Chapter 13 Soln - Chapter 13 Periodic Motion 13.1 a T = i =...

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Unformatted text preview: Chapter 13 Periodic Motion 13.1: a) T = i = 4.55x 10-3 s, w = 2,3 = 27rf = 1.38><103 rad/s. b) whim = 1.14 x 10—3 s, w = 20f = 5.53 x103 rad/s. 13.2: a) Since the glider is released from 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 = 1.610 s = 0.625 Hz. 13.3: The period is 0:203 z 1.14 x 10—35 and the angular frequency is w = 2T“ = 5.53 X 103 rad/s. 13.4: (a) From the graph of its motion, the object completes one full cycle in 2.0 5; its period is thus 2.0 s and its frequency = 1 / period = 0.5 5‘1. (b) The displacement varies from —0.20n1 to +0.20 In, so the amplitude is 0.20 m. (c) 2.0 8 (see part a) 13.5: This displacement is i of a period. 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)); 20 2 211' 2 _— 2 = — = —*‘-‘-‘-‘-‘—‘-‘-‘- :- k—w m (T) m (20.60 s)) (0.200 kg) 0.292 N/m. 13.7: a.) T = i = 0.167 s. b) w = 2m“ = 37.7 rad/s. c) m = wig = 0.034 kg. 13.8: Solving Eq. (13.12) for k, 27f 2 2w 2 3 k — m (?) — (0.600 kg) (0.150 s) — 1.05 x 10 N/m. 13.9: From Eq. (13.12) and Eq. (13.10), T = 27: 3mg = 0.375 s, f = i = 2.66 Hz, 0.: = 27rf =16.7 rad/s. 13.10: a) a.ac = $3 = —w2A sin(wt + ,6) = —w2:c, s0 $(t) is a. solution to Eq. (13.4) if (.02 = %. b) a. = 2Aw, a constant, so Eq. (13.4) is not satisfied. c) 0,, = % = ‘iwAe‘infil, a: = 99‘ = (040)2Aeflwt4'fll = ”(1)212, so 3:08) is a solution to Eq. (13.4) if (1)2 = k/m. 13.11: a) a: = (3.0 mm) cos ((211')(440 Hz)t) b) (3.0 x 10‘3 m)(21r)(440 Hz) = 8.29 m/s, (3.0 mm)(21r)2(440 Hz)2 = 2.29 x 104 m/s2. c) j(t) = (6.34 x 107 m/s3) sin((27r)(440 Hz)t), jm = 5.34 x 107 111/83. 13.12: a.) From Eq. (13.19), A = |£Q| = lfi: = 0.98 m. b) Equation (13.18) is indeterminant, but from Eq. (13.14), 05 = ig, and from Eq. (13.17), sin (,0 > 0, so 05 = +3 c) cos(wt + (tr/2)) = — sinwt, so :1: = (—0.98 m) sin((12.2 rad/s)t)). 331 332 Chapter 13 13.13: With the same value for 0;, Eq. (13.19) gives A = 0.383 m and Eq. (13.18) gives 05 = arctan ( « m(fl4'00 m/s) (0.200 m)\/300 N/m/2.00 kg and :L' = (0.383 m) cos ((12.2 rad/s)t + 1.02 rad). 13.14: For SHM, 0.3 = —w% = -(27rf)2x = —(21r(2.5 Hz))2(1.1 X 10‘2 m) = —2.71 m/s2. b) Fi‘om 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 21rf = 15.7 rad/s, so ) = 1.02 rad = 585°, .7: = (1.46 cm) cos((15.7 rad/s)t + 0.715 rad) v: = (F223 cm/s) sin((15.7 rad/s)t + 0.715 red) 0.: = (—359 cm/sz) cos((15.7 rad/s)t + 0.715 rad). 13.15: The equation describing the motion is :1: = Asinwt; this is best found from either inspection or from Eq. (13.14) (Eq. (13.18) involves an infinite argument of the arctangent). Even so, :5 is determined only up to the sign, but that does not affect the result of this exercise. The distance from the equilibrium position is Asin(21r(t/T)) = (0.600 m) sin(41r/5) = 0.353 m. - 13.16: Empty chair: T : 273/? 47rzm _ 402(425 kg) k = T2 _ (1.30 S)2 = 993 N/m With person in chair: T = 27rv‘m/k T2 . 