Baixe Solucionário Halliday 8ª edição vol. 2 - ch15 e outras Notas de aula em PDF para Física, somente na Docsity! Chapter 15 1. The textbook notes (in the discussion immediately after Eq. 15-7) that the acceleration amplitude is am = ω2xm, where ω is the angular frequency (ω = 2πf since there are 2π radians in one cycle). Therefore, in this circumstance, we obtain ( )( ) ( )22 2 2(2 ) 2 6.60 Hz 0.0220 m 37.8 m/s .m m ma x f xω π π= = = = 2. (a) The angular frequency ω is given by ω = 2πf = 2π/T, where f is the frequency and T is the period. The relationship f = 1/T was used to obtain the last form. Thus ω = 2π/(1.00 × 10–5 s) = 6.28 × 105 rad/s. (b) The maximum speed vm and maximum displacement xm are related by vm = ωxm, so = = 1.00 10 6.28 10 = 1.59 10 . 3 5 3x vm mω × × × − m / s rad / s m 3. (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 = ωxm, where ω is the angular frequency. Since ω = 2πf, where f is the frequency, ( ) ( )3= 2 = 2 120 Hz 1.0 10 m = 0.75 m/s.m mv fxπ π −× (c) The maximum acceleration is ( ) ( )( ) ( )222 3 2 2= = 2 = 2 120 Hz 1.0 10 m = 5.7 10 m/s .m m ma x f xω π π −× × 4. (a) The acceleration amplitude is related to the maximum force by Newton’s second law: Fmax = mam. The textbook notes (in the discussion immediately after Eq. 15-7) that the acceleration amplitude is am = ω2xm, where ω is the angular frequency (ω = 2πf since there are 2π radians in one cycle). The frequency is the reciprocal of the period: f = 1/T = 1/0.20 = 5.0 Hz, so the angular frequency is ω = 10π (understood to be valid to two significant figures). Therefore, = = 0.12 10 0.085 = 10 . 2 2F m xmmax kg rad / s m Nω b gb gb gπ 647 CHAPTER 15 (b) Using Eq. 15-12, we obtain ( ) ( ) 22 2 0.12kg 10 rad/s 1.2 10 N/m.k k m m ω ω π= ⇒ = = = × 5. (a) During simple harmonic motion, the speed is (momentarily) zero when the object is at a “turning point” (that is, when x = +xm or x = –xm). Consider that it starts at x = +xm and we are told that t = 0.25 second elapses until the object reaches x = –xm. To execute a full cycle of the motion (which takes a period T to complete), the object which started at x = +xm must return to x = +xm (which, by symmetry, will occur 0.25 second after it was at x = –xm). Thus, T = 2t = 0.50 s. (b) Frequency is simply the reciprocal of the period: f = 1/T = 2.0 Hz. (c) The 36 cm distance between x = +xm and x = –xm is 2xm. Thus, xm = 36/2 = 18 cm. 6. (a) Since the problem gives the frequency f = 3.00 Hz, we have ω = 2πf = 6π rad/s (understood to be valid to three significant figures). Each spring is considered to support one fourth of the mass mcar so that Eq. 15-12 leads to ( ) ( ) 2 5 car 1 1450kg 6 rad/s 1.29 10 N/m. / 4 4 k k m ω π= ⇒ = = × (b) If the new mass being supported by the four springs is mtotal = [1450 + 5(73)] kg = 1815 kg, then Eq. 15-12 leads to 5 new new total 1 1.29 10 N/m 2.68Hz. / 4 2 (1815 / 4) kg k f m ω π ×= ⇒ = = 7. (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 ω is ω = 2πf = 2π(2.00 Hz) = 12.6 rad/s. (d) The angular frequency is related to the spring constant k and the mass m by ω = k m . We solve for k and obtain k = mω2 = (0.500 kg)(12.6 rad/s)2 = 79.0 N/m. (e) Let xm be the amplitude. The maximum speed is vm = ωxm = (12.6 rad/s)(0.350 m) = 4.40 m/s. 136 ( )1 0.200 kg 0.600 kg| | 8.00 m/s 4.00 m/s 0.200 kg 0.600 kgf v −= = + . This becomes the initial speed v0 of the projectile motion of block 1. A variety of choices for the positive axis directions are possible, and we choose left as the +x direction and down as the +y direction, in this instance. With the “launch” angle being zero, Eq. 4-21 and Eq. 4-22 (with –g replaced with +g) lead to 0 0 0 2 2 2(4.90 m)(4.00 m/s) 9.8 m/s hx x v t v g − = = = . Since x – x0 = d, we arrive at d = 4.00 m. 15. (a) Eq. 15-8 leads to 2 2 123 m/s 35.07 rad/s . 0.100 m aa x x ω ω −= − ⇒ = = = Therefore, f = ω/2π = 5.58 Hz. (b) Eq. 15-12 provides a relation between ω (found in the previous part) and the mass: 2 400 N/m= 0.325kg. (35.07 rad/s) k m m ω ⇒ = = (c) By energy conservation, 12 2kxm (the energy of the system at a turning point) is equal to the sum of kinetic and potential energies at the time t described in the problem. 1 2 = 1 2 + 1 2 = + . 2 2 2 2 2kx mv kx x m k v xm m⇒ Consequently, 2 2(0.325 kg / 400 N/m)(13.6 m/s) (0.100 m) 0.400m.mx = + = 16. From highest level to lowest level is twice the amplitude xm of the motion. The period is related to the angular frequency by Eq. 15-5. Thus, x dm = 12 and ω = 0.503 rad/h. The phase constant φ in Eq. 15-3 is zero since we start our clock when xo = xm (at the highest point). We solve for t when x is one-fourth of the total distance from highest to lowest level, or (which is the same) half the distance from highest level to middle level (where we locate the origin of coordinates). Thus, we seek t when the ocean surface is at x x dm= =12 14 . With cos( )mx x tω φ= + , we obtain 139 CHAPTER 15 ( )1 1 1cos 0.503 0 cos(0.503 ) 4 2 2 d d t t = + ⇒ =   which has t = 2.08 h as the smallest positive root. The calculator is in radians mode during this calculation. 17. The maximum force that can be exerted by the surface must be less than µsFN or else the block will not follow the surface in its motion. Here, µs is the coefficient of static friction and FN is the normal force exerted by the surface on the block. Since the block does not accelerate vertically, we know that FN = 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 = mω2xm = m(2πf)2xm, where am is the magnitude of the maximum acceleration, ω is the angular frequency, and f is the frequency. The relationship ω = 2πf was used to obtain the last form. We substitute F = m(2πf)2xm and FN = mg into F < µsFN to obtain m(2πf)2xm < µsmg. The largest amplitude for which the block does not slip is = 2 = 0.50 9.8 2 2.0 0 0312 2 2x g fm sµ π πb g b gc h b g m / s Hz× = . .m 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. 18. They pass each other at time t, at x x xm1 2 12= = where x x t x x tm m1 1 2 2= + = +cos( ) cos( ).ω φ ω φand From this, we conclude that cos( ) cos( )ω φ ω φt t+ = + =1 2 12 , and therefore that the phases (the arguments of the cosines) are either both equal to π/3 or one is π/3 while the other is –π/3. Also at this instant, we have v1 = –v2≠0 where v x t v x tm m1 1 2 2= − + = − +ω ω φ ω ω φsin( ) sin( ).and This leads to sin(ωt + φ1) = – sin(ωt + φ 2). This leads us to conclude that the phases have opposite sign. Thus, one phase is π/3 and the other phase is –π /3; the wt term cancels if we take the phase difference, which is seen to be π /3 – (–π /3) = 2π /3. 19. (a) Let = 2 2 1x A t T cos πFHG IKJ 140 be the coordinate as a function of time for particle 1 and = 2 2 + 6 2x A t T cos π πFHG IKJ 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 2πt/T + π/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, 1 2 0.50 s= cos = 0.25 2 1.5 s Ax Aπ ×  −   and 2 2 0.50 s= cos + = 0.43 . 2 1.5 s 6 Ax Aπ π×  −   Their separation at that time is x1 – x2 = –0.25A + 0.43A = 0.18A. (b) The velocities of the particles are given by = = 2 1 1v dx dt A T t T π πsinFH IK and = = 2 + 6 . 2 2v dx dt A T t T π π πsinFH IK We evaluate these expressions for t = 0.50 s and find they are both negative-valued, indicating that the particles are moving in the same direction. 20. We note that the ratio of Eq. 15-6 and Eq. 15-3 is v/x = –ωtan(ωt + φ) where ω = 1.20 rad/s in this problem. Evaluating this at t = 0 and using the values from the graphs shown in the problem, we find φ = tan−1(–vo/xoω) = tan−1(+4.00/(2 × 1.20)) =1.03 rad (or –5.25 rad). One can check that the other “root” (4.17 rad) is unacceptable since it would give the wrong signs for the individual values of vo and xo. 21. Both parts of this problem deal with the critical case when the maximum acceleration becomes equal to that of free fall. The textbook notes (in the discussion immediately after Eq. 15-7) that the acceleration amplitude is am = ω2xm, where ω is the angular frequency; this is the expression we set equal to g = 9.8 m/s2. 141 CHAPTER 15 at equilibrium. The calculator is in degrees mode in the above calculation. The distance from the top of the incline is therefore (0.450 + 0.75) m = 0.525 m. (b) Just as with a vertical spring, the effect of gravity (or one of its components) is simply to shift the equilibrium position; it does not change the characteristics (such as the period) of simple harmonic motion. Thus, Eq. 15-13 applies, and we obtain 214.0 N 9.80 m/s2 0.686 s. 120 N/m T π= = 26. We wish to find the effective spring constant for the combination of springs shown in the figure. We do this by finding the magnitude F of the force exerted on the mass when the total elongation of the springs is ∆x. Then keff = 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 x= 2  . The left-hand spring exerts a force on the block and its magnitude is F k x= ∆  . Thus k k x x kreff = =∆ ∆ / /2 2 . The block behaves as if it were subject to the force of a single spring, with spring constant k/2. To find the frequency of its motion replace keff in f k m= 1 2/ /πa f eff with k/2 to obtain = 1 2 2 f k mπ . With m = 0.245 kg and k = 6430 N/m, the frequency is f = 18.2 Hz. 27. When the block is at the end of its path and is momentarily stopped, 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 2 = 1 2 1.3 10 0.024 = 3.7 10 . 2 2 2 2E kxm × × − N / m m Jc ha f 28. (a) The energy at the turning point is all potential energy: E kxm= 12 2 where E = 1.00 J and xm = 0.100 m. Thus, = 2 = 200 . 2k E xm N / m (b) The energy as the block passes through the equilibrium position (with speed vm = 1.20 m/s) is purely kinetic: = 1 2 = 2 = 1.39 . 2 2E mv m E vm m ⇒ kg 144 (c) Eq. 15-12 (divided by 2π) yields = 1 2 1 91f k mπ = . .Hz 29. The total energy is given by E kxm= 12 2 , where k is the spring constant and xm is the amplitude. We use the answer from part (b) to do part (a), so it is best to look at the solution for part (b) first. (a) The fraction of the energy that is kinetic is 1 3= =1 =1 = 0.75 4 4 K E U U E E E − − − = where the result from part (b) has been used. (b) When x xm= 12 the potential energy is U kx kxm= =12 2 1 8 2 . The ratio is 2 2 / 8 1 0.25. / 2 4 m m kxU E kx = = = (c) Since E kxm= 12 2 and U kx= 12 2 , U/E = x xm 2 2 . We solve x xm 2 2 = 1/2 for x. We should get x xm= / 2 . 30. The total mechanical energy is equal to the (maximum) kinetic energy as it passes through the equilibrium position (x = 0): mv2 = (2.0 kg)(0.85 m/s)2 = 0.72 J. Looking at the graph in the problem, we see that U(x=10)=0.5 J. Since the potential function has the form 2( )U x bx= , the constant is 3 25.0 10 J/cmb −= × . Thus, U(x) = 0.72 J when x = 12 cm. (a) Thus, the mass does turn back before reaching x = 15 cm. (b) It turns back at x = 12 cm. 31. (a) Eq. 15-12 (divided by 2π) yields = 1 2 1 2 1000 5 00 2 25f k mπ π = =N / m kg Hz . . . 145 CHAPTER 15 (b) With xo = 0.500 m, we have U kx0 12 0 2 125= = J . (c) With vo = 10.0 m/s, the initial kinetic energy is K mv0 12 0 2 250= = J . (d) Since the total energy E = Ko + Uo = 375 J is conserved, then consideration of the energy at the turning point leads to = 1 2 2 = 0.866 . 