Physics exercises_solution: Chapter 13

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Physics exercises_solution: Chapter 13

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Bộ tài liệu bài tập tham khảo môn vật lý bậc đại học bằng tiếng anh Chapter 13

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Nội dung Text: Physics exercises_solution: Chapter 13

1. 13.1: a) T 1 f  4.55  103 s, ω  2π T  2πf  1.38  103 rad s. b) 1 4 ( 220 Hz)  1.14  103 s, ω  2πf  5.53  103 rad s. 13.2: a) Since the glider is released form rest, its initial displacement (0.120 m) is the amplitude. b) The glider will return to its original position after another 0.80 s, so the period is 1.60 s. c) The frequency is the reciprocal of the period (Eq. (13.2)), f  1.60 s  0.625 Hz. 1 13.3: The period is 0.50 s 440  1.14  10 3 s and the angular frequency is ω 2π T  5.53  103 rad s. 13.4: (a) From the graph of its motion, the object completes one full cycle in 2.0 s; its period is thus 2.0 s and its frequency  1 period  0.5 s 1 . (b) The displacement varies from  0.20 m to  0.20 m, so the amplitude is 0.20 m. (c) 2.0 s (see part a) 13.5: This displacement is 1 of a period. 4 T  1 f  0.200 s, so t  0.0500 s. 13.6: The period will be twice the time given as being between the times at which the glider is at the equilibrium position (see Fig. (13.8)); 2 2  2π   2π  k ω m  m 2  2(2.60 s)  (0.200 kg)  0.292 N m.  T    13.7: a) T  1 f  0.167 s. b) ω  2πf  37.7 rad s. c) m  k ω2  0.084 kg. 13.8: Solving Eq. (13.12) for k, 2 2  2π   2π  k  m   (0.600 kg)    1.05  10 N m. 3 T   0.150 s  13.9: From Eq. (13.12) and Eq. (13.10), T  2π 0.500 kg 140 N m  0.375 s, f  T  2.66 Hz, 1 ω  2πf  16.7 rad s. 13.10: a) ax  d 2x dt 2  ω2 A sin(ωt  β )  ω2 x, so x(t ) is a solution to Eq. (13.4) if ω2  m . b) a  2 Aω a constant, so Eq. (13.4) is not satisfied. c) vx  k dx dt  iωi ( ωt  β ) ,  (iω) 2 Aei ( ωt  β )  ω2 x, so x(t ) is a solution to Eq. (13.4) if ω2  k m  dv x ax  dt
2. 13.11: a) x  (3.0 mm) cos ((2π )(440 Hz)t ) b) (3.0  103 m)(2π )(440 Hz)  8.29 m s, (3.0 mm)(2π ) 2 (440 Hz)2  2.29  104 m s 2 . c) j (t )  (6.34  107 m s 3 ) sin((2π )(440 Hz)t ), jmax  6.34  107 m s 3 . v0 v0 13.12: a) From Eq. (13.19), A  ω  k m  0.98 m. b) Equation (13.18) is indeterminant, but from Eq. (13.14),     , and from Eq. (13.17), sin   0, so    π . 2 2 c) cos (ωt  (π 2))   sin ωt , so x  (0.98 m) sin((12.2 rad s)t )). 13.13: With the same value for ω , Eq. (13.19) gives A  0.383 m and Eq. (13.18) gives and x  (0.383 m) cos (12.2 rad/s)t  1.02 rad .  (4.00 m/s)    arctan    (0.200 m) 300 N/m/2.00 kg   1.02 rad  58.5,   and x = (0.383 m) cos ((12.2 rad/s)t + 1.02 rad). 13.14: For SHM, a x  ω2 x  (2πf ) 2 x  2π (2.5 Hz)  (1.1  10 2 m)  2.71 m/s 2 . 2 b) From Eq. (13.19) the amplitude is 1.46 cm, and from Eq. (13.18) the phase angle is 0.715 rad. The angular frequency is 2πf  15.7 rad/s, so x  (1.46 cm) cos ((15.7 rad/s)t  0.715 rad) vx  (22.9 cm s) sin ((15.7 rad/s)t  0.715 rad) ax  (359 cm/s2 ) cos ((15.7 rad/s)t  0.715 rad) . 13.15: The equation describing the motion is x  A sin ωt; this is best found from either inspection or from Eq. (13.14) (Eq. (13.18) involves an infinite argument of the arctangent). Even so, x is determined only up to the sign, but that does not affect the result of this exercise. The distance from the equilibrium position is A sin 2π t T   0.600 m sin 4π 5  0.353 m.