2 m: 16.3 (2 54s) (993N/m) =162kg 4772 4N2 mpem =162kg — 42.5 kg = 120kg 13.17: T = 21n/m/k, m = 0.400 kg Use (1.3 = —2.70 m/s2 to calculate k: 2 491: = max gives 1:: : _T.E'f§_ = igflkgljm ___ +3.50 N/m 13.18: We have vx(t) = (3.60 cm/s) sin((4.71 s‘l)t—7r/ 2). Comparing this to the general form of the velocity for SHM: —wA = 3.60 cm/s w = 4.71 5“1 ¢ = —«/2 Periodic Motion 333 (a) T = 27r/w = 27f/4.718_1 = 1.335 3.6013n1/s 3.600m/s b 2 mm— = W ------ m = . 64 ( ) A a: 4.71 s~1 0 7 cm (‘3) omax 2 1112/1 = (4.71 s‘1)2(0.764cm) = 16.9 cm/s2 13.19: a) :1:(t) = (7.40 cm) cos((4.16 rad/s)t — 2.42 rad) When t = T, (4.16 rad/s)T = 271' so T = 1.51 s b) T = 2am so I: = m(21r/T)2 = 26.0 N/m 0) A = 7.40 em = 0.0740 111 11102 + §km2 = $16142 gives on,“ 2 AM = 0.308 m/s d) F = —k:1: so Fmax = kA = 1.92 N 3) f0!) evaluated at t = 1. 00 s gives :L'_ = —0. 0125)2 Hi Speed is 0.303 m/s. (1 = —k:1:/m = —(2a0/1.50)(—0.0125) m/s2 = +0215 m/s2 13.20: See Exercise 13.15; t: (arccos(-1.5/6))(0.3/(211')) = 0.0871 s. 13.21: a) Dividing Eq. (13.17) by w, Squaring and adding, which is the same as Eq. (13.19). b) At time t = 0, Eq. (13.21) becomes 121 11:21 1 —kA2 = 511100 + —kx§= 212.1200 + 219%, 2 2 where m = kw2 (Eq. (13.10)) has been used. Dividing by 16/2 gives Eq. (13.19). 13.22: a) Um = (271f)A = (27r(392 Hz))(0.60 x 10—3 111) = 1.48 m/s. b) Kmam = é'r11,(‘l/mx)2 = $9.7 x 10—5 kg)(1.48 m/s)2 = 2.96 x 10—5 J. 13. 23: a) Setting:J Jmo - —1k:1:2 in Eq. (13.21) and solving for .7: gives :13: :b— “‘2 .Elim- inating a: in favor of v with the same relation gives vz— — :bx/kA2/ 21%: if}; fib) This happens four times each cycle corresponding the four possible combinations of + and — in the results of parta (a ). The time between the occurrences is one-fourth of a period or 1741:2532 %_ c) U: %E1K=3E (U=k?f121K._.__..3k;12) 334 Chapter 13 13.24: a) From Eq. (13.23), 4....-.fi..-“ 12...... = 5 A = @130 040 m)— _ 1. 20 m/s 0 = ‘/WW- (0040 m)2 — (—0.015 m)2 = 1.11 m/s. 1:) The extreme; of acceleration occur at the extremes of motion, when :1: = :tA, and 12A _ (450 N/In) (0.040 m) _.._ _ _ 2 2 am“ _ m (0.500 kg) 36 ”1/5 d) From Eq. (13.4), a, = mm = 13.5 111/52. (0.500 kg) c) From Eq. (13.31), E = %(450 N/m)(0.040 m)2 = 0.36 J. 13.25: a) am = 1112A = (27rf)2A = (217(085 Hz))2(18.0 x 10‘2 m) = 5.13 [11/32. 1.1m,“ = wA = 21rfA = 0.961 m/s. b) a, = —(27rf)2:1: = —2.57 m/s2, 'v = (27rf)vA2 — 3:2 = (21r(0.85 Hz))\/(18.0 x 10-2 m)2 — [9.0 x 10:§"'fi{)2 = 0.833 m/s. 0) The fraction of one period is (1/211') arcsin(12.0/18.0), and so the time is (in/271') X arcsin(12. 0/18. 0)— — 1. 