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# f5ch16 - 442 CHAPTER 16 OSCILLATIONS CHAPTER 16 Answer to...

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Unformatted text preview: 442 CHAPTER 16 OSCILLATIONS CHAPTER 16 Answer to Checkpoint Questions 1. 2. 3. 4. 5. (a) xm ; (b) +xm ; (c) 0 a (a) 5 J; (b) 2 J; (c) 5 J all tie (in Eq. 16-32, m is included in I ) 1, 2, 3 (the ratio m=b matters; k does not) Answer to Questions 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. c (a) 2; (b) positive; (c) between 0 and +xm (a) 0; (b) between 0 and +xm ; (c) between xm and 0; (d) between xm and 0 a and b (a) toward xm ; (b) toward +xm ; (c) between xm and 0; (d) between xm and 0; (e) decreasing; (f) increasing (a) rad; (b) rad; (c) =2 rad (a) 3, 2, 1; (b) all tie all tie 3, 2, 1 half spring system with spring A (a) same; (b) same; (c) increase; (d) increase; (e) increase b (in nite period; does not oscillate), c, a (a) same; (b) same; (c) smaller; (d) smaller; (e) and (f) larger (T = 1) (a) same; (b) same; (c) same; (d) smaller; (e) smaller; (f) and (g) larger (T = 1) CHAPTER 16 OSCILLATIONS 443 16. one system: k = 1500 N/m, m = 500 kg other system: k = 1200 N/m, m = 400 kg the same ratio k=m = 3 gives resonance for both systems Solutions to Exercises & Problems (a) The period is T = 2(0:25 s) = 0:50 s: (b) The frequency is f = 1=T = 1=0:50 s = 2:0 Hz: (c) The amplitude is A = 36 cm=2 = 18 cm. (a) By de nition, T = 0:75 s. (b) f = 1=T = 1=0:75 s = 1:3 Hz: (c) ! = 2f = 2(1:3 Hz) = 8:4 rad/s. (a) The spring constant is 2 k = F = mg = (4:00 kg)(9:80 m/s ) = 245 N/m : x x 0:160 m 1E 2E 3E (b) The period is 0 0:500 kg T = 2 m = 2 245 N/m = 0:284 s : k r s (a) The motion repeats every 0:500 s so the period must be T = 0:500 s. (b) The frequency is the reciprocal of the period: f = 1=T = 1=(0:500 s) = 2:00 Hz. (c) The angular frequency ! is ! = 2f = 2(2:00 Hz) = 12:6 rad/s. p (d) The angular frequency is related to the spring constant k and the mass m by ! = k=m. Solve for k: k = m!2 = (0:500 kg)(12:6 rad/s)2 = 79:0 N/m. (e) Let xm be the amplitude. The maximum speed is vm = !xm = (12:57 rad/s)(0:350 m) = 4:40 m/s. 4E 444 CHAPTER 16 OSCILLATIONS (f) The maximum force is exerted when the displacement is a maximum and its magnitude is given by Fm = kxm = (79:0 N/m)(0:350 m) = 27:6 N. The eective spring constant ke satis es Solve for ke : e f = 21 km : r 5E ke 6E = 42 fm = 42 (1013 Hz) 0:108 kg = 708 N/m : 6:02 1023 The maximum acceleration is amax = !2 A = 42 f 2 A = 42 (6:60 Hz)2 (2:20 10 2 m) = 37:8 m/s2 : 7E The magnitude of the maximum acceleration is given by am = !2 xm , where ! is the angular frequency and xm is the amplitude. The angular frequency for which the maximum p acceleration is g is given by ! = g=xm and the corresponding frequency is given by 9:8 m=s2 = 500 Hz : ! = 1 g = 1 f = 2 2 x m 2 1:0 10 6 m For frequencies greater than 500 Hz the acceleration exceeds g for some part of the motion. r s (a) The spring constant is k = F=k = 32:0 lb=4:00 in. = 8:00 lb/in. = 96:0 lb/ft: p (b) Use f = (1=2) k=m and solve for the weight W : 2 kg W = mg = 42 f 2 = (8:00 lb/in.)(32::2 ft/s )(12 in./ft) = 19:6 lb : 42 (2 00 Hz)2 8E (a) The spring constant is 9E k = mg = 020 N = 100 N/m : x :20 m CHAPTER 16 OSCILLATIONS 445 (b) The period of oscillation is 5:0 N T = 2 m = 2 (100 N/m)(9:80 m/s2 ) = 0:45 s : k 10E r s (a) The spring constant k satis es T = 2 m=k; which gives 2 2 :0 3 k = 4 2m = 4 (50(0:g)(10 2 kg/g) = 7:90 N/m : T 500 s) p (b) Use vm = !A = (2=T )A to solve for A: (c) f = 1=T = 1=0:500 s = 2:00 Hz. 11E m A = v2T = (15:0 cm/s)(0:500 s) = 1:19 cm : 2 (a) The angular frequency ! is given by ! = 2f = 2=T , where f is the frequency and T is the period. The relationship f = 1=T was used to obtain the last form. Thus ! = 2=(1:00 10 5 s) = 6:28 105 rad/s. (b) The maximum speed vm and maximum displacement xm are related by vm = !xm , so :00 10 m/s m xm = v! = 6128 105 rad/s = 1:59 10 3 m : : 3 (a) and (b) The spring constant k satis es T = 2 m=k; or 12E p k = m 2 T 2 = (0:12 kg) 0:2 s 20 2 = 1:2 102 N/m : The maximum force is then Fm = kA = (1:2 102 N/m)(0:085 m) = 10 N: (a) The amplitude is half the range of the displacement, or xm = 1:0 mm. (b) The maximum speed vm is related to the amplitude xm by vm = !xm , where ! is the angular frequency. Since ! = 2f , where f is the frequency, vm = 2fxm = 2(120 Hz)(1:0 10 3 m) = 0:75 m/s. 13E 446 CHAPTER 16 OSCILLATIONS (c) The maximum acceleration is am = !2 xm = (2f )2 xm = [2(120 Hz)]2 (1:0 10 3 m) = 570 m/s2 . (a) The angular frequency is ! = 2f = 2(440 Hz) = 2:8 103 rad/s: (b) The maximum speed is vm = !A = (2:8 103 rad/s)(0:75 10 3 m) = 2:1 m/s: (c) The maximum acceleration is am = !2 A = (2:76 103 rad/s)2 (0:75 10 3 m) = 5:7 103 m/s2 : (a) The ecetive spring constant for the whole car of mass m is Ke = 4k, where k is the p spring constant of each spring. Solve f = (2) 1 ke =m for k: k = 1 ke = 2 f 2 m = 2 (3:00 Hz)2 (1450 kg) = 1:29 105 N/m : 4 (b) The vibraton frequency will be 5 4(1 f = 21 1450:29 10 N/m) = 2:68 Hz : kg + 5(73:0 kg) s 14E 15E (a) The displacement at t = 2:0 s is 16E xjt=2:0 s = (6:0 m) cos (3 rad/s)(2:0 s) + rad = 3:0 m : 3 (b) The velocity at t = 2:0 s is vjt=2:0 s = (6:0 m)(3 rad/s) sin (3 rad/s)(2:0 s) + rad = 49 m/s : 3 (c) The acceleration at t = 2:0 s is ajt=2:0 s = !2 xjt=2:0 s = (3 rad/s)2 (3:0 m) = 2:7 102 m/s2 : (d) The phase of the motion at t = 2:0 s is j = (3 rad/s)(2:0 s) + rad = 19 = 20 rad : t=2:0 s 3 3 (e) The frequency is rad/s f = 2! = 3 2 = 1:5 Hz : CHAPTER 16 OSCILLATIONS 447 (f) The period is T = 1=f = 1=1:5 Hz = 0:67 s: (a) The period is T = 1=f = 1=0:25 Hz = 4:0 s: (b) The angular frequency is ! = 2f = 2(0:25 Hz) = 1:57 rad/s: (c) The amplitude is A = 0:37 cm: (d) The displacement at time t is 17E x(t) = A cos(!t) = (0:37 cm) cos t : 2 (e) The velocity at time t is v(t) = dx(t) = !A sin !t dt = (1:57 rad/s)(0:37 cm) sin t = ( 0:58 cm/s) sin t : 2 2 (f) The maximum speed is vm = !A = (1:57 rad/s)(0:37 cm) = 0:58 cm/s: (g) The maximum acceleration is am = !2 A = (1:57 rad/s)2 (0:37 cm) = 0:91 cm/s2 : (h) The displacement at t = 3:0 s is xjt=3:0 s = (0:37 cm) cos (3:0) = 0 : 2 (i) The speed at t = 3:0 s is vjt=3:0 s = vm = 0:58 cm/s (since xjt=3:0 s = 0). 