3. 13.16: Empty chair: T  2π m k 4π 2 m 4π 2 (42.5 kg) k   993 N/m T2 (1.30 s) 2 With person in chair: T  2π m k T 2 k (2.54 s) 2 (993 N/m) m   162 kg 4π 2 4π 2 mperson  162 kg  42.5 kg  120 kg 13.17: T  2π m k , m  0.400 kg Use ax  2.70 m/s 2 to calculate k : max (0.400 kg)(2.70 m/s 2 )  kx  max gives k     3.60 N/m x 0.300 m T  2π m k  2.09 s 13.18: We have vx (t )  (3.60 cm/s)sin((4.71 s 1 ) t  π 2). Comparing this to the general form of the velocity for SHM:  ωA  3.60 cm/s ω  4.71 s 1   π 2 (a) T  2π ω  2π 4.71 s 1  1.33 s 3.60 cm s 3.60 cm s (b) A   0.764 cm ω 4.71 s 1 (c ) amax  ω2 A  (4.71 s 1 ) 2 (0.764 cm)  16.9 cm s 2
4. 13.19: a) x(t )  (7.40 cm) cos((4.16 rad s)t  2.42 rad) When t  T , (4.16 rad s)T  2π so T  1.51 s b) T  2π m k so k  m(2π T ) 2  26.0 N m c) A  7.40 cm  0.0740 m 1 2 mv 2  1 kx 2  1 kA2 gives vmax  A k m  0.308 m s 2 2 d) F  kx so Fmax  kA  1.92 N e) x (t ) evaluated at t  1.00 s gives x  0.0125 m v   k m A2  x 2   26.0 1.50 (0.0740) 2  (0.0125) 2 m s  0.303 m s Speed is 0.303 m s . a   kx m  (26.0 1.50)(0.0125) m s 2  0.216 m s 2 13.20: See Exercise 13.15; t  (arccos( 1.5 6))(0.3 (2π ))  0.0871 s. 13.21: a) Dividing Eq. (13.17) by ω , v0 x0  A cos ,   A sin . ω Squaring and adding, 2 v0 x  2  A2 , 2 0 ω which is the same as Eq. (13.19). b) At time t  0, Eq. (13.21) becomes 1 2 1 2 1 2 1 k 2 1 2 kA  mv0  kx0  v0  kx0 , 2 2 2 2 ω2 2 where m  kω (Eq. (13.10)) has been used. Dividing by k 2 gives Eq. (13.19). 2 13.22: a) vmax  (2πf ) A  (2π (392 Hz))(0.60  10 3 m)  1.48 m s. 1 1 b) K max  m(Vmax ) 2  (2.7  10 5 kg)(1.48 m s) 2  2.96  10 5 J. 2 2 13.23: a) Setting 1 2 mv 2  1 kx 2 in Eq. (13.21) and solving for x gives x   2 A 2 . Eliminating x in favor of v with the same relation gives vx   kA2 2m   ωA . b) This 2 happens four times each cycle, corresponding the four possible combinations of + and – in the results of part (a). The time between the occurrences is one-fourth of a period or T / 4  4 ω  2πω . c) U  1 E , K  3 E 2π 4 4  U  kA , K  3kA 8 2 8 2 
5. 13.24: a) From Eq. (13.23), k 450 N m vmax  A (0.040 m)  1.20 m/s. m 0.500 kg b) From Eq. (13.22), 450 N v (0.040 m) 2  (0.015 m) 2  1.11 m/s. 0.500 kg c) The extremes of acceleration occur at the extremes of motion, when x   A, and kA (450 N/m)(0.040 m) a max    36 m/s 2 m (0.500 kg) d) From Eq. (13.4), ax   ( 450 N/m)(kg) m)  13.5 m/s2 . (0.500 0.015 e) From Eq. (13.31), E  1 (450 N/m)(0.040 m) 2  0.36 J. 2 13.25: a) amax  ω2 A  (2f ) 2 A  2(0.85 Hz) (18.0  102 m)  5.13 m/s 2 . vmax  2 ωA  2πfA  0.961 m/s . b) ax  (2πf ) 2 x  2.57 m/s2 , v  (2πf ) A2  x 2  2π (0.85 Hz) (18.0  10 2 m)2  (9.0  10 2 m)2  0.833 m/s. c) The fraction of one period is (1 2π ) arcsin (12.0 18.0), and