37 x 10‘1 5 Note that this is also arcsin(:1:/A)/w d) The conservation of energy equation can be written 115142: 2—2mv + 136332. We are given amplitude frequency in Hz, and various values of :12. We could calculate velocity from this information if we use1 the relationship k/m— - 0.124713)” and rewrite the conservation equation as l[412:5 4—27»: + 1.112. Using energy principles is generally a good 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 : AIM; and now we want A2 = £3111. So % = (/Mf‘im, or m = 3M. For the energy, E2 = §kA§, but since A2 = $41, E2 = $131, or 331 is lost to heat. 13.27: a) Ema" + l1113;132:002841 b WWW 2+ (0—3003252 llllllllll l (/ngrZ—O: [ 012 m)+ FUSEWEE —0.014 m. = (/k/mA = 0.615 m/s. Periodic Motion 335 13.28: At the time in question we have: a: = Acos(wt + (:5) = 0.600 In 1) = —wA sin(wt + til) = 2.20 m/s a. = —w2A cos(wt + (:5) = —8.40n1/s2 Using the displacement and acceleration equations: —w2Acos(wt + gt) = —w2(0.600 m) = ——8.40 m/s2 L92 = 14.0 and w = 3.742 5’1 To find A, multiply the velocity equation by w: rw2A sin(wt + 03)) = (3.742 5‘1)(2.20 m/s) : 8.232 In/s2 Next square both this new equation and the acceleration equation and add them: w4A2 sin2(wt + (25) + (£1.42 0052(wt + (#2) = (8.232 m/82)2 + (—8.40 [11/32)2 = w4A2(sin2(wt + 915) + 0062(wt + (25)) (MA? = 67.771112/84 + 70.56 m2/s’;l = 138.3 [[12/s4 138.3 1112/54 138.3 m2/s4 2 A2 = w; ..... = ”(3.742 5“)“ = 0.7054m The object will therefore travel 0.8401n — 0.600n1 = 0.240 m to the right before stopping at its maximum amplitude. 13.29: cm : Auk/m Use T to find k/m: T = 27n/m/k so k/m = (21T/T)2 = 158 5’2 Use an,“ to find A: a ax = kA/m so A = emu/(Mm) = 0.0405 In Then 'vmx = Axfk/m = 0.509 m/s 13.30: Using k 2 £3 from the calibration data, m = Egg/L0) : (200 N) Kl} 25 x 10- m) _ _6_00 kg. (271' n2 (27r(2.60 Hz»2 13-31: a) k: ”9— (65' 0 $1980 )2m/S ) — _.5 31 x 103 N/m 0 120m T: = I . . b) 21r1/TE— =27r1/ég— 251/986 III/52 0695 8 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 336 Chapter 13 energy relative to the bottom is 2mgA = 2(4.00 kg) (9.80 111/82) x (0.050 m) = 3.92 J. This is the total energy, and is the same total for each part. b) Us.“ 2 0, K = 0, so Uspmg = 3.92 J. c) At equilibirum the spring is stretched half as much as it was for part (a), and so Uspring = fi(3.92 J) = 0.98 J, U3,“ = %(3.92 J) = 1.96 J, and so K = 0.98 J. 13.33: The elongation is the weight divided by the spring constant, 10 mg g? 2 A! = — = = — : . . k w2m 4713 3 97 cm 13.34: See Exercise 9.40. a) The mass would decrease by a factor of (1 /3)3 = 1/27 and so the moment of inertia would decrease by a factor of (1/2'.v’)(1/3)2 = (1/ 243), and for the same Spring constant, the frequency and angular frequency would increase by a factor of m = 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 = ng = 2.72x 10—8 kg-mz, or 2.7 X 10—8 kg-m2 to two figures b) a = (27rf)21 = (2W2 Hz)2(2.72 x10‘B kg-rn2) = 4.3 x 10—6 N-m/rad. ' 13.36: Solving Eq. (13.24) for a in terms of the period, 2% 2 = ( 2’” )2 ((1/2)(2.00 x 10*3 kg)(2.20 x 10‘2 no?) 1.00 s = 1.91 x 10—5 N-rn/rad. 13.37: H 0.450 N-m/rad 2 I = = = . a _ (2.71:)2 (2a(125)/(265 3))2 0 0512 kg m 13.38: The equation 9 = 8 cos(wt + a) describes angular SHM. In this problem, a = 0. a) fi = —w8 sin(wt) and ill—:3 = —w26 cos(wt). b) When the