2E kx x E km m ⇒ = m 32. We infer from the graph (since mechanical energy is conserved) that the total energy in the system is 6.0 J; we also note that the amplitude is apparently xm = 12 cm = 0.12 m. Therefore we can set the maximum potential energy equal to 6.0 J and solve for the spring constant k: k xm2 = 6.0 J ⇒ k = 8.3 ×102 N/m . 33. The textbook notes (in the discussion immediately after Eq. 15-7) that the acceleration amplitude is am = ω2xm, where ω is the angular frequency and xm = 0.0020 m is the amplitude. Thus, am = 8000 m/s2 leads to ω = 2000 rad/s. Using Newton’s second law with m = 0.010 kg, we have = = + = 80 2000 3 F ma m a t tm− − − FH IKcos N cosω φa fc h a f π where t is understood to be in seconds. (a) Eq. 15-5 gives T = 2π/ω = 3.1 × 10–3 s. (b) The relation vm = ωxm can be used to solve for vm, or we can pursue the alternate (though related) approach of energy conservation. Here we choose the latter. By Eq. 15- 12, the spring constant is k = ω2m = 40000 N/m. Then, energy conservation leads to 2 21 1= = 4.0 m/s. 2 2m m m m kkx mv v x m ⇒ = (c) The total energy is 12 2 1 2 2 0 080kx mvm m= = . J . (d) At the maximum displacement, the force acting on the particle is 4 3(4.0 10 N/m)(2.0 10 m)=80 N.F kx −= = × × (e) At half of the maximum displacement, 1.0 mmx = , and the force is 146 solve k m m k m/ ( ) /+ =∆ 12 for m. Square both sides of the equation, then take the reciprocal to obtain m + ∆m = 4m. This gives m = ∆m/3 = (300 g)/3 = 100 g = 0.100 kg. (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. We will need to find the value of the spring constant k. Use k = mω2 = m(2π f )2. Then ( ) ( ) ( ) ( ) ( ) ( ) 2 2 2 0.100 kg 0.300 kg 9.80 m/s+ = 0.200 m. 2 0.100 kg 2 2.24 Hz m m g y m fπ π +∆ = × This is measured from the initial position. 38. From Eq. 15-23 (in absolute value) we find the torsion constant: 0.20 N m 0.235 N m/rad . 0.85 rad τκ θ ⋅= = = ⋅ With I = 2mR2/5 (the rotational inertia for a solid sphere — from Chapter 11), Eq. 15–23 leads to ( ) ( ) 22 22 55 95 kg 0.15 m2 2 12 s. 0.235 N m/rad mRT π π κ = = = ⋅ 39. (a) We take the angular displacement of the wheel to be θ = θm cos(2πt/T), where θm is the amplitude and T is the period. We differentiate with respect to time to find the angular velocity: Ω = –(2π/T)θmsin(2πt/T). The symbol Ω is used for the angular velocity of the wheel so it is not confused with the angular frequency. The maximum angular velocity is ( ) ( )2 rad2 39.5 rad/s. 0.500 s m m T π ππ θΩ = = = (b) When θ = π/2, then θ/θm = 1/2, cos(2πt/T) = 1/2, and ( ) ( ) ( ) 22sin 2 1 cos 2 1 1 2 3 2t T t Tπ π= − = − = where the trigonometric identity cos2θ + sin2θ = 1 is used. Thus, = 2 2 = 2 0.500 3 2 = 34.2 . Ω − FH IK −FH IK FHG I KJ − π π π π T t Tm θ sin s rad rad / sa f 149 CHAPTER 15 During another portion of the cycle its angular speed is +34.2 rad/s when its angular displacement is π/2 rad. (c) The angular acceleration is ( ) 2 22 2 2 2cos 2 / .m d t T dt T T θ π πα θ π θ   = = − = −       When θ = π/4, 2 22 = 124 rad/s , 0.500 s 4 π πα    = − −       or 2| | 124 rad/s .α = 40. (a) Comparing the given expression to Eq. 15-3 (after changing notation x → θ ), we see that ω = 4.43 rad/s. Since ω = then we can solve for the length: L = 0.499 m. (b) Since vm = ωxm = ωLθm = (4.43 rad/s)(0.499 m)(0.0800 rad) and m = 0.0600 kg, then we can find the maximum kinetic energy: mvm2 = 9.40 × 10− 4 J. 41. (a) Referring to Sample Problem 15-5, we see that the distance between P and C is h L L L= − =23 12 16 . The parallel axis theorem (see Eq. 15–30) leads to = 1 12 + = 1 12 + 1 36 = 1 9 . 