so the time is (T 2π )  arcsin (12.0 18.0)  1.37  10 1 s. Note that this is also arcsin ( x A) ω . d) The conservation of energy equation can be written 1 kA 2  1 mv 2  1 kx 2 . We are 2 2 2 given amplitude, frequency in Hz, and various values of x . We could calculate velocity from this information if we use the relationship k m  ω2  4π 2 f 2 and rewrite the conservation equation as 1 2 A2  1 v2 2 4π 2 f 2  1 x 2 . Using energy principles is generally a good 2 approach when we are dealing with velocities and positions as opposed to accelerations and time when using dynamics is often easier. 13.26: In the example, A2  A1 M M m and now we want A2  1 A1 . So 1  2 2 M M m , or m  3M . For the energy, E 2  1 kA2 , but since A2  1 A1 , E 2  1 E1 , or 3 E1 is lost to 2 2 2 4 4 heat.
6. 13.27: a) 1 2 mv 2  1 kx 2  0.0284 J . 2 2 v0 (0.300 m/s)2 b) x0  2  (0.012 m)2   0.014 m. ω2 (300 N/m) (0.150 kg) c) ωA  k mA  0.615 m s  13.28: At the time in question we have x  A cos (ωt   )  0.600 m v  ωA sin(ωt  )  2.20 m s a  ω2 A cos (ωt  )  8.40 m s 2 Using the displacement and acceleration equations:  ω2 A cos (ωt  )  ω2 (0.600 m)  8.40 m s 2 ω2  14.0 and ω  3.742 s 1 To find A, multiply the velocity equation by ω :  ω2 A sin (ωt  )  (3.742 s 1 ) (2.20 m s)  8.232 m s 2 Next square both this new equation and the acceleration equation and add them: ω4 A2 sin 2 (ωt   )  ω4 A2 cos 2 (ωt  )  (8.232 m s 2 ) 2  (8.40 m s 2 ) 2  ω4 A2 sin 2 (ωt  )  cos 2 (ωt   ) ω4 A2  67.77 m 2 s 4  70.56 m 2 s 4  138.3 m 2 s 4 138.3 m 2 s 4 138.3 m 2 s 4 A2  4  1 4  0.7054 m 2 ω (3.742 s ) A  0.840 m The object will therefore travel 0.840 m  0.600 m  0.240 m to the right before stopping at its maximum amplitude. 13.29: vmax  A k m Use T to find k m : T  2π m k so k m  (2π T ) 2  158 s  2 Use amax to find A : amax  kA m so A  amax (k m)  0.0405 m. Then vmax  A k m  0.509 m s
7. F0 13.30: Using k  L0 from the calibration data, ( F0 L0 ) (200 N) (1.25  101 m) m   6.00 kg. (2πf ) 2 (2π (2.60 Hz))2 mg (650 kg) (9.80 m s 2 ) 13.31: a) k    531  103 N m. Δl (0.120 m) m l 0.120 m b) T  2π   2π  2π  0.695 s. k g 9.80 m s 2 13.32: a) At the top of the motion, the spring is unstretched and so has no potential energy, the cat is not moving and so has no kinetic energy, and the gravitational potential energy relative to the bottom is 2mgA  2(4.00 kg)(9.80 m/s 2 )  (0.050 m)  3.92 J . This is the total energy, and is the same total for each part. b) U grav  0, K  0, so U spring  3.92 J . c) At equilibrium the spring is stretched half as much as it was for part (a), and so U spring  1 (3.92 J)  0.98 J, U grav  1 (3.92 J)  1.96 J, and so K  0.98 J . 