angular displacement is 8, 9 = 8 cos(wt), and this occurs at t = 0, so 2 5:5. = 0 since 5mm) = 0. and g = #9, since cos(0)=1- When the angular diSplacement is 9/ 2, g = 6 cos(wt), or % = cos(wt). d9 —w8x/3 _ , J3 d26 —w29 . a — 2 . Since sm(wt) —— —2—, and d}? — —2—, SlIlCe cos(wt) —- 1/2. This corresponds to a displacement of 60°. Periodic Motion 13.39: Using the same procedure used to obtain Eq. (13.29), the potential may be expressed as U = U0 [(1 + m/Rorn — 2(1 + sneer] Note that at 'r — R0, U = —Uo. Using the appropriate forms of the binomial theorem for 196/ Rel << 1, -12 #13 (1 - mac/Bo) + l 2‘ lemon) U % U0 #6 -7 -2 (1 - fits/Rel + (gig—lama?) = U0 [*1 + £2172} 1 §k$2 '— U0. 2 from Eq. (13.28) must be kept; ion that R0 is an extreme (in Note that terms in u vanish is another indicat k = 721:)".3/R2 has been used. where order terms 13.41: T = 27rd” , so for a differ T’ = T g/g’ = (1.60 s)\X9.80 m/82/3.71 m/s2 = 2.60 s. n, the small-an which occurs after a qu (a), 0.25 s. T 13.42: a) To the given precisio d is at the bottom of the arc, b) The same as calculated in amplitude . spee proximating the pendulum motion as SHM, assume that the angle ength of the spring does not change While swi 'ng in the f the vertical motion as we = W;— = 131 and w' z are. Denote the angul fl = €er = “2%, which is solved for L :- 410/ 12. But L is the length of the stretched ' = L — w/k _- 3w/k = 3(1.00 N)/(1.50 N/m) = 338 Chapter 13 .4 : x 13 4 1'6 L r E L l0 L : \ : 1 i I ‘, I : : \T : r “/ I i a El” I I s , . .. H 1. a=~gsin6 1 . IvI=—gL(1—oosa) . I W (a) (b) (c) 13.45: The period of the pendulum is T = (136 s)/ 100 = 1.36 s. Then, _ 47r2 L _ 432(5 m) g T2 (1.36 3)2 13.46: From the parallel axis theorem, the moment of inertia of the hoop about the nail is I = MR2 + M3? = 2043?, so T = 29mg, with d = R in Eq. (13.39). Solving for R, R = gig/8'32 = 0.496 m. 13.47: For the situation described, I = 1011.2 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). 13.48: a) Solving Eq. (13.39) for I, = 10.6? m/s2. 2 2 1 = (g) mgd = (032:0 S) (1.30 kg)(9.80 m/s2)(0.250 m) = 0.0937 kg-mz. b) The small—angle approximation will not give three-figure accuracy for 9 = 0.400 rad. From energy considerations, mgd(1—- cos 9) = élflfimx. Expressing Qmax in terms of the period of small-angle oscillations, this becomes - ,fi(g)1.1...e)-,/2(fig,g,)zu_ 3.3... m.» = .0. 13.49: Using the given expression for I in Eq. (13.39), with d = R (and of course m = M), T = 2:r\/§R/3g = 0.58 3. 13.50: From Eq. (13.39), 2 2 I = mgd (23;?) = (1.80 kg)(9.80 m/s2)(0.200 m) (3921:100) = 0.129 kg-m2. 13.51: a) From Eq. (13.43), , fl (2.50 N/m) (0.90 kg/s)2 _ , _. a)" _ w — (0.300 kg) 4(0.300 kg)? — 2.47 rad/s, so f — 2—1; — 0.393 Hz. b) b = 2% = 2,/(2.50 N/m)(0.300"i<Ej = 1.73 kg/s. Periodic Motion 339 13.52: From Eq. (13.42), A2 = A] apogee. Solving for 0, 2m A1 2(0050 kg) 0300 m b = _1 — n ( ) In 0.100 m A2 t — ) = 0.0220 kg/s. As a check, note that the oscillation frequency is the same as the undamped frequency to 4.8 x 104%, so Eq. (13.42) is valid. 