2 2 2 2I mL mh mL mLFH IK Eq. 15-29 then gives T I mgh L gL L g = = =2 2 9 6 2 2 3 2 π π π/ / which yields T = 1.64 s for L = 1.00 m. (b) We note that this T is identical to that computed in Sample Problem 15-5. As far as the characteristics of the periodic motion are concerned, the center of oscillation provides a pivot which is equivalent to that chosen in the Sample Problem (pivot at the edge of the stick). 42. We require T L g I mgh o= =2 2π π 150 similar to the approach taken in part (b) of Sample Problem 15-5, but treating in our case a more general possibility for I. Canceling 2π, squaring both sides, and canceling g leads directly to the result; Lo = I/mh. 43. (a) A uniform disk pivoted at its center has a rotational inertia of 212 Mr , where M is its 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 2 21 12 2 ( )Mr M L r+ + . The rod is pivoted at one end and has a rotational inertia of mL2/3, where m is its mass. The total rotational inertia of the disk and rod is 2 2 2 2 2 2 2 1 1( ) 2 3 1 1(0.500kg)(0.100m) (0.500kg)(0.500m 0.100m) (0.270kg)(0.500m) 2 3 0.205kg m . I Mr M L r mL= + + + = + + + = ⋅ (b) We put the origin at the pivot. The center of mass of the disk is = + = 0.500 m + 0.100 m = 0.600 md L rl away and the center of mass of the rod is r L= = =/ ( . ) / .2 0 500 2 0 250m m away, on the same line. The distance from the pivot point to the center of mass of the disk-rod system is = + + = 0.500 0.600 + 0.270 0.250 0.500 + 0.270 = 0.477 . d M m M m d r  kg m kg m kg kg m a fa f a fa f (c) The period of oscillation is ( ) 2 2 0.205 kg m2 2 1.50 s . (0.500 kg 0.270 kg)(9.80 m/s )(0.447 m) IT M m gd π π ⋅= = = + + 44. We use Eq. 15-29 and the parallel-axis theorem I = Icm + mh2 where h = d, the unknown. For a meter stick of mass m, the rotational inertia about its center of mass is Icm = mL2/12 where L = 1.0 m. Thus, for T = 2.5 s, we obtain T mL md mgd L gd d g = + = +2 12 2 12 2 2 2 π π/ . Squaring both sides and solving for d leads to the quadratic formula: 151 CHAPTER 15 which yields / 12 (1.85 m)/ 12 0.53 mx L= = = as the value of x which should produce the smallest possible value of T. (b) With L = 1.85 m and x = 0.53 m, we obtain T = 2.1 s from the expression derived in part (a). 50. Consider that the length of the spring as shown in the figure (with one of the block’s corners lying directly above the block’s center) is some value L (its rest length). If the (constant) distance between the block’s center and the point on the wall where the spring attaches is a distance r, then rcosθ = d/ and rcosθ = L defines the angle θ measured from a line on the block drawn from the center to the top corner to the line of r (a straight line from the center of the block to the point of attachment of the spring on the wall). In terms of this angle, then, the problem asks us to consider the dynamics that results from increasing θ from its original value θo to θo + 3º and then releasing the system and letting it oscillate. If the new (stretched) length of spring is L′ (when θ = θo + 3º), then it is a straightforward trigonometric exercise to show that (L′)2 = r2 + (d/)2 – 2r(d/)cos(θo + 3º) = L2 + d2 – d2cos(3º)+ Ldsin(3º) . since θo = 45º. The difference between L′ (as determined by this expression) and the original spring length L is the amount the spring has been stretched (denoted here as xm). If one plots xm versus L over a range that seems reasonable considering the figure shown in the problem (say, from L = 0.03 m to L = 0.10 m) one quickly sees that xm ≈ 0.00222 m is an excellent approximation (and is very close to what one would get by approximating xm as the arc length of the path made by that upper block corner as the block is turned through 3º, even though this latter procedure should in principle overestimate xm). