Tài liệu Physics exercises_solution: Chapter 15 pdf

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 fv 15.2: f  15.3: v 1500 m s   1.5  106 Hz  0.001 m 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: a) Denoting the speed of light by c,   cf , and 0010 m s 54010 Hz  556 m 3.0010 m s b) 104  2.87 m .510 Hz 15.5: a)  max  (344 m s) (20.0 Hz)  17.2 m,   (344 m s) (20,000 Hz)  1.72 cm b)  max  (1480 m s) (20.0 Hz)  74.0 m,   (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, 1 c) f  T  0.0360  27.8 Hz and from Eq (15.1), d) v  (0.280 m)(27.8 Hz)  7.78 m s , s e)  x direction 15.7: a) f  v   (8.00 m s) (0.320 m)  25.0 Hz, T  f  (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  0.02 s 15.8: b) a) 15.9: and so b) and so a) 2 y x  y   Ak sin(kx  ωt ) x 2 y   Ak cos(kx  ωt ) x y   Aω sin (kx  ωt ) t 2 y   Aω2 cos(kx  ωt ), t k2  y ω t , and y ( x, t ) is a solution of Eq (15.12) with v  ω k y   Ak cos(kx  ωt ) x 2 y   Ak sin(kx  ωt ) x y   Aω cos(kx  ωt ) t 2 y   Aω2 sin (kx  ωt ), t 2 y x  k2  y ω t , 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 A sin (kx  ωt ) 15.10: a) The relevant expressions are y ( x, t )  A cos(kx  ωt ) y vy   ωA sin (kx  ωt ) t  y v a y   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 a y  ω2 A sin( π 4)  ω2 A , and the particle is moving up , 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 a y  ω2 A sin( 3π 4)  ω2 A , and the particle is moving down , 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 and the particle is moving down a y  ω2 A sin( 5π 4)  ω2 A 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 a y  ω2 A sin( 7π 4)   ω2 A opposite signs) , and the particle is moving upward and the particle is slowing down ( v y and a y have 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 15.12: a)   2π  x t  A cos 2π      A cos  x  t    T   T    A cos where   f  v has been used T 2 x  vt ,  y 2πv 2π  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 b) vy  15.13: a) t  : 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 15.14: Solving Eq (15.13) for the force F ,  0.120 kg  2 F  μv  μ f      ((40.0 Hz) (0.750 m))  43.2  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 v (1.50 kg)(9.80 m s ) F   16.3 m s (0.0550 kg m) μ 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 , 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 , and so for the same wavelength, the frequency must be multiplied by , 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 t L  v L  Mg (m L) (0.800 kg )(14.0 m) mL   0.390 s (7.50 kg )(9.80 m s ) Mg 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 )  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 )  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.) 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 )  1890 m s e) a y , max is much larger than g, so ok to ignore gravity 15.20: a) Using Eq.(15.25), Pave  μF ω2 A2  3.00  103 kg   (25.0 N) (2(120.0 Hz))2 (1.6  103 m) 0.80 m   = 0.223 W,  or 0.22 W to two figures W 15.21: b) Halving the amplitude quarters the average power, to 0.056 Fig 15.13 plots P ( x, t )  μF ω2 A2 sin (kx  ωt ) at x  For x  0, P ( x, t )  μF ω2 A2 sin (ωt )  Pmax sin (ωt ) When x   4, kx  (2π ) ( 4)  π sin (π  ωt )  cos ωt , so P ( 4, t )  Pmax cos ωt The graph is shifted by T but is otherwise the same The instantaneous power is still never negative and Pav  12 Pmax , the same as at x  15.22: r2  r1 Ι1 Ι2  (7.5 m) 0.11 W m 1.0 W m  2.5 m, so it is possible to move r1  r2  7.5 m  2.5 m  5.0 m closer to the source 15.23: a) Ι1r 12  Ι r 2 Ι  Ι1 (r1 r2 )  (0.026 W m )(4.3 m 3.1 m)  0.050 W m b) P  4πr Ι  4π (4.3m ) (0.026 W m )  6.04 W Energy = Pt  (6.04 W )(3600 s)  2.2  10 J 15.24: v ω k  (a) A  2.30 mm (b) f  742 rad s 6.98 rad m ω 2π  742 rad s 2 118 Hz (c)   2π k  2 6.98 rad m  0.90 m (d)  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  (2.504  103 kg m)(106.3 m s)  28.3 N (keeping an extra figure in v for accuracy) (g) Pav  μF ω2 A2  (2.50  103 kg m)(28.3 N) (742 rad s) (2.30  103 m)  0.39 W 15.25: I  0.250 W m at r  15.0 m P  4πr I  4π (15.0 m) (0.250 W m )  707 W 15.26: b) a) The wave form for the given times, respectively, is shown ... 103 m)  0.39 W 15. 25: I  0.250 W m at r  15. 0 m P  4πr I  4π (15. 0 m) (0.250 W m )  707 W 15. 26: b) a) The wave form for the given times, respectively, is shown 15. 27: b) 15. 28: a) The wave...   A cos 3π  ωt  e) See Exercise 15. 12; ωA  0. 315 m s f) From the result of part d , y  mm v y   0. 315 m s 15. 48: a) From comparison with Eq ? ?15. 4 , A  0.75 cm,   f  125 Hz, T... (15. 12) 15. 36: a) From Eq (15. 35), f1  b) 10 , 000 Hz 408 Hz 2L FL  2(0.400 m) m (800 N)(0.400 m) =408 Hz (3.00  10  kg )  24.5, so the 24 th harmonic may be heard, but not the 25 th 15. 37: