Physics exercises_solution: Chapter 15

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Physics exercises_solution: Chapter 15

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Bộ tài liệu bài tập tham khảo môn vật lý bậc đại học bằng tiếng anh Chapter 15

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Nội dung Text: Physics exercises_solution: Chapter 15

  1. 15.1: a) The period is twice the time to go from one extreme to the other, and v  f    T  (6.00 m) (5.0 s)  1.20 m s, or 1.2 m s to two figures. b) The amplitude is half the total vertical distance, 0.310 m. c) The amplitude does not affect the wave speed; the new amplitude is 0.150 m. d) For the waves to exist, the water level cannot be level (horizontal), and the boat would tend to move along a wave toward the lower level, alternately in the direction of and opposed to the direction of the wave motion. 15.2: fv v 1500 m s f    1.5  106 Hz  0.001 m 15.3: a)   v f  (344 m s) (784 Hz)  0.439 m. b) f  v   (344 m s) (6.55  105 m)  5.25  106 Hz. 15.4: Denoting the speed of light by c,   c , and f 3 .0010 8 m s 8 3.0010 a) 3 54010 Hz  556 m. b) 104.510 6 m s  2.87 m. Hz 15.5: a)  max  (344 m s) (20.0 Hz)  17.2 m,  min  (344 m s) (20,000 Hz)  1.72 cm. b)  max  (1480 m s) (20.0 Hz)  74.0 m,  min  (1480 m s) (20,000 Hz)  74.0 mm. 15.6: Comparison with Eq. (15.4) gives a) 6.50 mm, b) 28.0 cm, c) f  T  0.0360 s  27.8 Hz and from Eq. (15.1), d) v  (0.280 m)(27.8 Hz)  7.78 m s , 1 1 e)  x direction.
  2. 15.7: a) f  v   (8.00 m s) (0.320 m)  25.0 Hz, T  1 f  1 (25.0 Hz)  4.00  102 s, k  2π   (2π ) (0.320 m)  19.6 rad m.  x  b) y ( x, t )  (0.0700 m) cos 2π  t (25.0 Hz)   .  0.320 m   c) (0.0700 m) cos 2π ((0.150 s)(25.0 Hz)  (0.360 m) (0.320 m))  4.95 cm. d) The argument in the square brackets in the expression used in part (c) is 2π (4.875), and the displacement will next be zero when the argument is 10π; the time is then T (5  x )  (1 25.0 Hz)(5  (0.360 m) (0.320 m))  0.1550 s and the elapsed time is 0.0050 s, e) T 2  0.02 s. 15.8: a) b)
  3. y 2 y 15.9: a)   Ak sin(kx  ωt )   Ak 2 cos(kx  ωt ) x x 2 y 2 y   Aω sin (kx  ωt )   Aω2 cos(kx  ωt ), t t 2 2 y 2 k2  y and so x 2  2 ω t 2 , and y ( x, t ) is a solution of Eq. (15.12) with v  ω k . y 2 y b)   Ak cos(kx  ωt )   Ak 2 sin(kx  ωt ) x x 2 y 2 y   Aω cos(kx  ωt )   Aω2 sin (kx  ωt ), t t 2 2 y 2 k2  y and so x 2  2 ω t 2 , and y ( x, t ) is a solution of Eq. (15.12) with v  ω k . c) Both waves are moving in the  x -direction, as explained in the discussion preceding Eq. (15.8). d) Taking derivatives yields v y ( x, t )  ωA cos (kx  ωt ) and a y ( x, t )  ω 2 A sin (kx  ωt ).