18E Use vm = !xm = 2fxm . The frequency is 180=(60 s) = 3:0 Hz and the amplitude is half the stroke, or 0:38 m. Thus vm = 2(3:0 Hz)(0:38 m) = 7:2 m/s. 19P p (a) From T = 2 (m + M )=R; we solve for M to get M = (k=42 )T 2 m: (b) Since M = 0, kT 2 = (605:6 N/m)(0:90149 s) = 12:47 kg : m = 42 42 (c) The mass of the astronaut is M = (k=42 )T 2 m = (605:6 N/m)(2:08832 s) 12:47 kg = 54:43 kg : 42 448 CHAPTER 16 OSCILLATIONS (a) The spring constant is 3 2 k = F = (300 210 kg)(92:80 m/s ) = 147 N/m : x :00 10 m 20P (b) The period of motion is 2:00 kg T = 2 m = 2 147 N/m = 0:733 s : k 21P r s Since T / M , we get Solve for m to obtain m = 1:6 kg: (a) The maximum acceleration is 22P p r m 2 0s = 3::0 s : m + 2:0 kg am = !2 A = (2f )2 A = 42 (1000 Hz)2 (0:40 10 3 m) = 1:6 104 m/s2 : (b) The maximum speed is vm = !A = 2fA = 2(1000 Hz)(0:40 10 3 m) = 2:5 m/s : (c) When the displacement of the prong is x = 0:20 mm, its acceleration is ajx=0:20 mm = !2 x = 42 (1000 Hz)(0:20 10 3 m) = 7:9 103 m/s : (d) The speed of the prong at this moment is r vjx=0:20 mm = vm x2 1 A2 = (2:5 m/s) 1 s 0:20 mm = 2:2 m/s : 0:40 mm (a)f = (10 rad/s)=(2) = 1:6 Hz. (b) vm = !A = (10 rad/s)(10 cm) = 100 cm/s: (c) am = !2 A = (10 rad/s)2 (10 cm) = 1:0 103 cm/s2 : It occurs at x = A = 10 cm: 23P CHAPTER 16 OSCILLATIONS 449 (d) The force applied as a function of time is given by F = ma = m!2 x = (0:10 kg)(10 rad/s)2 x = ( 10 N/m)x : 24P The displacement x of the ocean surface from its equilibrium position can be written as x(t) = 1 d cos(2t=T ). As the wave falls a distance d=4 from its maximum height of d=2, 2 the displacement of the wave is x = d=2 d=4 = d=4. The corresponding time t satis es d=4 = (d=2) cos(2t=T ): Thus T t = 2 cos 25P 1 1 = 12:5 h cos 2 2 1 1 = 2:08 h : 2 Since the maximum horizontal force exerted on the block of mass m is fs;max = mg, the block is only capable of reaching a maximum acceleration am = fs;max =m = g. Since m is in SHM, its actual maximum acceleration is a0m = !2 A = kA=(m + M ). Thus the condition for m and M to be in contact is or a0m = m kAM am = g ; + 2 )(1:0 A g(mk+ M ) = (0:40)(9:80 m/s N/m kg + 10 kg) = 0:22 m : 200 The maximum force that can be exerted by the surface must be less than s N or else the block will not follow the surface in its motion. Here s is the coecient of static friction and N is the normal force exerted by the surface on the block. Since the block does not accelerate vertically you know that N = mg, where m is the mass of the block. If the block follows the table and moves in simple harmonic motion the magnitude of the maximum force exerted on it is given by F = mam = m!2 xm = m(2f )2 xm , where am is the magnitude of the maximum acceleration, ! is the angular frequency, and f is the frequency. The relationship ! = 2f was used to obtain the last form. Substitute F = m(2f )2 xm and N = mg into F < s N to obtain m(2f )2 xm < s mg, or 26P s g = (0:50)(9:8 m/s2 ) = 0:031 m : xm < (2f )2 (2 2:0 Hz)2 A larger amplitude requires a larger force at the end points of the motion. The surface cannot supply the larger force and the block slips. 450 CHAPTER 16 OSCILLATIONS (a) For SHM of amplitude A, am = !2 A = 42 A=T 2 . But the downward acceleration cannot exceed g if the block remains in contact with the piston. Thus 2 a = 4 A g ; m 27P or T2 2 2 2 A gT 2 = (9:80 m/s 2)(1:0 s) = 0:25 m : 4 4 2 f 2 A g . Thus (b) Use am = 4 r r 1 g = 1 9:80 m/s2 = 2:2 Hz : f 2 A 2 0:050 m (a) Use a = !2 x = 42 f 2 x to nd f : 2 f = 21 xa = 21 123 m/s = 5:58 Hz : 0:100 m p (b) The mass m of the block satis es f = (2) 1 k=m, or N/m m = 4kf 2 = 4400:58 Hz)2 = 0:325 kg : 2 2 (5 (c) The amplitude A of the oscillation satis es 1 kA2 = 1 kx2 + 1 mv2 , so 2 2 2 2 13: A = x2 + k = (0:100 m)2 + (0:325 kg)( N/m6 m/s) = 0:400 : 400 r r r 28P mv2 s (a) Use E = 1 kA2 = 1 kx2 + 1 mv2 to solve for A: 2 2 2 29P :415 A = x2 + k = (0:129 m)2 + (2:00 kg)(3N/m m/s) = 0:500 m : 100 (b) Write x(t) = A sin(!t + ) and v(t) = !A cos(!t + ). At t = t0 = 1:0 s we have !x(t0 )=v(t0 ) = tan(!t0 + ), which gives = tan 1 [!x(t0 )=v(t0 )] !t0 s "s # 100 N/m 0:129 m 100 N/m (1:0 s) = tan 1 2:00 kg 3:415 m/s 2:00 kg = 6:81 : 2 r mv2 s CHAPTER 16 OSCILLATIONS 451 Thus and xjt=0 = A sin = (0:500 m) sin( 6:81) = 0:251 m/s vjt=0 = !A cos = 100 N/m (0:500 m) cos( 6:81) = 3:06 m/s : 2:00 kg s (a) The object oscillates about its equilibrium point, where the downward force of gravity is balanced by the upward force of the spring. If ` is the elongation of the spring at equilibrium, then k` = mg, where k is the spring constant and m is the mass of the object. p p Thus k=m = g=` and f = !