: s s min 1 1 f v u f + = Substitute numerical values and evaluate f min : () () Hz474Hz500 m/s343 rev rad2 s rev 3.00m00.1 1 1 min = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ × + = π f Traveling Waves 1611 92 • Ocean waves move toward the beach with a speed of 8.90 m/s and a crest-to-crest separation of 15.0 m. You are in a small boat anchored off shore. (a) At what frequency do the wave crests reach you boat? (b) You now lift anchor and head out to sea at a speed of 15.0 m/s. At what frequency do the wave crests reach your boat now? Picture the Problem (a) The crest-to-crest separation of the waves is their wavelength. We can find the frequency of the waves from v = fλ. (b) When you lift anchor and head out to sea you’ll become a moving receiver and we can apply s s r r f uv uv f ± ± = to calculate the frequency you’ll observe. (a) The frequency of the ocean waves is the ratio of their speed to their wavelength: Hz0.593 m15.0 m/s8.90 0 === λ v f (b) Express the frequency of the waves in terms of their speed and the speed of a moving receiver: s r s r s s r r 1 f v u f v uv f uv uv f ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ += + = ± ± = Substitute numerical values and evaluate f r : () Hz1.59 Hz0.593 m/s8.90 m/s15.0 1 r = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ +=f 93 •• A 12.0-m-long wire that has an 85.0-g mass is stretched under a tension of 180 N. A pulse is generated at the left end of the wire, and 25.0 ms later a second pulse is generated at the right end of the wire. Where do the pulses first meet? Picture the Problem Let t be the time of travel of the left-hand pulse and the subscripts L and R refer to the pulse coming from the left and right, respectively. Because the pulse traveling from the right starts later than the pulse from the left, its travel time is t − Δt, where Δt = 25.0 ms. Both pulses travel at the same speed and the sum of the distances they travel is 12.0 m. Express the total distance the two pulses travel: ( )ttvvtddd Δ−+=+= RL Solve for vt to obtain: ( )tvdvt Δ+= 2 1 Chapter 15 1612 The speed of the pulse is given by: Lm FF v TT == μ Substituting for v yields: ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ += t Lm F dvt Δ T 2 1 Substitute numerical values and evaluate vt: ( )( ) ()m99.7s1025.0 kg0.085 m12.0N180 m12.0 3 2 1 = ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ ×+= − vt from the left end of the wire. 94 •• You are parked on the shoulder of a highway. Find the speed of a car in which the tone of the car’s horn drops by 10 percent as it passes you. (In other words, the total drop in frequency between the ″approach″ value and the ″recession″ value is 10%.) Picture the Problem Let the frequency of the car’s horn be f s , the frequency you hear as the car approaches be f r , and the frequency you hear as the car recedes be f r ′. We can use s s r r f uv uv f ± ± = to express the frequencies heard as the car approaches and recedes and then use these frequencies to express the fractional change in frequency as the car passes you. Express the fractional change in frequency as the car passes you: 10.0 Δ r = f f Relate the frequency heard as the car approaches to the speed of the car: s s s s s s r r 1 10 f v u f uv v f uv uv f − = − ± = ± ± = Express the frequency heard as the car recedes in terms of the speed of the car: s s r 1 1 f v u 'f + = Traveling Waves 1613 Divide the second of these frequency equations by the first to obtain: v u v u f 'f s s r r 1 1 + − = and 10.0 1 1 1 Δ s s rr r r r = + − −==− v u v u f f f 'f f f Solving for u s yields: vu 9.1 10.0 s = Substitute numerical values and evaluate u s : () km/h65 h s3600 10 km1 s m 05.18 m/s343 9.1 10.0 3 s = ××= =u 95 •• [SSM] A loudspeaker driver 20.0 cm in diameter is vibrating at 800 Hz with an amplitude of 0.0250 mm. Assuming that the air molecules in the vicinity have the same amplitude of vibration, find (a) the pressure amplitude immediately in front of the driver, (b) the sound intensity, and (c) the acoustic power being radiated by the front surface of the driver. Picture the Problem (a) and (b) The pressure amplitude can be calculated directly from , 00 vsp ρω= and the intensity from . 