Baixe Halliday 8 ediçao e outras Notas de estudo em PDF para Física, somente na Docsity! Chapter 16 1. (a) The angular wave number is (b) The speed of the wave is 2. The distance d between the beetle and the scorpion is related to the transverse speed and longitudinal speed as where and are the arrival times of the wave in the transverse and longitudinal directions, respectively. With and , we have . Thus, if then 3. (a) The motion from maximum displacement to zero is one-fourth of a cycle so 0.170 s is one-fourth of a period. The period is T = 4(0.170 s) = 0.680 s. (b) The frequency is the reciprocal of the period: (c) A sinusoidal wave travels one wavelength in one period: 4. (a) The speed of the wave is the distance divided by the required time. Thus, . (b) The width w is equal to the distance the wave has moved during the average time required by a spectator to stand and then sit. Thus, . 5. Let y1 = 2.0 mm (corresponding to time t1) and y2 = –2.0 mm (corresponding to time t2). Then we find kx + 600t1 + φ = sin−1(2.0/6.0) and 693 kx + 600t2 + φ = sin−1(–2.0/6.0) . Subtracting equations gives 600(t1 – t2) = sin−1(2.0/6.0) – sin−1(–2.0/6.0). Thus we find t1 – t2 = 0.011 s (or 1.1 ms). 6. Setting x = 0 in u = −ω ym cos(k x − ω t + φ) (see Eq. 16-21 or Eq. 16-28) gives u = −ω ym cos(−ω t+φ) as the function being plotted in the graph. We note that it has a positive “slope” (referring to its t-derivative) at t = 0: = = − ym ω² sin(−ω t + φ) > 0 at t = 0. This implies that – sinφ > 0 and consequently that φ is in either the third or fourth quadrant. The graph shows (at t = 0) u = −4 m/s, and (at some later t) umax = 5 m/s. We note that umax = ym ω. Therefore, u = − umax cos(− ω t + φ)|t = 0 F0DE φ = cos−1( ) = ± 0.6435 rad (bear in mind that cosθ = cos(−θ )), and we must choose φ = −0.64 rad (since this is about −37° and is in fourth quadrant). Of course, this answer added to 2nπ is still a valid answer (where n is any integer), so that, for example, φ = −0.64 + 2π = 5.64 rad is also an acceptable result. 7. Using v = fλ, we find the length of one cycle of the wave is λ = 350/500 = 0.700 m = 700 mm. From f = 1/T, we find the time for one cycle of oscillation is T = 1/500 = 2.00 × 10–3 s = 2.00 ms. (a) A cycle is equivalent to 2π radians, so that π/3 rad corresponds to one-sixth of a cycle. The corresponding length, therefore, is λ/6 = 700/6 = 117 mm. (b) The interval 1.00 ms is half of T and thus corresponds to half of one cycle, or half of 2π rad. Thus, the phase difference is (1/2)2π = π rad. 8. (a) The amplitude is ym = 6.0 cm. (b) We find λ from 2π/λ = 0.020π: λ = 1.0×102 cm. 694 which is a negative sine function. A plot of y(x,0) is depicted on the right. (b) From the figure we see that the amplitude is ym = 4.0 cm. (c) The angular wave number is given by k = 2π/λ = π/10 = 0.31 rad/cm. (d) The angular frequency is ω = 2π/T = π/5 = 0.63 rad/s. (e) As found in part (a), the phase is . (f) The sign is minus since the wave is traveling in the +x direction. (g) Since the frequency is f = 1/T = 0.10 s, the speed of the wave is v = fλ = 2.0 cm/s. (h) From the results above, the wave may be expressed as . Taking the derivative of y with respect to t, we find which yields u(0,5.0) = –2.5 cm/s. 14. From , we have 15. The wave speed v is given by v = , where τ is the tension in the rope and µ is the linear mass density of the rope. The linear mass density is the mass per unit length of rope: µ = m/L = (0.0600 kg)/(2.00 m) = 0.0300 kg/m. Thus, 16. The volume of a cylinder of height is V = πr2= πd2/4. The strings are