=2 = (1=2) k=m = (1=2) g=`. Now the equilibrium point is halfway between the points where the object is momentarily at rest. One of these points is where the spring is unstretched and the other is the lowest point, 10 cm below. Thus ` = 5:0 cm = 0:050 m and 30P : m/s2 f = 21 90850 m = 2:2 Hz : : (b) Use conservation of energy. Take the zero of gravitational potential energy to be at the initial position of the object, where the spring is unstretched. Then both the initial potential and kinetic energies are zero. Take the y axis to be positive in the downward direction and let y = 0:080 m. The potential energy when the object is at this point is U = 1 ky2 mgy. The energy equation becomes 0 = 1 ky2 mgy + 1 mv2 . Solve for v: 2 2 2 s k v = 2gy m y2 = 2gy g y2 ` v ! u u 9:8 m/s2 (0:080 m)2 = 0:56 m/s : 2 )(0:080 m) = t2(9:8 m/s 0:050 m (c) Let m be the original mass and m be the additional mass. The new angular frequency p p is !0 = k=(m + m). This should be half the original angular frequency, or 1 k=m. 2 p p Solve k=(m + m) = 1 k=m for m. Square both sides of the equation, then take the 2 reciprocal to obtain m + m = 4m. This gives m = m=3 = (300 g)=3 = 100 g. (d) The equilibrium position is determined by the balancing of the gravitational and spring forces: ky = (m + m)g. Thus y = (m + m)g=k. You will need to nd the value of the spring constant k. Use k = m!2 = m(2f )2 . Then 30 kg)(9 8 m/s2 y = (m + fm2)g = (0:10 kg + 0:kg)(2:24:Hz)2 ) = 0:20 m : m(2 ) 42 (0:10 This is measured from the initial position. r r 452 CHAPTER 16 OSCILLATIONS 31P (a) Let be the coordinate as a function of time for particle 1 and t x1 = A cos 2T 2 t x2 = A cos 2T + 2 6 be the coordinate as a function of time for particle 2. Here T is the period. Note that since the range of the motion is A the amplitudes are both A=2. The arguments of the cosine functions are in radians. Particle 1 is at one end of its path (x1 = A=2) when t = 0. Particle 2 is at A=2 when 2t=T + =6 = 0 or t = T=12. That is, particle 1 lags particle 2 by one-twelfth a period. We want the coordinates of the particles 0:50 s later; that is, at t = 0:50 s: 0 x1 = A cos 2 :5 :s50 s = 0:250A 2 1 and 0 x2 = A cos 2 :5 :s50 s + = 0:433A : 2 1 6 Their separation at that time is x1 x2 = 0:250A + 0:433A = 0:183A. (b) The velocities of the particles are given by t v1 = dx1 = A sin 2T dt T t v2 = dx2 = A sin 2T + : dt T 6 Evaluate these expressions for t = 0:50 s. You will nd they are both negative, indicating and that the particles are moving in the same direction. 32P Let the displacements of the two particles be x1 = A sin(!t) and x2 = A sin(!t + ): Then their velocities are v1 = !A cos(!t) and v2 = !A cos(!t + ): As they pass each other at t = t0 , we have x1 (t0 ) = A=2 so x1 (t0 ) = A sin(!t0 ) = A=2, which gives !t0 = 30 . (Here we take only the smallest value of t0 which satisfy this equation, but this will not aect our result.) Also for x2 (t0 ) we have x2 (t0 ) = A sin(!t0 + ) = A sin(30 + ) = A : 2 CHAPTER 16 OSCILLATIONS 453 The solutions to this equation within (0; 2) are = 0 or = 120 = 2=3. Since v1 and v2 have opposite signs at t0 , we need to choose = 2=3. Consider a displacement x (say, to the right) of the mass m. The force f1 exerted by the left spring on the mass is f1 = kx, while the force f2 exerted by the right one is also f2 = kx. Thus the net force on m is fnet = f1 + f2 = 2kx = ke x: So e f = 21 km = 21 2k : m r r 33P where we have used f1;2 = (1=2) k1;2 =m: 35P The eective spring constant of the system is ke = k1 + k2 . Thus the corresponding frequency is r r 1 ke = 1 k1 + k2 = qf 2 + f 2 ; f = 2 m 2 m m 1 2 p 34P We wish to nd the eective spring constant for the combination of springs shown in Fig. 28. We do this by nding the magnitude F of the force exerted on the mass when the total elongation of the springs is x. Then k e = F=x. Suppose the left-hand spring is elongated by x` and the right-hand spring is elongated by xr . The left-hand spring exerts a force of magnitude k x` on the right-hand spring and the right-hand spring exerts a force of magnitude k xr on the left-hand spring. By Newton's third law these must be equal, so x` = xr . The two elongations must be the same and the total elongation is twice the elongation of either spring: x = 2x` . The left-hand spring exerts a force on the mass and its magnitude is F = k x` . Thus k e = k x` =2xr = k=2. The mass behaves as if it were subject to the force of a single spring, with spring constant k=2. To nd the frequency of its motion replace k e in p f = (1=2) k e =m with k=2 to obtain f = 21 2k : m (a) As it reaches its equilibrium position, the displacement x0 of the block of mass m from the position when the spring is unstretched is given by mg sin = kx0 , or sin N)(sin 40 x0 = mg k = (14:0120 N/m :0 ) = 7:50 10 2 m : 36P r 454 CHAPTER 16 OSCILLATIONS The position of the block when it stops is therefore a distance 0:450+0:075 = 0:525 m from the top of the incline. (b) The net restoring force F as a function of the block's displacement x from its equilibrium position is F = kx. The period of the oscillation is then 14 0 N T = 2 m = 2 (9:80 m/s2:)(120 N/m) = 0:686 s : k 37P r s (a) First consider a single spring with spring constant k and unstretched length L. One end is attached to a wall and the other is attached to an object. If it is elongated by x the magnitude of the force it exerts on the object is F = k x. Now consider it to be two springs, with spring constants k1 and k2 , arranged so spring 1 is attached to the object. If spring 1 is elongated by x1 then the