  4. 15.10: a) The relevant expressions are y ( x, t )  A cos(kx  ωt ) y vy   ωA sin (kx  ωt ) t  2 y v a y  2  y  ω2 A cos (ωt  kx). t t b) (Take A, k and ω to be positive. At x t  0, the wave is represented by (19.7(a)); point (i) in the problem corresponds to the origin, and points (ii)-(vii) correspond to the points in the figure labeled 1-7.) (i) v y  ωA cos(0)  ωA, and the particle is moving upward (in the positive y-direction). a y  ω2 A sin(0)  0, and the particle is instantaneously not accelerating. (ii) v y  ωA cos( π 4)  ωA 2 , and the particle is moving up. a y  ω2 A sin( π 4)  ω2 A 2 , and the particle is speeding up. (iii) v y  ωA cos( π 2)  0, and the particle is instantaneously at rest. a y  ω2 A sin(  π 2)  ω2 A, and the particle is speeding up. (iv) v y  ωA cos( 3π 4)   ωA 2 , and the particle is moving down. a y  ω2 A sin( 3π 4)  ω2 A 2 , and the particle is slowing down ( v y is becoming less negative). (v) v y  ωA cos(π )  ωA and the particle is moving down. a y  ω2 A sin(π )  0, and the particle is instantaneously not accelerating. (vi) v y  ωA cos( 5π 4)   ωA 2 and the particle is moving down. a y  ω2 A sin( 5π 4)  ω2 A 2 and the particle is speeding up ( v y and a y have the same sign). (vii) v y  ωA cos( 3π 2)  0, and the particle is instantaneously at rest. a y  ω2 A sin(  3π 2)  ω2 A and the particle is speeding up. (viii) v y  ωA cos( 7π 4)  ωA 2 , and the particle is moving upward. a y  ω2 A sin( 7π 4)   ω2 A 2 and the particle is slowing down ( v y and a y have opposite signs).
  5. 15.11: Reading from the graph, a) A  4.0 mm, b) T  0.040 s. c) A displacement of 0.090 m corresponds to a time interval of 0.025 s; that is, the part of the wave represented by the point where the red curve crosses the origin corresponds to the point where the blue curve crosses the t-axis ( y  0) at t  0.025 s, and in this time the wave has traveled 0.090 m, and so the wave speed is 3.6 m s and the wavelength is vT  (3.6 m s)(0.040 s)  0.14 m . d) 0.090 m 0.015 s  6.0 m s and the wavelength is 0.24 m. d) No; there could be many wavelengths between the places where y (t ) is measured. x t  2π    15.12: a) A cos 2π      A cos  x  t   T    T  2   A cos x  vt ,   where  f  v has been used. T y 2πv 2π b) vy   A sin  x  vt . t λ λ c) The speed is the greatest when the cosine is 1, and that speed is 2πvA λ . This will be equal to v if A   2π , less than v if A   2π and greater than v if A   2π .
  6. 15.13: a) t  0 : ____________________________________________________________ x(cm) 0.00 1.50 3.00 4.50 6.00 7.50 9.00 10.50 12.00 ____________________________________________________________ y(cm) 0.000 0.212 0.300 0.212 0.000 0.212 0.300 0.212 0.000 ____________________________________________________________ b) i) t = 0.400 s: _______________________________________________________________ x(cm) 0.00 1.50 3.00 4.50 6.00 7.50 9.00 10.50 12.00 _______________________________________________________________ y(cm) 0.285 0.136 0.093 0.267 0.285 0.136 0.093 0.267 0.285 _______________________________________________________________ ii  t  0.800 s : ________________________________________________________________ x(cm) 0.00 1.50 3.00 4.50 6.00 7.50 9.00 10.50 12.00 ________________________________________________________________ y(cm) 0.176 0.296 0.243 0.047 0.176 0.296 0.243 0.047 0.176 ________________________________________________________________