2 0 2 2 1 vsI ρω= (c) The power radiated is the intensity times the area of the driver. (a) Relate the pressure amplitude to the displacement amplitude, angular frequency, wave velocity, and air density: 00 vsp ρω= Substitute numerical values and evaluate p 0 : ( ) ( )[ ] ()() 2 3 13 0 N/m6.55 m100.0250m/s343 s8002kg/m1.29 = ×× = − − πp (b) Relate the intensity to these same quantities: vsI 2 0 2 2 1 ρω= Chapter 15 1614 Substitute numerical values and evaluate I: ( ) ( )[ ] ()() 22 2 3 2 13 2 1 W/m49.3W/m494.3 m/s343m100.0250 s8002kg/m1.29 == ×× = − − πI (c) Express the power in terms of the intensity and the area of the driver: IrIAP 2 π== Substitute numerical values and evaluate P: ( ) ( ) W110.0 W/m3.494m100.0 22 = =πP 96 •• A plane, harmonic, sound wave in air has an amplitude of 1.00 μm and has an intensity of 10.0 mW/m 2 . What is the frequency of the wave? Picture the Problem The frequency of the wave is related to the density of the air, displacement amplitude, and velocity by . 2 0 2 2 1 vsI ρω= Relate the intensity of the wave to the density of the air, displacement amplitude, velocity, and angular frequency: vsI 2 0 2 2 1 ρω=⇒ v I s ρ ω 21 0 = Because fπω 2= : v I s f ρπ 2 2 1 0 = Substitute numerical values and evaluate f: () ( ) ()() kHz07.1 m/s343kg/m1.29 W/m1000.12 m1000.12 1 3 22 6 = × × = − − π f 97 •• Water flows at 7.0 m/s in a pipe of radius 5.0 cm. A plate with area equal to the cross-sectional area of the pipe is suddenly inserted to stop the flow. Find the force exerted on the plate. Take the speed of sound in water to be 1.4 km/s. Hint: When the plate is inserted, a pressure wave propagates through the water at the speed of sound v s . The mass of water brought to a stop in time Δt is the water in a length of pipe equal to v s Δt. Picture the Problem The force exerted on the plate is due to the change in momentum of the water. We can use Newton’s 2 nd law in the form F = Δp/Δt to relate F to the mass of water in a length of pipe equal to v s Δt and to the speed of Traveling Waves 1615 the water. This mass of water, in turn, is given by the product of its density and the volume of water in a length of pipe equal to v s Δt. Relate the force exerted on the plate to the change in momentum of the water: t mv t p F Δ Δ = Δ Δ = w Express Δm in terms of the mass of water in a length of pipe equal to v s Δt: tAvVm Δ=Δ=Δ s ρρ Substitute for Δm to obtain: ws AvvF ρ= Substitute numerical values and evaluate F: ()( ) ( )[ ]( ) kN77m/s7.0m0.050km/s1.4kg/m011.00 233 =×= πF 98 •• A high-speed flash photography setup to capture a picture of a bullet exploding a soap bubble is shown in Figure 15-33. The shock wave from the bullet is to be detected by a microphone that will trigger the flash. The microphone is placed on a track that is parallel to, and 0.350 m below, the path of the bullet. The track is used to adjust the position of the microphone. If the bullet is traveling at 1.25 times the speed of sound, and the distance between the lab bench and the track is 0.350 m, how far back from the soap bubble must the microphone be set to trigger the flash? (Assume that the flash itself is instantaneous once the microphone is triggered.) Picture the Problem Let d be the horizontal distance from the soap bubble to the position of the microphone. The angle θ of the shock wave is related to the speed of sound in air u and the speed of the bullet c according to .sin vu=θ We can determine θ from the given information and then use this angle to find d. Express d in terms of the angle of the shock wave and the distance from the soap bubble to the laboratory bench: θtan m350.0 =d (1) Relate the speed of the bullet to the angle of the shock-wave cone: v u =θsin ⇒ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ = − v u 1 sinθ Chapter 15 1616 Substitute for θ in equation (1) to obtain: ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ = − v u d 1 sintan m350.0 Substitute numerical values and evaluate d: cm3.26 25.1 sintan m350.0 1 = ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ = − u u d 99 •• A column of precision marchers keeps in step by listening to the band positioned at the head of the column. The beat of the music is for 100 paces/min. A television camera shows that only the marchers at the front and the rear of the column are actually in step. The marchers in the middle section are striding forward with the left foot when those at the front and rear are striding forward with the right foot. The marchers are so well trained, however, that they are all certain that they are in proper step with the music. How long is the column? Picture the Problem The source of the problem is that it takes a finite time for the sound to travel from the front of the line of marchers to the back. We can use the given data to determine the time required for the beat to reach the marchers in the back of the column and then use this time and the speed of sound to find the length of the column. Express the length of the column in terms of the speed of sound and the time required for the beat to travel the length of the column: tvL Δ= Calculate the time for the sound to travel the length of the column: s600.0min 100 1 Δ ==t