magnitude of the force exerted on the object is F = k1 x1 . This must be the same as force of the single spring, so k x = k1 x1 . We must determine the relationship between x and x1 . The springs are uniform so equal unstretched lengths are elongated by the same amount and the elongation of any portion of the spring is proportional to its unstretched length. This means spring 1 is elongated by x1 = CL1 and spring 2 is elongated by x2 = CL2 , where C is a constant of proportionality. The total elongation is x = x1 + x2 = C (L1 + L2 ) = CL2 (n + 1), where L1 = nL2 was used to obtain the last form. Since L2 = L1 =n, this can also be written x = CL1 (n + 1)=n. Substitute x1 = CL1 and x = CL1 (n + 1)=n into k x = k1 x1 and solve for k1 . The result is k1 = k(n + 1)=n. Now suppose the object is placed at the other end of the composite spring, so spring 2 exerts a fore on it. Now k x = k2 x2 . Substitute x2 = CL2 and x = CL2 (n + 1), then solve for k2 . The result is k2 = k(n + 1). p (b) To nd the frequency when spring 1 is attached to mass m, replace k in (1=2) k=m with k(n + 1)=n to obtain + 1) f1 = 21 (n nm k = n + 1 f ; n p where the substitution f = (1=2) k=m was made. To nd the frequency when spring 2 is attached to the mass, replace k with k(n + 1) to obtain r r p f2 = 21 (n + 1)k = n + 1f ; m where the same substitution was made. r (a) Let the mass of each of the cars be m. The restoring force of the cable before it breaks is given by f = 3mg sin 30 = 3(10; 000 kg)(9:80 m/s2 )(sin 30 ) = 1:47 105 N: The eective 38P CHAPTER 16 OSCILLATIONS 455 spring constant of the cable is ke = 1:47 105 N=0:15 m = 9:8 105 N/m: The frequency of oscillation is then s r 1 ke = 1 9:8 105 N/m = 1:1 Hz : f = 2 2m 2 2(10; 000 kg) (b) The dierence in the equilibrium length of the cable with two and three cars attached is x = F = mg sin 30 which is also the amplitude of the oscillation. 39P ke ke 2 = (10; 000 kg)(9:805m/s )(sin 30 ) = 0:050 m = 5:0 cm ; 9:8 10 N/m (a) Measure the position of the train with the variable x shown on the diagram to the right. If x the train is to the left of the center of its path x it negative; if the train is to the right x is positive. When the train is at the point shown the gravitational force due to all of the Earth's mass outside R a sphere of radius r is 0. The mass within the sphere is given by M = Me (r=R)3 , where Me is the total mass of the Earth and R is the radius of the Earth. The magnitude of the gravitation force on the train is F = GMm=r2 = GMe mr=R3 , where m is the mass of the train. We are interested in the component of the gravitational force along the train's path. This is the component that produces the acceleration of the train. The component perpendicular to the path is balanced by the normal force of the track on the train. The component along the path is Fx = F sin = (GMe m=R3 )r sin . now x = r sin , so Fx = (GMe m=R3 )x. Note that this force is toward the center of the train's path. That is, it is to the right (positive) if x is negative and to the left (negative) if x is positive. It is a restoring force that is proportional to the displacement. We conclude that the train is in simple harmonic motion. Since the train is at rest at the beginning and end of the path, a one-way trip takes half a period. (b) Newton's second law yields or, when m is canceled, GMe m x = m dx2 R3 dt2 GMe x = dx2 : R3 dt2 456 CHAPTER 16 OSCILLATIONS By comparison with the analogous equation for a spring-mass system, !2 x = dx2 =dt2 , p we can identify the angular frequency: ! = GMe =R3 . The period is 6 m)3 R3 (6:37 T = 2! = 2 GM = 2 (6:67 10 11 m3 = 102 )(5:98 1024 kg) = 5:06 103 s : kg s e s s The time for a trip is T=2 = 2:53 103 s = 42 min. 40E When the block is at the end of its path and is momentarily at rest, its displacement is equal to the amplitude and all the energy is potential in nature. If the spring potential energy is taken to be zero when the block is at its equilibrium position, then 1 1 E = 2 kx2 = 2 (1:3 102 N/m)(0:024 m)2 = 3:7 10 2 J : m 41E 1 2 Use E = 1 mvm = 2 kA2 and vm = 2fA. 2 (a) k = 2E=A2 = 2(1:00 J)=(0:100 m)2 = 200 N/m: 2 (b) m = 2E=vm = 2(1:00 J)=(1:2 m/s)2 = 1:39 kg: (c) f = vm =2A = (1:20 m/s)=[2(0:100 m)] = 1:91 Hz. (a) The frequency is 42E k f = 21 m = 21 1000 N/m = 2:25 Hz : 5:00 kg (b) The initial potential energy is 1 1 Ui = 2 kx2 = 2 (1000 N/m)(0:500 m)2 = 125 J : i (c) The initial kinetic energy is 1 1 Ki = 2 mvi2 = 2 (5:00 kg)(10:0 m/s)2 = 250 J : (d) Use 1 kA2 = Ui + Ki to nd the amplitude of oscillation: 2 r s A= r 2(Ui + Ki ) = s k 2(125 J + 250 J) = 0:866 m = 86:6 cm : 1000 N/m CHAPTER 16 OSCILLATIONS 457 (a) The spring stretches until the magnitude of its upward force on the block equals the magnitude of the downward force of gravity: ky = mg, where y is the elongation of the spring at equilibrium, k is the spring constant, and m is the mass of the block. Thus k = mg=y = (1:3 kg)(9:8 m/s2 )=(0:096 m) = 130 N/m. p p (b) The period is given by T = 1=f = 2=! = 2 m=k = 2 (1:3 kg)=(133 N/m) = 0:62 s. (c) The frequency is f = 1=T = 1=0:62 s = 1:6 Hz. (d) The block oscillates in simple harmonic motion about the equilibrium point determined by the forces of the spring and gravity. It is started from rest 5:0 cm below the equilibrium point so the amplitude is 5:0 cm. (e) The block has maximum speed as it passes the equilibrium point. Then the spring is elongated by y = 9:6 cm and the spring potential energy is 1 ky2 = 1 (133 N/m)(0:096 m)2 = 