13.53: a) With 05 = 0, 33(0) 2 A. d b b) 113 = d? = 14112“("/2""‘”t [—fi cosw’t - w’sinw’t] , and at t = 0, v = —Ab/2m; the graph of 3: versus t near it = 0 slopes down. d r 52 rb c) as,» = 7:? = A€_(b/2m)t [(HE - w'z) cosw’t + gr; sin 0ft] , and att=0, (Note that this is (—600 — kxo)/m.) This will be negative if b < v2km, zero if b = 1/ 216111 and positive if b > v2km. 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 = Fm/bwd. a) fit b) 2A1. Note that the resonance frequency is independent of the value of 0 (see Fig. (13.27)). 13.55: a) The damping constant has the same units as force divided by speed, or [kg-m/s2]/[rn/s] = [kg/s]. b) The units of x/% are the same as [[kg/s2][kg]]1(2 = [kg/s], the same as those for b. c) Lodz = k/m. (i) baud = 0.21:, so A = Finn/(0.21%) = 5me/k. (ii) bud = 0.41%, so A = Jinn/(0.4.19) = 2.5Fmax/k, as shown in Fig. (13.27). 13.56: The resonant frequency is y/k/m = (/(21 QTfiS—NymTfigk—g) = 139 rad/s = 22.2 Hz, and this package does not meet the criterion. 13.57: a) 2 a = Aw? = (”“1020 H1) ((3500 rev/min) (0% $341)) = 6.72 x 103 m/S2. ll 30 revfmin b) ma = 3.02 x103 N. c) 10.4 = (3500 rev/min)(.05 m) (1 JEL) = 18.3 m/s. K §m112 = (%)(.45 kg)(18.3 m/s)2 = 75.6 J. d) At the midpoint of the stroke, cos(wt) = 0 and so not = 1r/2, thus 15 = 7r/2w. w = (3500 rev/min)(%§fn%) = % rad/s, so 3 = 29350) s. Then P = AK/At, or P = 75.6 J/(2{3350) s) = 1.76 X 10“ W. e) If the frequency doubles, the acceleration and hence the needed force will quadruple (12.1 X 103 N). The 340 Chapter 1‘3 maximum speed increases by a factor of 2 since 11 ocw, 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 constant is then k = mg/Al. The period of oscillation of the empty car is TE = 27n/M/k and the period of the loaded car is Al T, = 2.r M? = 13+ (2702—. so ___________ w 9 TE _ {1"121—(21r)2égE — 1003 3 13.59: a) For SHM, the period, frequency and angular frequency are indpendent 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' 18 decreased by a factor of21. (1) Initially, the speed at 141/4 was J4CwA1; after the amplitude is reduced the speed is :..;.J\/(1411/2)2 /4)21— 34510141, so the speed is decreased by a factor of 7 (this result is valid at a: — ——A1/4 as well) e) The potential energy depends on position and is unchanged. From the result of part ((1), the kinetic energy is decreased by a factor of g. 13.60: The distance L is L : mg/k; the period of the oscillatory motion is T=2rr E=2W¢E, V16 9 which is the period of oscillation of a simple pendulum of length L. 13.61: a) Rewriting Eq. (13.22) in terms of the period and solving, c) If the block is just on the verge of slipping, the friction force is its maximum, f = pan = 1131119. Setting this equal to me. = mA(211'/T)2 gives 11,, = A(27r/T)2/g = 0.143. 