  7. 15.14: Solving Eq. (15.13) for the force F ,  0.120 kg  F  μv 2  μ f     2  ((40.0 Hz) (0.750 m))  43.2 . 2  2.50 m  15.15: a) Neglecting the mass of the string, the tension in the string is the weight of the pulley, and the speed of a transverse wave on the string is F (1.50 kg)(9.80 m s 2 ) v   16.3 m s. μ (0.0550 kg m) b)   v f  (16.3 m s) (120 Hz)  0.136 m. c) The speed is proportional to the square root of the tension, and hence to the square root of the suspended mass; the answers change by a factor of 2 , to 23.1 m s and 0.192 m. 15.16: a) v  F μ  (140.0  ) (10.0 m) (0.800 kg )  41.8 m s. b)   v f  (41.8 m s) (1.20 Hz)  34.9 m. c) The speed is larger by a factor of 2 , and so for the same wavelength, the frequency must be multiplied by 2 , or 1.70 Hz. 15.17: Denoting the suspended mass by M and the string mass by m, the time for the pulse to reach the other end is L L mL (0.800 kg )(14.0 m) t     0.390 s. v Mg (m L) Mg (7.50 kg )(9.80 m s 2 ) 15.18: a) The tension at the bottom of the rope is due to the weight of the load, and the speed is the same 88.5 m s as found in Example 15.4 b) The tension at the middle of the rope is (21.0 kg ) (9.80 m s 2 )  205.8 N (keeping an extra figure) and the speed of the rope is 90.7 m s. c) The tension at the top of the rope is (22.0 kg)(9.80 m s 2 )  215.6 m s and the speed is 92.9 m s . (See Challenge Problem (15.80) for the effects of varying tension on the time it takes to send signals.)
  8. 15.19: a) v  F μ  (5.00 N) (0.0500 kg m)  10.0 m s b)   v f  (10.0 m s) (40.0 Hz)  0.250 m c) y ( x, t )  A cos(kx  ωt ) (Note : y (0.0)  A, as specified.) k  2π   8.00π rad m; ω  2πf  80.0π rad s - y ( x, t )  (3.00 cm)cos[π (8.00 rad m) x  (80.0π rad s)t ] d) v y   Aω sin( kx  ωt ) and a y   Aω2cos(kx  ωt ) a y , max  Aω2  A(2πf ) 2  1890 m s 2 e) a y , max is much larger than g, so ok to ignore gravity. 15.20: a) Using Eq.(15.25), 1 Pave  μF ω2 A2 2 1  3.00  103 kg     (25.0 N) (2(120.0 Hz))2 (1.6  103 m) 2  2  0.80 m  = 0.223 W, or 0.22 W to two figures. b) Halving the amplitude quarters the average power, to 0.056 W. 15.21: Fig. 15.13 plots P ( x, t )  μF ω2 A2 sin 2 (kx  ωt ) at x  0. For x  0, P ( x, t )  μF ω2 A2 sin 2 (ωt )  Pmax sin 2 (ωt ) When x   4, kx  (2π ) ( 4)  π 2. sin (π 2  ωt )  cos ωt , so P ( 4, t )  Pmax cos 2 ωt The graph is shifted by T 4 but is otherwise the same. The instantaneous power is still never negative and Pav  1 Pmax , the same as at x  0. 2 Ι1 0.11 W m 2 15.22: r2  r1 Ι2  (7.5 m) 1.0 W m 2  2.5 m, so it is possible to move r1  r2  7.5 m  2.5 m  5.0 m closer to the source.