2 2 0:612 J and the gravitational potential energy is mgy = (1:3 kg)(9:8 m/s2 )(0:096 m) = 1:22 J. The total potential energy is 0:612 J 1:22 J = 0:61 J. At the initial position the block is not moving but it has potential energy Ui = 0:44 J. Write the equation for conservation of energy as Ui = Uf + 1 mv2 and solve for v: 2 43E v= r 2(Ui Uf ) = m s 2( 0:44 J + 0:61 J) = 0:51 m/s : 1:3 kg (a) Use 1 kA2 = 1 mv2 to nd k: 2 2 44E v k=m A 2 103 = (0:130 kg) 11:21:50 m m/s 2 = 7:25 106 N/m : (b) The number of people required is 6 N/m)(1 kA n = 220 N = (7:25 10220 N :50 m) = 49; 400 : (a) and (b) The total energy is given by E = 1 kx2 , where k is the spring constant and xm 2 m is the amplitude. When x = 1 xm the potential energy is U = 1 kx2 = 1 kx2 . The ratio is 2 2 8 m 45E U = kx2 =8 = 1 : m E kx2 =2 4 m 458 CHAPTER 16 OSCILLATIONS K = E U =1 U =1 1 = 3: E E E 4 4 (c) Since E = 1 kx2 and U = 1 kx2 , U=E = x2 =x2 . Solve x2 =x2 = 1=2 for x. You should m m 2 p2 m get x = xm = 2. (a) Use mv = MV + m for the velocity V of the block: V = mv=(M + m): (b) Use 1 (M + m)V 2 = 1 kA2 to solve for A: 2 2 46E The fraction of the energy that is kinetic is : A = V M + m = p mv k k(M + m) (a) Let the value fo p be x0 when U (x) = 1 E . Thus U (x) = 1 kx2 = 1 E = 1 kA2 , or x 2 2 0 2 4 p x0 = A= 2 = 5:0 m= 2 = 3:5 m: (b) The phase dierence between the two positions (the equilibrium position and the position x = x0 ) is cos 1 3:5 m = 0:775 : = 2 5:0 m The time t it takes is then 0: 1 t = f = (2)(=775 =2) = 0:75 s : 2 3)(1 r 47P (a) Let the displacement of the particle be x(t) = xm cos(!t ), then the correpsonding F force F (t) is given by F (t) = ma(t) = m!2 x(t) = pm (cos !t ), where Fm = mam = 3 kg)(8:0 103 m/s2 ) = 80 N; ! = pam =A = (8:0 103 m/s2 )=(2:0 10 3 m) = (10 10 2:0 103 rad/s, and = =3 rad: Thus F (t) = (80 N) cos[(2:0 103 rad/s)t =3]: (b) T = 2=! = 2=(2:0 103 rad/s) = 3:1 10 3 s. (c) vm = !A = (2:0 103 rad/s)(2:0 10 3 m) = 4:0 m/s: 2 (d) E = 1 mvm = 1 (10 10 3 kg)(4:0 m/s)2 = 8:0 10 2 J: 2 2 (a) Suppose that the spring stretches by an amount ` before the body stops momentarily. According to the work-energy theorem, the total work done on the body is W = 1 k`2 + 2 mgl = K = 0, or 2 l = 2mg = 2(0:20 kg)(9:80 m/s ) = 0:21 m : k 19 N/m 49P 48P CHAPTER 16 OSCILLATIONS 459 (b) The frequency is k N/m f = 21 m = 21 1920 kg = 1:6 Hz : 0: r s (c) The amplitude A is equal to half of the distance between the two turning points of the oscillator, i.e., A = l=2 = 0:21 m=2 = 0:10 m. 50P (a) Assume that the bullet becomes imbedded and moves with the block before the block moves a signi cant distance. Then the momentum of the bullet-block system is conserved during the collision. Let m be the mass of the bullet, M be the mass of the block, v0 be the initial speed of the bullet, and v be the nal speed of the block and bullet. Conservation of momentum yields mv0 = (m + M )v, so v = mmv0 = (00::050 kg)(1500m/s) = 1:85 m/s : +M 050 kg + 4: kg After the collision the block with the bullet embedded oscillates with simple harmonic motion. The energy is all kinetic and is given by E = 1 (m + M )v2 . When the block is at 2 2 the end of its path the energy is all potential and is given by E = 1 kym , where k is the 2 2 spring constant and ym is the amplitude. Since energy is conserved, 1 (m + M )v2 = 1 kym 2 2 and s r (m + M )v2 = (0:050 kg + 4:0 kg)(1:85 m/s)2 = 0:167 m : ym = k 500 N/m 2 (b) The original energy of the bullet is E0 = 1 mv0 = 1 (0:050 kg)(150 m/s)2 = 563 J. The 2 2 energy after the collision is E = 1 (m + M )v2 = 1 (0:050 kg + 4:0 kg)(1:85 m/s)2 = 6:94 J. 2 2 The ratio is E=E0 = (6:94 J)=(563 J) = 0:0123 or 1:23%. (a) Use conservation of energy. Initially the energy is all potential in nature and is given by E = 1 kx2 , where k is the spring constant and xm is the initial elongation of the 2 m spring. When the disk is at the equilibrium point (and the kinetic energy is maximum) 2 2 the energy is all kinetic in nature and is given by E = 1 Mvm + 1 I!m , where M is the 2 2 mass of the disk, vm is the speed of its center of mass, I is its rotational inertia, and !m is its angular speed (about the center of mass). The rst term in the energy is the translational kinetic energy and the second is the rotational kinetic energy. Since the wheel rolls without slipping !m = vm =R, where R is its radius. Furthermore the rotational inertia of a disk is given by I = 1 MR2 . When these substitutions are made the expression for the 2 2 2 2 energy at the equilibrium point becomes E = 1 Mvm + 1 Mvm = 3 Mvm . Conservation of 2 4 4 2 2 energy yields 1 kx2 = 3 Mvm . Thus Mvm = 2 kx2 and the translational kinetic energy is 2 m 4 3 m 1 Mv 2 = 1 kx2 = 1 (3:0 N/m)(0:25 m)2 = 0:0625 J. m 3 m 3 2 51P 460 CHAPTER 16 OSCILLATIONS 2 (b) The rotational kinetic energy at the equilibrium point is 1 I!m . Replace I with 1 MR2 2 2 1 Mv 2 . Now replace Mv 2 with 2 kx2 to obtain 1 kx2 = and !m with vm =R to obtain 4 m m m 3 6 m 1 (3:0 N/m)(0:25 m/s)2 = 0:031 J. 6 (c) Suppose the center of mass executes simple harmonic motion with angular frequency ! Use 2 2 (not to be confused withp angular speed of rotation). Then vm = !xm .p 1 Mvm = the p 1 kx2 to nd that vm = 2k=3Mxm . Thus 2k=3Mxm = !xm and ! = 2k=3M . The 3 m period is r 2 = 2 3M : T= ! 