long, narrow cylinders, one of diameter d1 and the other of diameter d2 (and corresponding linear densities µ1 and µ2). The mass is the (regular) density multiplied by the volume: m = ρV, so that the mass-per-unit length is and their ratio is Therefore, the ratio of diameters is 697 CHAPTER 16 17. (a) The amplitude of the wave is ym=0.120 mm. (b) The wave speed is given by v = , where τ is the tension in the string and µ is the linear mass density of the string, so the wavelength is λ = v/f = /f and the angular wave number is (c) The frequency is f = 100 Hz, so the angular frequency is ω = 2πf = 2π(100 Hz) = 628 rad/s. (d) We may write the string displacement in the form y = ym sin(kx + ωt). The plus sign is used since the wave is traveling in the negative x direction. In summary, the wave can be expressed as 18. We use to obtain 19. (a) The wave speed is given by v = λ/T = ω/k, where λ is the wavelength, T is the period, ω is the angular frequency (2π/T), and k is the angular wave number (2π/λ). The displacement has the form y = ym sin(kx + ωt), so k = 2.0 m–1 and ω = 30 rad/s. Thus v = (30 rad/s)/(2.0 m–1) = 15 m/s. (b) Since the wave speed is given by v = , where τ is the tension in the string and µ is the linear mass density of the string, the tension is 20. (a) Comparing with Eq. 16-2, we see that k = 20/m and ω = 600/s. Therefore, the speed of the wave is (see Eq. 16-13) v = ω/k = 30 m/s. (b) From Eq. 16–26, we find 21. (a) We read the amplitude from the graph. It is about 5.0 cm. (b) We read the wavelength from the graph. The curve crosses y = 0 at about x = 15 cm and again with the same slope at about x = 55 cm, so λ = (55 cm – 15 cm) = 40 cm = 0.40 m. 698 (c) The wave speed is where τ is the tension in the string and µ is the linear mass density of the string. Thus, (d) The frequency is f = v/λ = (12 m/s)/(0.40 m) = 30 Hz and the period is T = 1/f = 1/(30 Hz) = 0.033 s. (e) The maximum string speed is um = ωym = 2πfym = 2π(30 Hz) (5.0 cm) = 940 cm/s = 9.4 m/s. (f) The angular wave number is k = 2π/λ = 2π/(0.40 m) = 16 m–1. (g) The angular frequency is ω = 2πf = 2π(30 Hz) = 1.9×102 rad/s (h) According to the graph, the displacement at x = 0 and t = 0 is 4.0 × 10–2 m. The formula for the displacement gives y(0, 0) = ym sin φ. We wish to select φ so that 5.0 × 10–2 sin φ = 4.0 × 10–2. The solution is either 0.93 rad or 2.21 rad. In the first case the function has a positive slope at x = 0 and matches the graph. In the second case it has negative slope and does not match the graph. We select φ = 0.93 rad. (i) The string displacement has the form y (x, t) = ym sin(kx + ωt + φ). A plus sign appears in the argument of the trigonometric function because the wave is moving in the negative x direction. Using the results obtained above, the expression for the displacement is 22. (a) The general expression for y (x, t) for the wave is y (x, t) = ym sin(kx – ωt), which, at x = 10 cm, becomes y (x = 10 cm, t) = ym sin[k(10 cm – ωt)]. Comparing this with the expression given, we find ω = 4.0 rad/s, or f = ω/2π = 0.64 Hz. (b) Since k(10 cm) = 1.0, the wave number is k = 0.10/cm. Consequently, the wavelength is λ = 2π/k = 63 cm. (c) The amplitude is (d) In part (b), we have shown that the angular wave number is k = 0.10/cm. (e) The angular frequency is ω = 4.0 rad/s. 699 CHAPTER 16 (b) Converting to radians, we have φ = 1.45 rad. (c) In terms of wavelength (the length of each cycle, where each cycle corresponds to 2π rad), this is equivalent to 1.45 rad/2π = 0.230 wavelength. 