Substitute numerical values and evaluate L: ( )( ) m206s0.600m/s343 ==L 100 •• A bat flying toward a stationary obstacle at 12.0 m/s emits brief, high- frequency sound pulses at a repetition frequency of 80.0 Hz. What is the interval between the arrival times of the reflected pulses heard by the bat? Picture the Problem The interval between the arrival times of the reflected pulses heard by the bat is the reciprocal of the frequency of the reflected pulses. We can use s s r r f uv uv f ± ± = to relate the frequency of the reflected pulses to the speed of the bat and the frequency it emits. Traveling Waves 1617 Relate the interval between the arrival times of the echo pulses heard by the bat to frequency of the reflected pulses: r 1 f t =Δ Relate the frequency of the pulses received by the bat to its speed and the frequency it emits: s s r s s r s s r r 1 1 f v u v u f uv uv f uv uv f − + = − + = ± ± = Substitute for f r to obtain: s r s 1 1 Δ f v u v u t ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ + − = Substitute numerical values and evaluate Δt: () ms7.11 s80.0 m/s343 m/s12.0 1 m/s343 m/s12.0 1 Δ 1 = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ + − = − t 101 •• Laser ranging to the moon is done routinely to accurately determine the Earth–moon distance. However, to determine the distance accurately, corrections must be made for the average speed of light in Earth’s atmosphere, which is 99.997 percent of the speed of light in vacuum. Assuming that Earth’s atmosphere is 8.00 km high, estimate the length of the correction. Picture the Problem Let d be the distance to the moon, h be the height of Earth's atmosphere, and v be the average speed of light in Earth’s atmosphere. We can express d ′, the distance measured when Earth’s atmosphere is ignored, in terms of the time for a pulse of light to make a round-trip from Earth to the moon and solve this equation for the length of correction d ′ − d. Express the roundtrip time for a pulse of light to reach the moon and return: c hd v h ttt − += += 22 atmosphere searth' ofout atmosphere searth' Express the "measured" distance d ′ when we do not account for the atmosphere: hdh v c c hd v h cctd' −+= ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ − +== 22 2 1 2 1 Chapter 15 1618 Solve for the length of correction d ′ − d: ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ −=− 1 v c hdd' Substitute numerical values and evaluate d ′ − d: () m2.0 1 99997.0 km00.8 ≈ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ −=− c c dd' Remarks: This is larger than the accuracy of the measurements, which is about 3 to 4 cm. 102 •• A tuning fork attached to a taut string generates transverse waves. The vibration of the fork is perpendicular to the string. Its frequency is 400 Hz and the amplitude of its oscillation is 0.50 mm. The string has a linear mass density of 0.010 kg/m and is under a tension of 1.0 kN. Assume that there are no waves reflected at the far end of the string. (a) Find the period and frequency of waves on the string. (b) What is the speed of the waves? (c) What are the wavelength and wave number? (d) Write a suitable wave function for the waves on the string. (e) Calculate the maximum speed and acceleration of a point on the string. (f) At what minimum average rate must energy be supplied to the fork to keep it oscillating at a steady amplitude? Picture the Problem (a) The frequency of the waves on the string is the same as the frequency of the tuning fork and their period is the reciprocal of the frequency. (b) We can find the speed of the waves from the tension in the string and its linear density. (c) The wavelength can be determined from the frequency and the speed of the waves and the wave number from its definition. (d) The general form of the wave function for waves on a string is ( )()tkxAx,ty ω±= sin , so, once we know k and ω, because A is given, we can write a suitable wave function for the waves on this string. (e) The maximum speed and acceleration of a point on the string can be found from the angular frequency and amplitude of the waves. (f) Finally, we can use vAP 22 2 1 av μω= to find the minimum average rate at which energy must be supplied to the tuning fork to keep it oscillating with a steady amplitude. (a) The frequency of the waves on the string is the same as the frequency of the tuning fork: Hz400=f The period of the waves on the wire is the reciprocal of their frequency: ms2.50 s400 11 1 === − f T Traveling Waves 1619 (b) Relate the speed of the waves to the tension in the string and its linear density: km/s0.32 m/s316.2 kg/m0.010 kN1.0 T = === μ F v (c) Use the relationship between the wavelength, speed and frequency of a