2k (a) I = 1 mR2 = 1 (3:00 kg)(70:0 10 2 m)2 = 0:735 kg m2 : 2 2 (b) = = = (0:p N m)=2:50 rad = 0:0240 N m: 0600 p (c) ! = =I = (0:0240 N m)=(0:735 kg m2 ) = 0:181 rad/s: The torque applied to the wire is related to the turning angle by = , where = j=j = 0:20 Nm=0:85 rad = 0:235 Nm/rad. The period of oscillation of sphere of mass m and radius R is then 53P 52E I T = 2 = 2 54P r r 2mR2 2(95 = 2(0:15 m) 5(0:235 N kg) 5 m/rad) = 12 s : s Use T = 2 I= to solve for I : p I = T = (50 s=20)4(02:50 N m) = 7:9 10 2 kg m2 : 4 2 2 2 (a) Take the angular displacement of the wheel to be = m cos(2t=T ), where m is the amplitude and T is the period. Dierentiate with respect to time to nd the angular velocity: = (2=T )m sin(2t=T ). The symbol is used for the angular velocity of the wheel so it is not confused with the angular frequency. The maximum angular velocity is m = 2m = (2:)( rad) = 39:5 rad/s : T 0 500 s 55P CHAPTER 16 OSCILLATIONS 461 (b) When = =2, then =m = 1=2, cos(2t=T ) = 1=2, and p p sin(2t=T ) = 1 cos2 (2t=T ) = 1 (1=2)2 = 3=2 ; where the trigonometric identity cos2 A + sin2 A = 1 was used. Thus p t = 2 m sin 2T = T 2 ( rad) 0:500 s p 3 = 34:2 rad/s : 2 ! The minus sign is not signi cant. During another portion of the cycle its angular speed is +34:2 rad/s when its angular displacement is =2 rad. (c) The angular acceleration is 2 = d 2 = dt 2 2 cos(2t=T ) = T m 2 2 : T 2 = 0:500 s Again the minus sign is not signi cant. When = =4, 2 = 124 rad/s2 : 4 Use T = 2 L=g to solve for L: 2g 2 (32 2 L = T2 = (1:00 s) 42 :2 ft/s ) = 9:79 in: : 4 56E p (a) The period of the swing of length L is given by 17 m T = 2 L = 2 9:80 m/s2 = 8:3 s : g (b) No. T depends only on L and g. 58E s s 57E The period of a simple pendulum is given by T = 2 L=g, where L is its length. Thus L = T 2 g=42 = (2:0 s)2 (9:8 m/s2 )=(42 ) = 0:99 m. p 462 CHAPTER 16 OSCILLATIONS Use T = 2 L=g = 2 1:50 m=g = 180 s=72:0 and solve for g: 2 42 g = 4 2L = (180(1=:50:m)2 = 9:47 m/s2 : T s 72 0) 59E p p Since the spring constant k is given by k = mg=h, 60E m T spring = 2 m = 2 mg=h = 2 h = T pendulum : k g r r s Let the eective length of the simple pendulum be L1 when the performer is seated on the p trapeze. Then T1 = 8:85 s = 2 L1 =g; or 61E T1 2 85 2 L1 = 2 g = 8:2 s (9:80 m/s2 ) = 19:4 m : Thus as the performer stands up, the new period T2 is 0:350 T2 = 2 L2 = 2 19:49m m/s2 m = 8:77 s : g :80 s s Since T / L, the new period T 0 is 62E p T T 0 = T L=2 = p : L 2 63E 2 12 + 2 I T = 2 mgh = 2 mL =mgx mx s s r Use CHAPTER 16 OSCILLATIONS 463 to solve for x: 1 T 2 g 2 L2 x= 2 42 3 s (2:5 s)2 (9:80 m/s2 ) 1 (2:5 s)2 (9:80 m/s2 ) 2 = 82 2 42 = 0:056 m : T 2g 82 s (1:00 m)2 3 64E (a) The period of the pendulum is given by s according to the parallel-axis theorem, its rotational inertia when it is pivoted a distance d from the center is I = mL2 =12 + md2 . Thus 2 =12 2 2 + 12 2 T = 2 m(L mgd+ d ) = 2 L 12gd d : s s I T = 2 mgd ; where I is its rotational inertia, m is its mass, and d is the distance from the center of mass to the pivot point. The rotational inertia of a rod pivoted at its center is mL2 =12 and, so (b) (L2 + 12d2 )=12gd, considered as a function of d, has a minimum at d = L= 12, p the p period increases as d decreases if d < L= 12 and decreases as d decreases if d > L= 12. (c) L occurs only in the numerator of the expression for the period, so T increases as L increases. (d) The period does not depend on the mass of the pendulum, so T does not change when m increases. As the pendulum makes a small angular displacement as shown, the torque applied to it about its pivot is 65E p = mgx = mgd sin mgd = : Thus its period of oscillation is pivot d x mg I T = 2 = 2 r s 1 mR2 + md2 2 mgd 2 2 = 2 R 2+ 2d ; gd s 464 CHAPTER 16 OSCILLATIONS where we have made use of the parallel axis theorem to nd I . (a) The period is 2 2 I T = 2 mgh = 2 mR =2 + mR mgR s 2 2 = 2 (12:5 cm) =22+ (12:5 cm) = 0:869 s : (980 cm/s )(12:5 cm) s s 66E (b) Let 2 2 = 2 mR =2 + mR mgr mgR and solve for r: r = R=2 = 12:5 cm=2 = 6:25 cm: s T = 2 mR2 =2 + mr2 s 67E (a) A uniform disk pivoted at its center has a rotational inertia of 1 MR2 , where M is its 2 mass and R is its radius. The disk of this problem rotates about a point that is displaced from its center by R + L, where L is the length of the rod, so, according to the parallel-axis theorem, its rotational inertia is 1 MR2 + M (L + R)2 . The rod is pivoted at one end and has 2 a rotational inertia of mL2 =3, where m is its mass. The total rotational inertia of the disk and rod is I = 1 MR2 + M (L + R)2 + 1 mL2 = 1 (0:500 kg)(0:100 m)2 +(0:500 kg)(0:500 m+ 2 3 2 0:100 m)2 + 1 (0:270 kg)(0:500 m)2 = 0:205 kg m2 . 3 (b) Put the origin at the pivot. The center of mass of the disk is `d = L + R = 0:500 m + 0:100 m = 0:600 m away and the center of mass of the rod is `r = L=2 = (0:500 m)=2 = 0:250 m away, on the same line. The distance from the pivot point to the center of mass of the disk-rod system is + m` + (0:270 kg)(0 d = M`d + m r = (0:500 kg)(0::600 m) + 0:270 kg :250 m) = 0:477 m : M 0 500 kg (c) The period of oscillation is 0:205 kg m2 T = 2 (M + m)gd = 2 = 1:50 s : (0:500 kg + 0:270 kg)(9:8 m/s2 )(0:447 m) s I s (a) and (b) If the pendulum is suspended at point P then in Eq. 16-32 we must change h 1 to h0 = L0 h = 2 L 1 L = 1 L and I to I 0 = I0 + mh02 = 12 mL2 + m( 1 L)2 = 1 mL2 . 