33. (a) The amplitude of the second wave is , as stated in the problem. (b) The figure indicates that λ = 40 cm = 0.40 m, which implies that the angular wave number is k = 2π/0.40 = 16 rad/m. (c) The figure (along with information in the problem) indicates that the speed of each wave is v = dx/t = (56.0 cm)/(8.0 ms) = 70 m/s. This, in turn, implies that the angular frequency is ω = k v =1100 rad/s = 1.1F 0B 410 3 rad/s. (d) The figure depicts two traveling waves (both going in the –x direction) of equal amplitude ym. The amplitude of their resultant wave, as shown in the figure, is yF 0A 2m = 4.00 mm. Eq. 16-52 applies: yF 0A 2m = 2 ym cos( φ2) F 0 D E φ2 = 2 cos−1(2.00/9.00) = 2.69 rad. (e) In making the plus-or-minus sign choice in y = ym sin(k x ± ω t + φ), we recall the discussion in section 16-5, where it shown that sinusoidal waves traveling in the –x direction are of the form y = ym sin(k x + ω t + φ). Here, φ should be thought of as the phase difference between the two waves (that is, φ1 = 0 for wave 1 and φ2 = 2.69 rad for wave 2). In summary, the waves have the forms (with SI units understood): y1 = (0.00900)sin(16 x +1100 t) and y2 = (0.00900)sin(16 x + 1100 t + 2.7 ) . 34. (a) We use Eq. 16-26 and Eq. 16-33 with µ = 0.00200 kg/m and ym = 0.00300 m. These give v = = 775 m/s and Pavg = µv ω2ym2 = 10 W. (b) In this situation, the waves are two separate string (no superposition occurs). The answer is clearly twice that of part (a); P = 20 W. (c) Now they are on the same string. If they are interfering constructively (as in Fig. 16-16(a)) then the amplitude ym is doubled which means its square ym2 increases by a factor of 4. Thus, the answer now is four times that of part (a); P = 40 W. (d) Eq. 16-52 indicates in this case that the amplitude (for their superposition) is 2 ymcos(0.2π) = 1.618 times the original amplitude ym. Squared, this results in an increase in the power by a factor of 2.618. Thus, P = 26 W in this case. 702 (e) Now the situation depicted in Fig. 16-16(b) applies, so P = 0. 35. The phasor diagram is shown below: y1m and y2m represent the original waves and ym represents the resultant wave. The phasors corresponding to the two constituent waves make an angle of 90° with each other, so the triangle is a right triangle. The Pythagorean theorem gives . Thus ym = 5.0 cm. 36. (a) As shown in Figure 16-16(b) in the textbook, the least-amplitude resultant wave is obtained when the phase difference is F 07 0 rad. (b) In this case, the amplitude is (8.0 mm – 5.0 mm) = 3.0 mm. (c) As shown in Figure 16-16(a) in the textbook, the greatest-amplitude resultant wave is obtained when the phase difference is 0 rad. (d) In the part (c) situation, the amplitude is (8.0 mm + 5.0 mm) = 13 mm. (e) Using phasor terminology, the angle “between them” in this case is F 07 0/2 rad (90º), so the Pythagorean theorem applies: = 9.4 mm . 37. The phasor diagram is shown on the right. We use the cosine theorem: We solve for cos φ : The phase constant is therefore F 06 6 = 84°. 38. We see that y1 and y3 cancel (they are 180º) out of phase, and y2 cancels with y4 because their phase difference is also equal to π rad (180º). There is no resultant wave in this case. 39. (a) Using the phasor technique, we think of these as two “vectors” (the first of “length” 4.6 mm and the second of “length” 5.60 mm) separated by an angle of φ = 0.8π radians (or 144º). Standard techniques for adding vectors then lead to a resultant vector of length 3.29 mm. 703 CHAPTER 16 (b) The angle (relative to the first vector) is equal to 88.8º (or 1.55 rad). (c) Clearly, it should in “in phase” with the result we just calculated, so its phase angle relative to the first phasor should be also 88.8º (or 1.55 rad). 40. (a) The wave speed is given by (b) The wavelength of the wave with the lowest resonant frequency f1 is λ1 = 2L, where L = 125 cm. Thus, 41. Possible wavelengths are given by λ = 2L/n, where L is the length of the wire and n is an integer. The corresponding frequencies are given by f = v/λ = nv/2L, where v is the wave speed. The wave speed is given by where τ is the tension in the wire, µ is the linear mass density of the wire, and M is the mass of the wire. µ = M/L was used to obtain the last form. Thus (a) The lowest frequency is (b) The second lowest frequency is (c) The third lowest frequency is 42. The nth resonant frequency of string A is while for string B it is (a) Thus, we see f1,A = f4,B. That is, the fourth harmonic of B matches the frequency of A’s first harmonic. (b) Similarly, we find f2,A = f8,B. (c) No harmonic of B would match 43. (a) The wave speed is given by where τ is the tension in the string and µ is the linear mass density of the string. Since the mass density is the mass per unit length, µ = M/L, where M is the mass of the string and L is its length. Thus 704 51. (a) The amplitude of each of the traveling waves is half the maximum displacement of the string when the standing wave is present, or 0.25 cm. (b) Each traveling wave has an angular frequency of ω = 40π rad/s and an angular wave number of k = π/3 cm–1. The wave speed is v = ω/k = (40π rad/s)/(π/3 cm–1) = 1.2×102 cm/s. (c) The distance between nodes is half a wavelength: d = λ/2 = π/k = π/(π/3 cm–1) = 3.0 cm. Here 2π/k was substituted for λ. (d) The string speed is given by u(x, t) = ∂y/∂t = –ωymsin(kx)sin(ωt). For the given coordinate and time, 52. The nodes are located from vanishing of the spatial factor sin 5πx = 0 for which the solutions are (a) The smallest value of x which corresponds to a node is x = 0. (b) The second smallest value of x which corresponds to a node is x = 0.20 m. (c) The third smallest value of x which corresponds to a node is x = 0.40 m. (d) Every point (except at a node) is in simple harmonic motion of frequency f = ω/2π = 40π/2π = 20 Hz. Therefore, the period of oscillation is T = 1/f = 0.050 s. (e) Comparing the given function with Eq. 16–58 through Eq. 16–60, we obtain for the two traveling waves. Thus, we infer from these that the speed is v = ω/k = 40π/5π = 8.0 m/s. (f) And we see the amplitude is ym = 0.020 m. (g) The derivative of the given function with respect to time is which vanishes (for all x) at times such as sin(40πt) = 0. Thus, 707 CHAPTER 16 Thus, the first time in which all points on the string have zero transverse velocity is when t = 0 s. (h) The second time in which all points on the string have zero transverse velocity is when t = 1/40 s = 0.025 s. (i) The third time in which all points on the string have zero transverse velocity is when t = 2/40 s = 0.050 s. 53. (a) The waves have the same amplitude, the same angular frequency, and the same angular wave number, but they travel in opposite directions. We take them to be y1 = ym sin(kx – ωt), y2 = ym sin(kx + ωt). The amplitude ym is half the maximum displacement of the standing wave, or 5.0 × 10–3 m. (b) Since the standing wave has three loops, the string is three half-wavelengths long: L = 3λ/2, or λ = 2L/3. With L = 3.0m, λ = 2.0 m. The angular wave number is k = 2π/λ = 2π/(2.0 m) = 3.1 m–1. (c) If v is the wave speed, then the frequency is The angular frequency is the same as that of the standing wave, or ω = 2π f = 2π(50 Hz) = 314 rad/s. (d) The two waves are and Thus, if one of the waves has the form , then the other wave must have the form . The sign in front of ω for is minus. 