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. Periodic Motion 341 b) :c = +fl/(27Tf)2 = +(9.80 m/s2)/(27r(1.50 Hz))2 2 0.110 m. c) The cowboy’s speed will be the saddle’s speed, 0 = (27rf)‘\/A2 —§;~2 = 2.11m/s. (1) Taking t = 0 at the time when the cowboy leaves, the position of the saddle as a function of time is given by Eq. (13.13), with cosqb = ——;¥—; this is checked by setting it = 0 and finding that :1: = “-5; = —w—“;. The cowboy’s position is .rc = :50 + cot — (g/2)t2. Finding the time at which the cowboy and the saddle are again in contact involves a transcendental equatiou which must be solved numerically; specifically, (0.110 m) + (2.11 m/s)t — (4.90 In/S2)t2 = (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(wt+¢5) = 1.72 m/s and the cowboy’s speed is (2.11 m/s) — (9.80 m/s2) x (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 am = kA/(m + M), and so the maximum force on the top block is (milk!) RA = psmg, and so the maximum amplitude is Am“ = #5011 + M)g/k. 13.64: (a) Momentum conservation during the collision: mvo = (2m)V 1 l V = 500 = -2-(2.00n1/s)=1.00m/s Energy conservation after the collision: 1 2_1 2 2MV—2kx _ A??? _ (20.0kg)(1.00 inn/s)2 _ . :1: _ (f k _ WW _ 0.500111 (amphtude) w=21rf= k/M 1 ,..__ 1 80.0N/m T=l— 1 =3.14s f _ 0.318Hz (b) It takes 1/2 period to first return: %(3.14 s) = 1.573 13.65: a) m —+ m/2 Splits at :5 = 0 where energy is all kinetic energy, E : %mv2, so E 1., E/ 2 it stays same E = gm? so A = ,/2E/k Then E —> E/2 means A —> A/x/i T = 27n/m/k so am —> m/2 means T .1, T/fl 342 Chapter 13 b) m —+ m/ 2 Splits at .7: 2 A where all the energy is potential energy in the spring, so E doesn’t change. E = fife/12 so A stays the same. T = ZWN/m/k so T —> T/x/Z 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 .7: = :tA, the energy and amplitude remian 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=—,/ k (10. 0 mm) -0200 m. c) ’2" = 0.050 .1. d) If U = 3,13, .1: = $5 = 0.141 m. e) From Eq. (13.18), using on = — 2—KQJalld120: “$1, /2K v —" _ m» =\/U. 429 013:0 —\/: /2U= and a = arctan(\/(l:12—9) = 0.530 rad. 13.67: a) The quantity Al 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 A! — a: (see Fig. (13.16(c))) and the elastic potential energy is U91 = (1/2)};(Al — 3:)2. Periodic Motion 343 b) E U=Uel+mg(a:—xo)= 2 193:2 + $— (A!)2 — kAl a: + mga: — mgxo. Since A! = mgr/k, the two terms proportional to a: cancel, and 1 U = ékfiCQ + 5k (A02 — mga‘o. c) An additive constant to the mechanical energy does not change the dependence of the force on 3:, F: = 4%, and so the relations expressing Newton’s laws and the resulting equations of motion are unchanged. 13.68: The “spring constant” for this wire is k = fih so 1 k 1 g 1 9.80m/s2 f:—— -——=_ ___.._:___ =1. . 211' m 27r A! 211' 2.00x10—3m “Hz 13.69: a) ¥ 2 0.1501n/s. b) a = ~(27r/T)23; = —0.112m/52. c) The time to go from equilibrium to half the amplitude is sinwt = (1/2), or cat = 7r/6 rad, or one- twelfth of a period. The needed time is twice this, or one-sixth o...
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