0 3 2 6 6 9 68E CHAPTER 16 OSCILLATIONS 465 Here we made use of the parallel axis theorem to calculate the moment of inertia I 0 about point P . Thus the new period of oscillation is I0 mL2 = T 0 = 2 mgh0 = 2 mg(L=9 = 2 2L : 6) 3g This is the same as the period we obtained before in Sample Problem 16-6. The numerical value is once again 1:64 s. Solve for L0 from to obtain L0 = I=mh. Let the frequency of oscillation of the meter stick of length L be f0 = c g=L, where c is a constant which you don't need to calculate for this problem. p L is reduced to L=2, then If p p p the new frequency f must be f = c g=(L=2) = 2c g=L = 2f0 : (a) Since I0 = mL2 =12 + mx2 the period of the pendulum is 71P 70E p s s s 69E I T = 2 mgh = 2 L0 g s s I0 T = 2 mgx = 2 (b) Set s s mL2 =12 + mx2 mgx 2 + 12 2 = 2 L 12gx x : s L2 + 1 d L2 + 12x2 = dx 12gx 12gx2 g x=x0 p p and solve for x: x0 = L= 12 = 1:00 m= 12 = 0:289 m. (c) s x=x0 =0 2+ 2 Tmin = T jx=x0 = 2 (1:00 :m) m/s12(0::289 m) = 1:53 s : 12(9 80 2 )(0 289 m) (a) The linear acceleration a due to the force F exerted at point O is given by F = ma, or a = F=m. 72P 466 CHAPTER 16 OSCILLATIONS (b) The angular acceleration about point C is =6) 2F = = F (L20=12 = mL : I mL0 0 (c) The actual linear acceleration at point O is given by F 2F a0 = a L0 = m mL L0 = 0 : 2 0 2 (d) Since a0 = 0 no force is felt at point O, so point P is indeed the \sweet spot." Denote Paris with subscript p and Cayenne with c. Then (1 da)(86400 s/da) Tc = r gp = Tp gc (1 da)(86400 s/da) (2:5 min)(60 s/min) = 1:0017 : Thus 2 gp gc = (1:0017)2 = 9:81 m/s2 = 9:78 m/s2 : (1:0017) The dierence in the measured values of the acceleration of gravity is due to the dierent apparent angular speed of the Earth's rotation as viewed from the submarine. Let the angular speed of the Earth viewed from a stationary observer on Earth be !e , then its apparent angular speed as viewed from the submarine is given by ! = !e Vs =Re , where Vs is the speed of the submarine and Re the radius of the Earth. Since the Earth rotates from west to east, the plus sign corresponds to the case when the submarine moves westward (i.e., in the opposite direction to the spin of the Earth). If we denote the acceleration of gravity on the eauator to be ag if the Earth were to stop rotating, then the measured 2 acceleration of gravity satis es g = ag ! Re , i.e., 2 jg j = jg+ g j = j(ag !+ R) (ag !2 Re )j = 4!e Vs : 73P 74P Thus The fractional error g=g for either travel direction is therefore g=g = 1 g =g = 1:3 2 10 4 =2 = 6:6 10 5 . 75P g = 4!e Vs = 4(2=86; 400 s)(16 103 m=3600 s) = 1:3 10 4 : g g 9:8 m/s2 If the torque exerted by the spring on the rod is proportional to the angle of rotation of the rod and if the torque tends to pull the rod toward its equilibrium orientation, then the rod CHAPTER 16 OSCILLATIONS 467 will oscillate in simple harmonic motion. If = C, where is the torque, is the angle of rotation, and C is constant of proportionality, then the angular frequency of oscillation p p is ! = C=I and the period is T = 2=! = 2 I=C , where I is the rotational inertia of the rod. The plan is to nd the torque as a function of and identify the constant C in terms of given quantities. This immediately gives the period in terms of given quantities. Let `0 be the distance from the pivot point to the wall. This is also the equilibrium length of the spring. Suppose the rod turns through the angle , with the left end moving away from the wall. This end is now (L=2) sin further from the wall and has moved (L=2)(1 cos ) p to the right. The length of the spring is now (L=2)2 (1 cos )2 + [`0 + (L=2) sin ]2 . If the angle is small we may approximate cos with 1 and sin with in radians. Then the length of the spring is given by `0 + L=2 and its elongation is x = L=2. The force it exerts on the rod has magnitude F = k x = kL=2. Since is small we may approximate the torque exerted by the spring on the rod by = FL=2, where the pivot point was taken as the origin. Thus = (kL2 =4). The constant of proportionality C that relates the torque and angle of rotation is C = kL2 =4. The rotational inertia for a rod pivoted at it center is I = mL2 =12, where m is its mass. Thus the period of oscillation is I 4mL2 m T = 2 C = 2 12kL2 = 2 3k : r r r (a) The frequency for small amplitude oscillations is f = (1=2) g=L, where L is the q length of the pendulum. This gives f = (1=2) (9:80 m/s2 )=(2:0 m) = 0:35 Hz. (b) The forces acing on the pendulum are the tension T in the rod and the force of gravity mg. Newton's second law yields T+mg = ma, where m is the mass and a is the acceleration of the pendulum. Let a = ae + a0 , where ae is the acceleration of the elevator and a0 is the acceleration of the pendulum relative to the elevator. Newton's second law can then be written m(g ae ) + T = ma0 . Relative to the elevator the motion is exactly the same as it would be in an inertial frame where the acceleration due to gravity is g ae . Since g and ae are along the same line and in opposite directions we can nd the frequency for p small amplitude oscillations by replacing g with g + ae in the expression f = (1=2) g=L. Thus s r 1 g + ae = 1 9:8 m/s2 + 2:0 m/s2 = 0:39 Hz : f = 2 L 2 2:0 m (c) Now the acceleration due to gravity and the acceleration of the elevator are in the same direction and have the same magnitude. That is g ae = 0. p nd the frequency for To small amplitude oscillations, replace g with zero in f = (1=2) g=L. The result is zero. The pendulum does not oscillate. 