54. From the x = 0 plot (and the requirement of an anti-node at x = 0), we infer a standing wave function of the formwhere , with length in meters and time in seconds. The parameter k is determined by the existence of the node at x = 0.10 (presumably the first node that one encounters as one moves from the origin in the positive x direction). This implies k(0.10) = π/2 so that k = 5π rad/m. (a) With the parameters determined as discussed above and t = 0.50 s, we find 708 (b) The above equation yields (c) We take the derivative with respect to time and obtain, at t = 0.50 s and x = 0.20 m, . d) The above equation yields u = –0.13 m/s at t = 1.0 s. (e) The sketch of this function at t = 0.50 s for 0 ≤ x ≤ 0.40 m is shown below: 55. (a) The angular frequency is ω = 8.00π/2 = 4.00π rad/s, so the frequency is f = ω/2π = (4.00π rad/s)/2π = 2.00 Hz. (b) The angular wave number is k = 2.00π/2 = 1.00π m–1, so the wavelength is λ = 2π/k = 2π/(1.00π m–1) = 2.00 m. (c) The wave speed is (d) We need to add two cosine functions. First convert them to sine functions using cos α = sin (α + π/2), then apply Letting α = kx and β = ωt, we find Nodes occur where cos(kx) = 0 or kx = nπ + π/2, where n is an integer (including zero). Since k = 1.0π m–1, this means . Thus, the smallest value of x which corresponds to a node is x = 0.500 m (n=0). (e) The second smallest value of x which corresponds to a node is x = 1.50 m (n=1). (f) The third smallest value of x which corresponds to a node is x = 2.50 m (n=2). (g) The displacement is a maximum where cos(kx) = ±1. This means kx = nπ, where n is an integer. Thus, x = n(1.00 m). The smallest value of x which corresponds to an anti- node (maximum) is x = 0 (n=0). 709 CHAPTER 16 Thus, we see that the block mass is inversely proportional to the harmonic number squared. Thus, if the 447 gram block corresponds to harmonic number n then = = = 1 + . Therefore, – 1 = 0.5624 must equal an odd integer (2n + 1) divided by a squared integer (n2). That is, multiplying 0.5624 by a square (such as 1, 4, 9, 16, etc) should give us a number very close (within experimental uncertainty) to an odd number (1, 3, 5, …). Trying this out in succession (starting with multiplication by 1, then by 4, …), we find that multiplication by 16 gives a value very close to 9; we conclude n = 4 (so n2 = 16 and 2n + 1 = 9). Plugging m = 0.447 kg, n = 4, and the other values given in the problem, we find µ = 0.000845 kg/m = 0.845 g/m. 61. (a) The phasor diagram is shown here: y1, y2, and y3 represent the original waves and ym represents the resultant wave. The horizontal component of the resultant is ymh = y1 – y3 = y1 – y1/3 = 2y1/3. The vertical component is ymv = y2 = y1/2. The amplitude of the resultant is (b) The phase constant for the resultant is (c) The resultant wave is The graph below shows the wave at time t = 0. As time goes on it moves to the right with speed v = ω/k. 62. Setting x = 0 in y = ym sin(k x − ω t + φ) gives y = ym sin(−ω t + φ) as the function being plotted in the graph. We note that it has a positive “slope” (referring to its t-derivative) at t = 0: = = – ymω cos(−ω t+ φ) > 0 at t = 0. This implies that – cos(φ) > 0 and consequently that φ is in either the second or third quadrant. The graph shows (at t = 0) y = 2.00 mm, and (at some later t) ym = 6.00 mm. Therefore, 712 y = ym sin(−ω t + φ)|t = 0 F0DE φ = sin−1( ) = 0.34 rad or 2.8 rad (bear in mind that sin(θ) = sin(π − θ)), and we must choose φ = 2.8 rad because this is about 161° and is in second quadrant. Of course, this answer added to 2nπ is still a valid answer (where n is any integer), so that, for example, φ = 2.8 – 2π = −3.48 rad is also an acceptable result. 