4 2 13.33: The elongation is the weight divided by the spring constant, w mg gT 2  l   2  2  3.97 cm . k ω m 4π 13.34: See Exercise 9.40. a) The mass would decrease by a factor of (1 3) 3  1 27 and so 2 the moment of inertia would decrease by a factor of (1 27)(1 3)  (1 243) , and for the same spring constant, the frequency and angular frequency would increase by a factor of 243  15.6 . b) The torsion constant would need to be decreased by a factor of 243, or changed by a factor of 0.00412 (approximately). 13.35: a) With the approximations given, I  mR 2  2.72  108 kg  m 2 , or 2.7  108 kg  m 2 to two figures. b) κ  (2πf ) 2 I  (2π 2 Hz)2 (2.72  108 kg  m 2 )  4.3  106 N  m rad . 13.36: Solving Eq. (13.24) for  in terms of the period, 2  2π    I T  2  2π  3 2   ((1 2)(2.00  10 kg)(2.20  10 m) ) 2  1.00 s   1.91  10 5 N  m/rad.
8. 13.37:  0.450 N  m/rad I   0.0152 kg  m 2 . (2πf ) 2 2π (125) (265 s)2 13.38: The equation θ  cos (ωt  φ) describes angular SHM. In this problem, φ  0.  ω  sin(ω t ) and d 2θ  ω2  cos(ω t ). 2 a) dθ dt dt b) When the angular displacement is ,    cos(ω t ) , and this occurs at t  0, so dθ d 2θ  0 since sin(0)  0, and 2  ω2, since cos(0)  1. dt dt When the angular displacement is  2, 2   cos(ω t ), or 1  cos(ω t ).  2 dθ  ω 3 3 d 2θ  ω2  since sin(ω t )  , and 2  , since cos(ω t )  1 2. dt 2 2 dt 2 This corresponds to a displacement of 60 . 13.39: Using the same procedure used to obtain Eq. (13.29), the potential may be expressed as U  U 0 [(1  x R0 ) 12  2(1  x R0 ) 6 ]. Note that at r  R0 , U  U 0 . Using the appropriate forms of the binomial theorem for | x R0 |
9. 13.42: a) To the given precision, the small-angle approximation is valid. The highest speed is at the bottom of the arc, which occurs after a quarter period, T   L  0.25 s. 4 2 g b) The same as calculated in (a), 0.25 s. The period is independent of amplitude. 13.43: Besides approximating the pendulum motion as SHM, assume that the angle is sufficiently small that the length of the spring does not change while swinging in the arc. Denote the angular frequency of the vertical motion as 0  k m  kg  and   g L  1 ω0  2 kg 4w , which is solved for L  4w k . But L is the length of the stretched spring; the unstretched length is L0  L  w k  3 w k  31.00 N  1.50 N/m   2.00 m. 13.44: 13.45: The period of the pendulum is T  136 s  100  1.36 s. Then, 4π 2 L 4π 2 .5 m  g 2   10.67 m s 2 . T 1.36 s  2 13.46: From the parallel axis theorem, the moment of inertia of the hoop about the nail is I  MR 2  MR 2  2 MR 2 , so T  2π 2 R g , with d  R in Eq.13.39. Solving for R, 2 R  gT 2 8π  0.496 m. 13.47: For the situation described, I  mL2 and d  L in Eq. (13.39); canceling the factor of m and one factor of L in the square root gives Eq. (13.34).
10. 13.48: a) Solving Eq. (13.39) for I, 2 2 T  I    mgd    0.940 s     1.80 kg  9.80 m s 0.250 m   0.0987 kg  m . 2 2  2π   2π  b) The small-angle approximation will not give three-figure accuracy for Θ  0.400 rad. From energy considerations, 1 mgd 1  cos    IΩ 2 . max 2 Expressing  max in terms of the period of small-angle oscillations, this becomes 2 2  2π   2π  max  2  1  cos    2  1  cos0.40 rad   2.66 rad s . T   0.940 s  13.49: Using the given expression for I in Eq. (13.39), with d=R (and of course m=M), T  2π 5 R 3 g  0.58 s. 13.50: From Eq. (13.39), 2 2 T     120 s 100  I  mgd    1.80 kg  9.80 m s 2 0.200 m   2π  2π   0.129 kg.m . 2    13.51: a) From Eq. (13.43), ω  2.50 N m   0.90 kg s 2  2.47 rad s , so f   ω  0.393 Hz. 0.300 kg  40.300 kg 2 2π b) b  2 km  2 2.50 N m 0.300 kg   1.73 kg s . 13.52: From Eq. (13.42) A2  A1 exp  2bm t . Solving for b, 2m  A1  2(0.050 kg )  0.300 m  b ln   A  ln  0.100 m   0.0220 kg s.  t  2 (5.00 s)   As a check, note that the oscillation frequency is the same as the undamped frequency to 4.8  10 3%, so Eq. (13.42) is valid.
11. 13.53: a) With   0, x(0)  A. dx  b  b) vx   Ae  ( b 2 m ) t  cos ωt  ω sin ωt , dt  2m  and at t  0, v   Ab 2m ; the graph of x versus t near t  0 slopes down. dvx  b 2  ωb  c) ax   Ae  ( b 2 m )t  2  ω2  cos  ' t   4m  sin ωt , dt   2m  and at t  0,  b2   b2 k a x  A 2  ω2   A 2  .  4m   2m    m  (Note that this is ( bv0  kx0 ) m.) This will be negative if b  2km , zero if b  2km and positive if b  2km . The graph in the three cases will be curved down, not curved, or curved up, respectively. 13.54: At resonance, Eq. (13.46) reduces to A  Fmax bd . a) 31 . b) 2 A1. Note that A the resonance frequency is independent of the value of b (see Fig. (13.27)). 13.55: a) The damping constant has the same units as force divided by speed, or   kg  m s 2 m s   kg s.  b)The units of km are the same as [[kg s 2 ][kg]]1 2  kg s , the same as those for b. c) ωd  k m. (i) bωd  0.2 k , so A  Fmax 0.2k   5 Fmax k . 2 (ii) bd  0.4k , so A  Fmax (0.4 k )  2.5Fmax k , as shown in Fig.(13.27). 13.56: The resonant frequency is k m  (2.1  106 N m) 108 kg )  139 rad s  22.2 Hz, and this package does not meet the criterion.
12. 13.57: a) 2  0.100 m    π rad s   2 a  Aω   2   (3500 rev min )     6.72  10 m s . 3  2   30 rev min    π rad s  b) ma  3.02  10 3 N. c) ωA  (3500 rev min ) (.05 m)   18.3 m s .  30 rev min  K  1 mv 2  ( 1 )(.45 kg )(18.3 m s) 2  75.6 J. d) At the midpoint of the stroke, cos( ω t)=0 2 2 π rad s and so ωt  π 2, thus t  π 2ω. ω  (3500 rev min)( 30 rev min )  350π 3 rad s, so t 3 2 ( 350 ) s. Then P  K t , or P  75.6 J ( 2(350) s)  1.76  10 4 W. 3 e) If the frequency doubles, the acceleration and hence the needed force will quadruple (12.1  10 3 N). The maximum speed increases by a factor of 2 since v α ω, so the speed will be 36.7 m/s. Because the kinetic energy depends on the square of the velocity, the kinetic energy will increase by a factor of four (302 J). But, because the time to reach the midpoint is halved, due to the doubled velocity, the power increases by a factor of eight (141 kW). 13.58: Denote the mass of the passengers by m and the (unknown) mass of the car by M. The spring cosntant is then k  mg l . The period of oscillation of the empty car is TE  2π M k and the period of the loaded car is M m 2 l TL  2π  TE2  2π  , so k g l TE  TL2  2π  2  1.003 s. g 13.59: a) For SHM, the period, frequency and angular frequency are independent of amplitude, and are not changed. b) From Eq. (13.31), the energy is decreased by a factor of 1 . c) From Eq. (13.23), the maximum speed is decreased by a factor of 1 d) Initially, 4 2 15 the speed at A1 4 was 4 ωA1; after the amplitude is reduced, the speed is ω  A1 22   A1 42  4 3 ωA1 , so the speed is decreased by a factor of 1 5 (this result is valid at x   A1 4 as well). e) The potential energy depends on position and is unchanged. From the result of part (d), the kinetic energy is decreased by a factor of 1 5 . 13.60: This distance L is L  mg k ; the period of the oscillatory motion is m L T  2π  2 , k g which is the period of oscillation of a simple pendulum of lentgh L.
13. 13.61: a) Rewriting Eq. (13.22) in terms of the period and solving, 2π A2  x 2 T  1.68 s. v b) Using the result of part (a), 2  vT  x A 2     0.0904 m.  2  c) If the block is just on the verge of slipping, the friction force is its maximum, f  μs n  μs mg. Setting this equal to ma  mA2π T  gives μs  A2π T  g  0.143. 2 2 13.62: a) The normal force on the cowboy must always be upward if he is not holding on. He leaves the saddle when the normal force goes to zero (that is, when he is no longer in contact with the saddle, and the contact force vanishes). At this point the cowboy is in free fall, and so his acceleration is  g ; this must have been the acceleration just before he left contact with the saddle, and so this is also the saddle’s acceleration. b) x   a (2π f ) 2   (9.80 m s 2 ) 2π (1.50 Hz)) 2  0.110 m. c) The cowboy’s speed will be the saddle’s speed, v  (2πf ) A2  x 2  2.11 m s. d) Taking t  0 at the time when the cowboy leaves, the position of the saddle as a function of time is given by Eq. g (13.13), with cos    2 ; this is checked by setting t  0 and finding that ω A x  ω 2   ω 2 . The cowboy’s position is xc  x0  v0t  ( g 2)t 2 . Finding the time at which g a the cowboy and the saddle are again in contact involves a transcendental equation which must be solved numerically; specifically, (0.110 m)  (2.11m s)t  (4.90 m s 2 )t 2  (0.25 m) cos ((9.42 rad s)t  1.11 rad), which has as its least non-zero solution t  0.538 s. e) The speed of the saddle is (2.36 m s) sin (ω t  )  1.72 m s , and the cowboy’s speed is (2.11 m s)  (9.80 m s 2 )  (0.538 s)  3.16 m s, giving a relative speed of 4.87 m s (extra figures were kept in the intermediate calculations). 13.63: The maximum acceleration of both blocks, assuming that the top block does not slip, is amax  kA (m  M ), and so the maximum force on the top block is  m  M  kA  μs mg , and so the maximum amplitude is Amax  μs (m  M ) g k . m
14. 13.64: (a) Momentum conservation during the collision: mv0  (2m)V 1 1 V  v0  (2.00 m s)  1.00 m s 2 2 Energy conservation after the collision: 1 1 MV 2  kx 2 2 2 MV 2 (20.0 kg)(1.00 m s) 2 x   0.500 m (amplitude) k 80.0 N m ω  2πf  k M 1 1 80.0 N m f  k M   0.318 Hz 2π 2π 20.0 kg 1 1 T   3.14 s f 0.318 Hz (b) It takes 1 2 period to first return: 1 (3.14 s)  1.57 s 2
15. 13.65: a) m  m 2 Splits at x  0 where energy is all kinetic energy, E  1 mv 2 , so E  E 2 2 k stays same E  1 kA2 so A  2 E k 2 Then E  E 2 means A  A 2 T  2π m k so m  m 2 means T  T 2 b) m  m 2 Splits at x  A where all the energy is potential energy in the spring, so E doesn’t change. E  1 kA2 so A stays the same. 2 T  2π m k so T  T 2, as in part (a). c) In example 13.5, the mass increased. This means that T increases rather than decreases. When the mass is added at x  0, the energy and amplitude change. When the mass is added at x   A, the energy and amplitude remain the same. This is the same as in this problem. 13.66: a) For space considerations, this figure is not precisely to the scale suggested in the problem. The following answers are found algebraically, to be used as a check on the graphical method. 2E 2(0.200 J) b) A   0.200 m. k (10.0 N/m) c) E 4  0.050 J. d) If U  1 E , x  2 A 2  0.141 m. e) From Eq. (13.18), using 2 K0 2U 0 v0   m and x0  k , 2K0 v K0  0  m   0.429 ω x0 k 2U 0 U0 m k and   arctan   0.429  0.580 rad .
16. 13.67: a) The quantity l is the amount that the origin of coordinates has been moved from the unstretched length of the spring, so the spring is stretched a distance l  x (see Fig. (13.16 ( c ))) and the elastic potential energy is U el  (1 2)k (l  x) 2 . 1 2 1 U  U el  mg  x  x0   kx   l   k lx  mgx  mgx0 . 2 b) 2 2 Since l  mg k , the two terms proportional to x cancel, and 1 2 1 kx  k l   mgx0 . 2 U 2 2 c) An additive constant to the mechanical energy does not change the dependence of the force on x, Fx   dU , and so the relations expressing Newton’s laws and the dx resulting equations of motion are unchanged. 13.68: The “spring constant” for this wire is k  mg l , so 1 k 1 g 1 9.80 m s 2 f     11.1 Hz. 2π m 2π l 2π 2.00  10 3 m a) 2T  0.150 m s. b) a  2π T  x  0.112 m s 2 . The time to go from 2 13.69: πA equilibrium to half the amplitude is sin ωt  1 2, or ωt  π 6 rad, or one-twelfth of a period. The needed time is twice this, or one-sixth of a period, 0.70 s. d) l  mg  ωg2  2πgr 2  4.38 m. k
17. 13.70: Expressing Eq. (13.13) in terms of the frequency, and with   0, and taking two derivatives,  2πt  x  0.240 m  cos   1.50 s      2π 0.240 m    2πt   2πt   1.50 s   sin  1.50 s   1.00530 m s  sin  1.50 s  v x              2  2π   2πt   2πt         1.50 s  0.240 m  cos 1.50 s    4.2110 m s cos 1.50 s . a x   2        a) Substitution gives x  0.120 m, or using t  T gives x  A cos 120  3 A 2 . b) Substitution gives ma x   0.0200 kg 2.106 m s 2   4.21  10 2 N, in the  x - direction.   c) t  2Tπ arccos 3A 4  0.577 s. A d) Using the time found in part (c) , v  0.665 m s (Eq.(13.22) of course gives the same result). 13.71: a) For the totally inelastic collision, the final speed v in terms of the initial speed V  2 gh is v V M mM    2 9.80 m s 2 0.40 m  2..2   2.57 m s, or 2.6 m s to two figures. b) When 24 Mg the steak hits, the pan is k above the new equilibrium position. The ratio is v 2 k m  M   2 ghM 2 k m  M , and so the amplitude of oscillation is 2 v0 2 ω 2  Mg  2 ghM 2 A     k  k m  M      2.2 kg  9.80 m/s 2   2 2(9.80 m s 2 )(0.40 m)(2.2 kg) 2  400 N m    (400 N m)(2.4 kg)    0.206 m. (This avoids the intermediate calculation of the speed.) c) Using the total mass, T  2π (m  M ) k  0.487 s.
18. 13.72: f  0.600 Hz, m  400 kg; f  1 m gives k  5685 N/m. 2 k This is the effective force constant of the two springs. a) After the gravel sack falls off, the remaining mass attached to the springs is 225 kg. The force constant of the springs is unaffected, so f  0.800 Hz. To find the new amplitude use energy considerations to find the distance downward that the beam travels after the gravel falls off. Before the sack falls off, the amount x0 that the spring is stretched at equilibrium is given by mg  kx0 , so x0  mg k  400 kg  9.80 m/s 2  5685 N/m   0.6895 m. The maximum upward displacement of the beam is A  0.400 m. above this point, so at this point the spring is stretched 0.2895 m. With the new mass, the mass 225 kg of the beam alone, at equilibrium the spring is stretched mg k  (225 kg) (9.80 m s 2 ) (5685 N m)  0.6895 m. The new amplitude is therefore 0.3879 m  0.2895 m  0.098 m. The beam moves 0.098 m above and below the new equilibrium position. Energy calculations show that v  0 when the beam is 0.098 m above and below the equilibrium point. b) The remaining mass and the spring constant is the same in part (a), so the new frequency is again 0.800 Hz. The sack falls off when the spring is stretched 0.6895 m. And the speed of the beam at this point is v  A k m  0.400 m   5685 N/m  400 kg   1.508 m/s. . Take y  0 at this point. The total energy of the beam at this point, just after the sack falls off, is 2   E  K  U el  U g  1 225 kg  1.508 m/s2  1 5695 N/m 0.6895 m   0  1608 J. Let this 2 2 be point 1. Let point 2 be where the beam has moved upward a distance d and where v  0 . E2  1 k 0.6985 m  d   mgd . E1  E2 gives d  0.7275 m . At this end point of 2 2 motion the spring is compressed 0.7275 m – 0.6895 m =0.0380 m. At the new equilibrium position the spring is stretched 0.3879 m, so the new amplitude is 0.3789 m + 0.0380 m = 0.426 m. Energy calculations show that v is also zero when the beam is 0.426 m below the equilibrium position. 13.73: The pendulum swings through 1 cycle in 1.42 s, so T  2.84 s. L  1.85 m. 2 Use T to find g: T  2π L g so g  L2π T   9.055 m/s 2 2 Use g to find the mass M p of Newtonia: g  GM p / Rp 2 2πRp  5.14  107 m, so Rp  8.18  106 m 2 gRp mp   9.08  1024 kg G
19. 13.74: a) Solving Eq. (13.12) for m , and using k  F l 2 2  T  F  1  40.0 N m     4.05 kg.  2  l  2  0.250 m b) t  (0.35)T , and so x   Asin2π (0.35)  0.0405 m. Since t  T , the mass has 4 already passed the lowest point of its motion, and is on the way up. c) Taking upward forces to be positive, Fspring  mg   kx, where x is the displacement from equilibrium , so Fspring  (160 N m)(0.030 m)  (4.05 kg)(9.80 m/s 2 )  44.5 N. 13.75: Of the many ways to find the time interval, a convenient method is to take   0 in Eq. (13.13) and find that for x  A 2, cos ω t  cos(2πt / T )  2 and so t  T / 6 . 1 The time interval available is from  t to t , and T / 3  1.17 s. 13.76: See Problem 12.84; using x as the variable instead of r , dU GM E m GM E g F ( x)    3 x, so ω2  3  . dx RE RE RE The period is then 2π R 6.38  106 m T  2π E  2π  5070 s, ω g 9.80 m/s 2 or 84.5 min.
20. 13.77: Take only the positive root (to get the least time), so that dx k  A2  x 2 , or dt m dx k  dt A x 2 2 m A dx k t1 k  0 A2  x 2  m 0 dt  m (t1 ) k arcsin(1)  t1 m π k  t1 , 2 m where the integral was taken from Appendix C. The above may be rearranged to show that t1   k  T , which is expected. 4 2 m