76P p 468 CHAPTER 16 OSCILLATIONS Consider the motion of the pendulum in the frame of reference which is xed on the moving car. A ctitious centrifugal force (pseudoforce) Fc = mv2 =R must be intruduced to account for the fact that the frame of reference we are using is non-inertial in nature. Equivalently, as depicted in the gure to the right, we can regard this situation as if the pendulum were subject to an eective gravitational eld ge which satis es 77P mge = mg + Fc ; where 2 ge = g2 + v R s 2 : T Fc (=mv2 /R ) Thus the frequency of oscillation is L f = 21 g0 = 21 s s p L : g2 + (v2 =R)2 mg' mg The approximation we made in deriving T = 2 L=g was to replace sin by for small jj. Thus the angular amplitude m must satisfy 78P p sin m m = 0:01 : sin m 3 Substitute sin m ' m m =3 into the equation above and solve for m . You should get m ' 14:0 : (a) The loss of gravitational potential energy of the bob in the Earth's gravitational eld as it swings down is mgh = mgR(1 cos m ). Thus its speed v at the bot1 tom of the swing satis es 2 mv2 = mgR(1 cos m ), 2 or mv2 = 2mgR(1 cos m ) mgRm , where we have 2 used the expansion cos m ' 1 m =2 for small m . Thus the tension T at that position satis es T mg = 2 Fc = mv2 =R ' mgm , or 2 T ' mg(1 + m ) : 79P m T y mg CHAPTER 16 OSCILLATIONS 469 (b) The tension T at any other position (where m is replaced by a smaller value ) satis es 2 2 T mg cos = mv =R or T = mg cos + mv =R. Note that 0 < cos < 1 and v < v, so T < T . (a) Suppose that the wheel is rotated through a small angle . The corresopnding change in the length of the spring is x ' r. The torque applied by the spring to the wheel about its center is then = Fr = kxr ' kr2 = . Thus the angular frequency of the small oscillation is r r r = kr2 = r k : != I mR2 R m (b) If r = R then ! reaches its maximum value of k=m. (c) If r = 0 then ! = 0 (i.e., no oscillation). Denote the length of the rod as L and the diameter of the disk of mass m as d. (a) 2 2 I m( T = 2 mgx = 2 md =8 +L +L + d=2) mg( d=2) s 2 + (76 cm + 42 cm=2)2 = 2 (42 cm) =8 2 )(76 cm + 42 cm=2) = 2:0 s : (980 cm/s s s p 80P 81P (b) Let T 0 = 2 I=(mgh + ) and solve for : p =I 2 T0 mgh =[(42 cm)2 (2:5 kg)=8 + (2:5 kg)(76 cm + 42 cm=2)2 ] (2:5 kg)(9:80 m/s2 )(76 cm + 42 cm=2) =18:5 N m/rad : 82P 2 2:0 s 0:50 s 2 Let the distances from points A and B p the center of mass be LA and LB , respectively. to p Then T = 2 IA =(mgLA ) and T = 2 IB =(mgLB ), where IA and IB are the moments of inertia of the pendulum about points A and B , respectively. Solve for LA and LB : LA = 42 IA =(mgT 2 ); LB = 42 IB =(mgT 2 ). 470 CHAPTER 16 OSCILLATIONS Let the distance from the pivot to the center of mass be x. Then 83P* I + mx2 T (x) = 2 mgx = 2 I0 mgx : Set and solve for x0 : s s d I0 + mx2 dx mgx r r I0 1 = mgx2 + g = 0 0 x=x0 2 x0 = I0 = mL = pL 0:29L : m 12m 12 Since E = 1 kA2 / A2 , the fraction of the energy lost in each full oscillation is 2 frac = A 85E 2 84E [(1 3:0%)A]2 = 5:9% : A2 Let A = xm e bt=2m . You want to evaluate A=xm = e bt=2m for t = 20T , where T is the period. Values for b (70 g/s), and m (250 g) are given in Sample Problem 10 and T is found in that problem to be 0:34 s. Thus bt=2m = (0:070 kg/s)(6:82 s)=2(0:250 kg) = 0:955 and A = e 0:955 = 0:385 : xm Note the conversions to SI units. (a) You want to solve e bt=2m = 1=3 for t. Take the natural logarithm of both sides to obtain bt=2m = ln(1=3). Now solve for t: t = (2m=b) ln(1=3) = (2m=b) ln 3, where the sign was reversed when the argument of the logarithm was replaced by its reciprocal. Thus 2(1:50 kg) t = 0:230 kg/s ln 3 = 14:3 s : (b) The angular frequency is 2 k g=s)2 !d = m 4b 2 = 8:100 N/m (0:230 kkg)2 = 2:31 rad/s : m :50 kg 4(1:50 r s 86E CHAPTER 16 OSCILLATIONS 471 The period is T = 2=!d = (2)=(2:31 rad/s) = 2:72 s and the number of oscillations is t=T = (14:3 s)=(2:72 s) = 5:27. (a) Use A(t) = xm e bt=2m . At t0 = 4T = 8 m=k = 8 2:00 kg=(10:0 N/m) = 11:24 s we have A(t0 ) = 3xm =4 . Thus 3 e bt0 =2m = 4 : Solve for b : 4 4 b = 2tm ln 3 = 2(2:00 kg) ln 3 = 0:102 kg/s : 11:24 s 0 87P p p (b) The amount of energy lost is 1 9 jE j = 1 kjx2 A2 (t0 )j = 2 (10:0 N/m) 1 16 (0:250 m)2 = 0:137 J : m 2 (a) Use x(t) = xm e bt=2m cos(!0 t + ). The maximum value of ( bdx=dt) is attained when dx(t)=dt reaches its maximum. Since 0 < e bt=2m = e (70 g/s)(0:34 s)=[2(250 g)] = 0:95 ' 1; we have dx(t)=dt ' !d xm e bt=2m sin(!d t + ): Thus [dx(t)=dt]max ' !d xm . The maximum value of the damping force is then 88P fm = b dx dt max ' b !d xm : r The ratio in question is thus (b) Since ' b= km, it should not change appreciably. The value of k is 89P p b k fm ' kx ' b!d xm = b!d ' b! = k m kxm k k m 3 kg/s = 0:015 : = p 70 10 (0:25 kg)(85 N/m) k = m g = (2000 kg)(980 cm/s ) = 490 N/cm ; x 4(10 m) m 0 2 472 CHAPTER 16 OSCILLATIONS where m0 = m=4 (since there are four wheels). From A(T ) = xm e bT=2m = 0:50xm ; we get e bT=2m = 0:50. Solve for b: 0 0 k m b = 2T ln 2:0 = 2m m0 ln 2:0 2 p = 2000 kg (49000 N/m)(500 kg) ln 2:0 = 1100 kg/s : (a) Since ! = !0 at resonance, the amplitude is xm j!=!0 = Fm =(b!0 ): (b) The velocity amplitude is vm j!=!0 = (!xm )j!=!0 = !0 [Fm =(b!0 )] = Fm =b: The natural frequency of the car is 91P 90E r !0 = 2v = 2(10 mi/h)(1 h=3600 s)(5279 ft/mi) = 7:1 rad/s : x 13 ft Let the mass of the car be m1 and that of the passengers be m2 . Then !0 = k=(m1 + m2 ), which gives 2 k = (m1 + m2 )!0 = p 2200 lb + 4(180 lb) (7:1 rad/s)2 = 4:6 103 lb/ft : 32:2 ft/s2 Thus the car rises by lb)(12 in./ft) 2 x = F = mk g = 4(180 103 lb/ft = 1:9 in. : k 4:6 92 (a) m1 = 288 g, m2 = 120 g; (b) 477 g; (c) 8:0 cm (a) ym = 8:8 10 4 m, T = 0:18 s, ! = 35 rad/s; (b) ym = 5:6 10 2 m, T = 0:48 s, ! = 13 rad/s; (c) ym = 3:3 10 2 m, T = 0:31 s, ! = 20 rad/s 93 ...
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