  9. 15.23: a) Ι1r 1  Ι 2 r 2 2 2 Ι 2  Ι1 (r1 r2 ) 2  (0.026 W m 2 )(4.3 m 3.1 m) 2  0.050 W m 2 b) P  4πr 2 Ι  4π (4.3m ) 2 (0.026 W m 2 )  6.04 W Energy = Pt  (6.04 W )(3600 s)  2.2  10 4 J 742 rad s 15.24: (a) A  2.30 mm. (b) f  ω 2π  2 118 Hz. (c)   2π k  2 6.98 rad m  0.90 m. (d) v ω k  742 rad s 6.98 rad m  106 m s. (e) The wave is traveling in the –x direction because the phase of y (x,t) has the form kx  ωt. (f) The linear mass density is μ  (3.38  10 3 kg ) (1.35 m )  2.504  10 3 kg m , so the tension is F  μv 2  (2.504  103 kg m)(106.3 m s) 2  28.3 N (keeping an extra figure in v for accuracy). (g) Pav  1 2 μF ω2 A2  1 2 (2.50  103 kg m)(28.3 N) (742 rad s) 2 (2.30  103 m) 2  0.39 W. 15.25: I  0.250 W m 2 at r  15.0 m P  4πr 2 I  4π (15.0 m) 2 (0.250 W m 2 )  707 W
  10. 15.26: a) The wave form for the given times, respectively, is shown. b)
  11. 15.27: a) The wave form for the given times, respectively, is shown. b) 15.28:
  12. 15.29: 15.30: Let the wave traveling in the  x direction be y1 ( x, t )  A cos (kx  ωt). The wave traveling in the  x direction is inverted due to reflection from the fixed end of the string at x  0, so it has the form y2 ( x, t )   A cos(kx  ωt ). The wave function of the resulting standing wave is then y ( x, t )  y1 ( x, t )  y2 ( x, t ) , where A  2.46 mm, ω  2π T  2π (3.65  103 s)  1.72  103 rad s, k  ω v  (1.72  103 rad s)(111 m s)  15.5 rad m. 15.31: a) The nodes correspond to the places where y  0 for all t in Eq. (15.1); that is, sin kxnode  0 or kxnode  nπ , n an integer . With k  0.75π rad m, xnode  (1.333 m)n and for n  0, 1, 2, ..., xnode  0, 1.333 m, 2.67 m, 4.00 m, 5.33 m, 6.67 m,... b) The antinodes correspond to the points where cos kx  0, which are halfway between any two adjacent nodes, at 0.667 m, 2.00 m, 3.33 m, 4.67 m, 6.00 m, ... 2 y 2 y 15.32: a)  k  Asw sin ωt sin kx, 2  ω 2  Asw sin ωt sin kx, x 2 t 2  ω2 ω so for y ( x, t ) to be a solution of Eq. (15.12),  k 2  2 , and v  . v k b) A standing wave is built up by the superposition of traveling waves, to which the relationship v  ω k applies.
  13. 15.33: a) The amplitude of the standing wave is Asw  0.85 cm, the wavelength is twice the distance between adjacent antinodes, and so Eq. (15.28) is y ( x, t )  (0.85 cm) sin((2π 0.075 s)t ) sin(2πx 30.0 cm). b) v  λ f  λ T  (30.0 cm) (0.0750 s)  4.00 m/s. c) (0.850 cm) sin(2π (10.5 cm) (30.0 cm))  0.688 cm. 15.34: y1  y2  A [ cos(kx  ωt )  cos(kx  ωt )]  A [ cos kx cos ωt  sin kx sin ωt  cos kx cos ωt  sin kx sin ωt ]  2 A sin kx sin ωt. 15.35: The wave equation is a linear equation, as it is linear in the derivatives, and differentiation is a linear operation. Specifically, y  ( y1  y2 ) y1 y2    . x x x x Repeating the differentiation to second order in both x and t, 2 y  2 y1  2 y2 2 y  2 y1  2 y 2   2 ,   . x 2 x 2 x t 2 t 2 t 2 The functions y1 and y2 are given as being solutions to the wave equation; that is, 2 y  2 y1  2 y2  1   2 y1  1   2 y 2   2  2  2 x 2 x 2 x  v  t 2  v  t 2  1    y  y2  2 2   2   21    v   t t 2   1 y 2  2 2  v  t and so y  y1  y 2 is a solution of Eq. (15.12).