63. We compare the resultant wave given with the standard expression (Eq. 16–52) to obtain , and . (a) Therefore, λ = 2π/k = 0.31 m. (b) The phase difference is φ = 1.64 rad. (c) And the amplitude is ym = 2.2 mm. 64. Setting x = 0 in ay = –ω² y (see the solution to part (b) of Sample Problem 16-2) where y = ym sin(k x − ω t + φ) gives ay = –ω² ym sin(−ω t + φ) as the function being plotted in the graph. We note that it has a negative “slope” (referring to its t-derivative) at t = 0: = = ym ω3 cos(− ω t + φ) < 0 at t = 0. This implies that cosφ < 0 and consequently that φ is in either the second or third quadrant. The graph shows (at t = 0) ay = −100 m/s², and (at another t) amax = 400 m/s². Therefore, ay = −amax sin(−ω t + φ)|t = 0 F0DE φ = sin−1( ) = 0.25 rad or 2.9 rad (bear in mind that sinθ = sin(π − θ)), and we must choose φ = 2.9 rad because this is about 166° and is in the second quadrant. Of course, this answer added to 2nπ is still a valid answer (where n is any integer), so that, for example, φ = 2.9 – 2π = −3.4 rad is also an acceptable result. 65. We note that dy/dt = −ωcos(kx – ωt + φ), which we will refer to as u(x,t). so that the ratio of the function y(x,t) divided by u(x,t) is – tan(kx − ωt + φ)/ω. With the given information (for x = 0 and t = 0) then we can take the inverse tangent of this ratio to solve for the phase constant: φ = tan−1 = tan−1= 1.2 rad. 713 CHAPTER 16 66. (a) Recalling the discussion in §16-5, we see that the speed of the wave given by a function with argument x – 5.0t (where x is in centimeters and t is in seconds) must be . (b) In part (c), we show several “snapshots” of the wave: the one on the left is as shown in Figure 16–48 (at t = 0), the middle one is at t = 1.0 s, and the rightmost one is at . It is clear that the wave is traveling to the right (the +x direction). (c) The third picture in the sequence below shows the pulse at 2.0 s. The horizontal scale (and, presumably, the vertical one also) is in centimeters. (d) The leading edge of the pulse reaches x = 10 cm at t = (10 – 4.0)/5 = 1.2 s. The particle (say, of the string that carries the pulse) at that location reaches a maximum displacement h = 2 cm at t = (10 – 3.0)/5 = 1.4 s. Finally, the trailing edge of the pulse departs from x = 10 cm at t = (10 – 1.0)/5 = 1.8 s. Thus, we find for h(t) at x = 10 cm (with the horizontal axis, t, in seconds): 67. (a) The displacement of the string is assumed to have the form y(x, t) = ym sin (kx – ωt). The velocity of a point on the string is u(x, t) = ∂y/∂t = –ω ym cos(kx – ωt) and its maximum value is um = ωym. For this wave the frequency is f = 120 Hz and the angular frequency is ω = 2πf = 2π (120 Hz) = 754 rad/s. Since the bar moves through a distance of 1.00 cm, the amplitude is half of that, or ym = 5.00 × 10–3 m. The maximum speed is um = (754 rad/s) (5.00 × 10–3 m) = 3.77 m/s. (b) Consider the string at coordinate x and at time t and suppose it makes the angle θ with the x axis. The tension is along the string and makes the same angle with the x axis. Its transverse component is τtrans = τ sin θ. Now θ is given by tan θ = ∂y/∂x = kym cos(kx – ωt) and its maximum value is given by tan θm = kym. We must calculate the angular wave number k. It is given by k = ω/v, where v is the wave