wave to find λ: cm79 cm79.05 s400 m/s316.2 1 = === − f v λ Using its definition, express and evaluate the wave number: 1 1 2 m9.7 m95.7 m1005.79 22 − − − = = × == π λ π k (d) Determine the angular frequency of the waves: ( ) 131 s102.51s40022 −− ×=== ππω f Substitute for A, k, and ω in the general form of the wave function to obtain: () ( ) ( )[ ( ) ]txx,ty 131 s1051.2m9.7sinmm50.0 −− ×−= (e) Relate the maximum speed of a point on the string to the amplitude of the waves and the angular frequency of the tuning fork: ()() m/s1.3 s102.51m100.50 133 max = ××= = −− ωAv Express the maximum acceleration of a point on the string in terms of the amplitude of the waves and the angular frequency of the tuning fork: ()() 2 2 133 2 max km/s2.3 s102.51m100.50 = ××= = −− ωAa (f) Express the minimum average power required to keep the tuning fork oscillating at a steady amplitude in terms of the linear density of the string, the amplitude of its vibrations, and the speed of the waves on the string: vAP 22 2 1 av μω= Chapter 15 1620 Substitute numerical values and evaluate P av : ()()( ) ( ) W25m/s316m100.50s102.51kg/m0.10 2 3 2 13 2 1 av =××= −− P 103 ••• A long rope with a mass per unit length of 0.100 kg/m is under a constant tension of 10.0 N. A motor drives one end of the rope with transverse simple harmonic motion at 5.00 cycles per second and an amplitude of 40.0 mm. (a) What is the wave speed? (b) What is the wavelength? (c) What is the maximum transverse linear momentum of a 1.00-mm segment of the rope? (d) What is the maximum net force on a 1.00-mm segment of the rope? Picture the Problem Let Δm represent the mass of the segment of length Δx = 1.00 mm. We can find the wave speed from the given data for the tension in the rope and its linear density. The wavelength can be found from v = fλ. We’ll use the definition of linear momentum to find the maximum transverse linear momentum of the 1-mm segment and apply Newton’s 2 nd law to the segment to find the maximum net force on it. (a) Find the wave speed from the tension and linear density: m/s10.0 kg/m0.100 N10.0 T === μ F v (b) Express the wavelength in terms of the speed and frequency of the wave: m2.00 s5.00 m/s10.0 1 === − f v λ (c) Relate the maximum transverse linear momentum of the 1.00-mm segment to the maximum transverse speed of the wave: xAfxAmvp Δ=Δ=Δ= μπωμ 2 maxmax Substitute numerical values and evaluate p max : ( )( ) ()() m/skg1026.1 m/skg10257.1 m0.0400m101.00 kg/m0.100s5.002 4 4 3 1 max ⋅×= ⋅×= ×× = − − − − πp (d) The maximum net force acting on the segment is the product of the mass of the segment and its maximum acceleration: max max 2 maxmax 2 ΔΔ fp pxAmaF π ωωμ = === Traveling Waves 1621 Substitute numerical values and evaluate F max : ( )( ) mN95.3 m/skg101.257s5.002 41 max = ⋅×= −− πF 104 ••• In this problem, you will derive an expression for the potential energy of a segment of a string carrying a traveling wave (Figure 15-34). The potential energy of a segment equals the work done by the tension in stretching the string, which is ΔU = F T Δnull −Δx(), where F T is the tension, Δnull is the length of the stretched segment, and Δx is its original length. From the figure we see that ()() () 21 222 1 xyxyx ΔΔ+Δ=Δ+Δ=Δnull . (a) Use the binomial expansion to show that ( ) xxyx ΔΔΔΔΔ 2 2 1 ≈−null , and therefore x x y FU Δ Δ Δ Δ 2 T2 1 ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ≈ . (b) Compute dy/dx from the wave function () ( )tkxAtxy ω−= sin, (Equation 15.-15) and show that ()xtkxkAFU Δ−≈ ω 222 T2 1 cosΔ . Picture the Problem We can follow the step-by-step instructions outlined above to obtain the given expressions for ΔU. (a) Express the potential energy of a segment of the string: ( )xFU ΔΔΔ T −= null For Δy/Δx << 1: ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎣ ⎡ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ += 2 2 1 Δ Δ 1ΔΔ x y xnull and x x y x x y xx Δ Δ Δ Δ Δ Δ 1ΔΔΔ 2 2 1 2 2 1 ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ = − ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎣ ⎡ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ +=−null Substitute to obtain: x x y F x y FU Δ Δ Δ Δ Δ Δ 2 T2 1 2 2 1 T ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ = ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎣ ⎡ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ = (b) Differentiate () ( )tkxAx,ty ω−= sin to obtain: ()tkxkA dx dy ω−= cos Chapter 15 1622 Approximate Δy/Δx by dy/dx and substitute in our result from Part (a): ( )( ) ()tkxxkAF xtkxkAFU ω ω −Δ= Δ−= 222 T2 1 2 T2 1 cos cosΔ Traveling Waves 1623 Chapter 15 1624 kpettus Microsoft Word - Ch15 ISM 061031.doc

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