  14. 15.36: a) From Eq. (15.35), 1 FL 1 (800 N)(0.400 m) f1   =408 Hz. 2L m 2(0.400 m) (3.00  10  3 kg ) b) 10 , 000 Hz 408 Hz  24.5, so the 24 th harmonic may be heard, but not the 25 th . 15.37: a) In the fundamental mode,   2 L  1.60 m and so v  f   (60.0 Hz)(1.60 m)  96.0 m s. b) F  v 2 μ  v 2 m L  (96.0 m s) 2 (0.0400 kg) (0.800 m)  461 . 15.38: The ends of the stick are free, so they must be displacement antinodes. 1st harmonic: 1 L  1  1  2 L  4.0 m 2 2nd harmonic: L  1 2   2  L  2.0 m rd 3 harmonic: 3 2L L  3  3   1.33 m 2 3
  15. 15.39: a) b) Eq. (15.28) gives the general equation for a standing wave on a string: y ( x, t )  ( Asw sin kx ) sinωt Asw  2 A, so A  ASW 2  (5.60 cm) 2  2.80 cm c) The sketch in part (a) shows that L  3( 2) k  2π ,   2π k Comparison of y ( x, t ) given in the problem to Eq.(15.28) gives k  0.0340 rad cm. So,   2π (0.0340 rad cm)  184.8 cm L  3( 3)  277 cm d)   185 cm, from part (c) ω  50.0 rad s so f  ω 2π  7.96 Hz period T  1 f  0.126 s v  f   1470 cm s e) v y  dy dt  ωAsw sin kx cos ωt v y , max  ωASW  (50.0 rad s)(5.60 cm)  280 cm s f) f 3  7.96 Hz  3 f 1 , so f 1  2.65 Hz is the fundamental f 8  8 f1  21.2 Hz; ω8  2πf 8  133 rad s   v f  (1470 cm s) (21.2 Hz)  69.3 cm and k  2π   0.0906 rad cm. y ( x, t )  (5.60 cm) sin ([0.0906 rad cm]x) sin ([133 rad s]t )
  16. 15.40: (a) A  1 ASW  1 (4.44 mm)  2.22 mm. (b)   2 2 2π k  2π 32.5 rad m  0.193 m. 754 rad m 754 rad m ( c) f  ω 2  2  120 Hz. (d) v  ω k  32.5 rad m  23.2 m s. (e) If the wave traveling in the  x direction is written as y1 ( x, t )  A cos(kx  t ), then the wave traveling in the  x direction is y2 ( x, t )   A cos(kx  ωt ), where A  2.22 mm from (a), and k  32.5 rad m and ω  754 rad s. (f) The harmonic cannot be determined because the length of the string is not specified. 15.41: a) The traveling wave is y ( x, t )  (2.30 m) cos ([6.98 rad m]x)  [742 rad s]t ) A  2.30 mm so ASW  4.60 mm; k  6.98 rad m and ω  742 rad s The general equation for a standing wave is y ( x, t )  ( ASW sin kx) sin t , so y ( x, t )  (4.60 mm) sin([6.98 rad m]x)sin([742 rad s]t ) b) L  1.35 m (from Exercise 15.24)   2π k  0.900 m L  3( 2), so this is the 3rd harmonic c) For this 3rd harmonic, f  ω 2π  118 Hz f 3  3 f 1 so f 1  (118 Hz) 3  39.3 Hz 15.42: The condition that x  L is a node becomes kn L  nπ. The wave number and wavelength are related by kn  n  2π , and so  n  2 L n. 15.43: a) The product of the frequency and the string length is a constant for a given string, equal to half of the wave speed, so to play a note with frequency 587 Hz, x  (60.0 cm) (440 Hz) (587 Hz)  45.0 cm. b) Lower frequency requires longer length of string free to vibrate. Full length of string gives 440 Hz, so this is the lowest note possible. 15.44: a) (i) x   2 is a node, and there is no motion. (ii) x   4 is an antinode, and vmax  A(2πf )  2πfA, amax  (2πf )vmax  4π 2 f 2 A. (iii) cos π  4 1 2 , and this factor multiplies the results of (ii), so vmax  2π fA, amax  2 2π 2 f 2 A . b) The amplitude is A sin kx, or (i)0, (ii) A, (iii) A 2. c) The time between the extremes of the motion is the same for any point on the string (although the period of the zero motion at a node might be considered indeterminate) and is 21f .