speed. The wave speed is given by where τ is the tension in the rope and µ is the linear mass density of the rope. Using the data given, and Thus, and θ = 7.83°. The maximum value of the transverse component of the tension in the string is τ trans = (90.0 N) sin 7.83° = 12.3 N. 714 The results may be summarized as 72. We orient one phasor along the x axis with length 3.0 mm and angle 0 and the other at 70° (in the first quadrant) with length 5.0 mm. Adding the components, we obtain (a) Thus, amplitude of the resultant wave is (b) And the angle (phase constant) is tan–1 (4.70/4.71) = 45°. 73. (a) Using v = fλ, we obtain (b) Since frequency is the reciprocal of the period, we find 74. By Eq. 16–66, the higher frequencies are integer multiples of the lowest (the fundamental). (a) The frequency of the second harmonic is f2 = 2(440) = 880 Hz. (b) The frequency of the third harmonic is and f3 = 3(440) = 1320 Hz. 75. We make use of Eq. 16–65 with L = 120 cm. (a) The longest wavelength for waves traveling on the string if standing waves are to be set up is (b) The second longest wavelength for waves traveling on the string if standing waves are to be set up is (c) The third longest wavelength for waves traveling on the string if standing waves are to be set up is The three standing waves are shown below: 76. (a) At x = 2.3 m and t = 0.16 s the displacement is 717 CHAPTER 16 (b) We choose ym = 0.15 m, so that there would be nodes (where the wave amplitude is zero) in the string as a result. (c) The second wave must be traveling with the same speed and frequency. This implies , (d) and . (e) The wave must be traveling in –x direction, implying a plus sign in front of ω. Thus, its general form is y´ (x,t) = (0.15 m)sin(0.79x + 13t). (f) The displacement of the standing wave at x = 2.3 m and t = 0.16 s is 77. (a) The wave speed is (b) For the one-loop standing wave we have λ1 = 2L = 2(1.50 m) = 3.00 m. (c) For the two-loop standing wave λ2 = L = 1.50 m. (d) The frequency for the one-loop wave is f1 = v/λ1 = (144 m/s)/(3.00 m) = 48.0 Hz. (e) The frequency for the two-loop wave is f2 = v/λ2 = (144 m/s)/(1.50 m) = 96.0 Hz. 78. We use (a) If the tension is quadrupled, then (b) If the frequency is halved, then 79. We use Eq. 16-2, Eq. 16-5, Eq. 16-9, Eq. 16-13, and take the derivative to obtain the transverse speed u. (a) The amplitude is ym = 2.0 mm. (b) Since ω = 600 rad/s, the frequency is found to be f = 600/2π ≈ 95 Hz. (c) Since k = 20 rad/m, the velocity of the wave is v = ω/k = 600/20 = 30 m/s in the +x direction. (d) The wavelength is λ = 2π/k ≈ 0.31 m, or 31 cm. (e) We obtain 718 so that the maximum transverse speed is um = (600)(2.0) = 1200 mm/s, or 1.2 m/s. 80. (a) Since the string has four loops its length must be two wavelengths. That is, λ = L/ 2, where λ is the wavelength and L is the length of the string. The wavelength is related to the frequency f and wave speed v by λ = v/f, so L/2 = v/f and L = 2v/f = 2(400 m/s)/(600 Hz) = 1.3 m. (b) We write the expression for the string displacement in the form y = ym sin(kx) cos(ωt), where ym is the maximum displacement, k is the angular wave number, and ω is the angular frequency. The angular wave number is k = 2π/λ = 2πf/v = 2π(600 Hz)/(400 m/s) = 9.4m–1 and the angular frequency is ω = 2πf = 2π(600 Hz) = 3800 rad/s. With ym = 2.0 mm, the displacement is given by 81. To oscillate in four loops means n = 4 in Eq. 16-65 (treating both ends of the string as effectively “fixed”). Thus, λ = 2(0.90 m)/4 = 0.45 m. Therefore, the speed of the wave is v = fλ = 27 m/s. The mass-per-unit-length is µ = m/L = (0.044 kg)/(0.90 m) = 0.049 kg/m. Thus, using Eq. 16-26, we obtain the tension: τ = v2 µ = (27 m/s)2(0.049 kg/m) = 36 N. 82. (a) This distance is determined by the longitudinal speed: (b) Assuming the acceleration is constant (justified by the near-straightness of the curve a = 300/40 × 10–6) we find the stopping distance d: which gives d = 6.0×10–3 m. This and the radius r form the legs of a right triangle (where r is opposite from θ = 60°). Therefore, 719 CHAPTER 16 where C is an integration constant (which we will assume to be zero). The sketch of this function at t = 2.0 s for 0 ≤ x ≤ 0.20 m is shown below. 89. (a) The wave speed is (b) The time required is Thus if , then and if , then 90. (a) The wave number for each wave is k = 25.1/m, which means λ = 2π/k = 250.3 mm. The angular frequency is ω = 440/s; therefore, the period is T = 2π/ω = 14.3 ms. We plot the superposition of the two waves y = y1 + y2 over the time interval 0 ≤ t ≤ 15 ms. The first two graphs below show the oscillatory behavior at x = 0 (the graph on the left) and at x = λ/8 ≈ 31 mm. The time unit is understood to be the millisecond and vertical axis (y) is in millimeters. The following three graphs show the oscillation at x = λ/4 =62.6 mm ≈ 63 mm (graph on the left), at x = 3λ/8 ≈ 94 mm (middle graph), and at x = λ/2 ≈ 125 mm. (b) We can think of wave y1 as being made of two smaller waves going in the same direction, a wave y1a of amplitude 1.50 mm (the same as y2) and a wave y1b of amplitude 1.00 mm. It is made clear in §16-12 that two equal-magnitude oppositely-moving waves form a standing wave pattern. Thus, waves y1a and y2 form a standing wave, which leaves y1b as the remaining traveling wave. Since the argument of y1b involves the subtraction kx – ωt, then y1b travels in the +x direction. (c) If y2 (which travels in the –x direction, which for simplicity will be called “leftward”) had the larger amplitude, then the system would consist of a standing wave plus a leftward moving wave. A simple way to obtain such a situation would be to interchange the amplitudes of the given waves. (d) Examining carefully the vertical axes, the graphs above certainly suggest that the largest amplitude of oscillation is ymax = 4.0 mm and occurs at x = λ/4 = 62.6 mm. 722 (e) The smallest amplitude of oscillation is ymin = 1.0 mm and occurs at x = 0 and at x = λ/2 = 125 mm. (f) The largest amplitude can be related to the amplitudes of y1 and y2 in a simple way: ymax = y1m + y2m, where y1m = 2.5 mm and y2m = 1.5 mm are the amplitudes of the original traveling waves. (g) The smallest amplitudes is ymin = y1m – y2m, where y1m = 2.5 mm and y2m = 1.5 mm are the amplitudes of the original traveling waves. 91. Using Eq. 16-50, we have with length in meters and time in seconds (see Eq. 16-55 for comparison). (a) The amplitude is seen to be (b) Since k = 5π and ω = 200π, then (using Eq. 16-12) (c) k = 2π/λ leads to λ = 0.40 m. 92. (a) For visible light and (b) For radio waves and (c) For X rays and 93. (a) Centimeters are to be understood as the length unit and seconds as the time unit. Making sure our (graphing) calculator is in radians mode, we find 723 CHAPTER 16 (b) The previous graph is at t = 0, and this next one is at t = 0.050 s. And the final one, shown below, is at t = 0.010 s. (c) The wave can be written as , where is the speed of propagation. From the problem statement, we see that and . This yields (d) These graphs (as well as the discussion in the textbook) make it clear that the wave is traveling in the –x direction. 724