  17. ( 48.0 m s ) 15.45: a) 1  2 L  3.00 m, f1  v 2L  2 (1.50 m )  16.0 Hz. b)  3  1 3  1.00 m, f 2  3 f1  48.0 Hz. c)  4  1 4  0.75 m, f 3  4 f1  64.0 Hz. 15.46: a) For the fundamental mode, the wavelength is twice the length of the string, and v  f   2 fL  2(245 Hz)(0.635 m)  311 m s. b) The frequency of the fundamental mode is proportional to the speed and hence to the square root of the tension; (245 Hz) 1.01  246 Hz. c) The frequency will be the same, 245 Hz. The wavelength will be  air  vair f  (344 m s) (245z)  1.40 m, which is larger than the wavelength of standing wave on the string by a factor of the ration of the speeds. 15.47: a) f  v   (36.0 m s) (1.80 m)  20.0 Hz, ω  2πf  126 rad s, k  ω v  2π   3.49 rad m. b) y ( x, t )  A cos (kx  ωt )  (2.50 mm)cos (3.49 rad m) x  (126 rad s)t . c)At x  0, y (0, t )  A cos ωt  (2.50 mm) cos (126 rad s)t . From this form it may be seen that at x  0, t  0, y  0. d) At x  1.35 m  3 4, kx  3π 2 and t y 3 4, t   A cos 3π 2  ωt . e) See Exercise 15.12; ωA  0.315 m s. f) From the result of part d , y  0 mm. v y   0.315 m s.
  18. 15.48: a) From comparison with Eq. 15.4 , A  0.75 cm,   2 0.400 cm  5.00 cm, f  125 Hz, T  1 f  0.00800 s and v  f  6.25 m s. b) c) To stay with a wavefront as t increases, x and so the wave is moving in the  x - direction. d) From Eq. 15.13, the tension is F  μv 2  (0.50 kg m) (6.25 m s) 2  19.5 N. e) Pav  1 2 μF ω2 A2  54.2 W. 15.49: a) Speed in each segment is v  F μ. The time to travel through a segment is μ1 4 μ1 μ1 t  L v. The travel times then, are t1  L F , t2  L F , and t3  L 4F . Adding gives μ1 μ1 μ1 μ1 t total  L F  2L F 1L 2 F  7L 2 F . b) No, because the tension is uniform throughout each piece.
  19. 15.50: The amplitude given is not needed, it just ensures that the wave disturbance is small. Both strings have the same tension F , and the same length L  1.5 m. The wave takes different times t1 and t2 to travel along each string, so the design requirements is t1  t 2  0.20 s. Using t  L v and v  F μ  FL m   gives m1  m2 L F  0.20 s, with m1  90  10 3 kg and m1  10  10 3 kg . Solving for F gives F  6.0 N. 15.51: a) y ( x, t )  A cos(kx  ωt ) v y  dy dt   Aω sin(kx  ωt ) v y , max  Aω  2πfA v F  1  FL f  and v  , so f     m L   M  2πA  FL v y , max       M b) To double v y , max increase F by a factor of 4 15.52: The maximum vertical acceleration must be at least g. Because a  ω A, g  ω2 Amin and thus Amin  g ω2. Using ω  2πf  2πv  and v  F μ, this 2 g2 μ becomes Amin  4π 2 F .
  20. 2 y 15.53: a) See Exercise 15.10; a y  t 2  ω2 y, and so k   mω 2  xμ ω 2 . 2  2πv  4π 2 F ω2  2πf    2 b)    λ  μλ 2 and so k   (4π 2 F 2 )x. The effective force constant k  is independent of amplitude, as for a simple harmonic oscillator, and is proportional to the tension that provides the restoring force. The factor of 1 2 indicates that the curvature of the string creates the restoring force on a segment of the string. More specifically, one factor of 1  is due to the curvature, and a factor of 1 (μ) represents the mass in one wavelength, which determines the frequency of the overall oscillation of the string. The mass m  μx also contains a factor of μ, and so the effective spring constant per unit length is independent of μ.
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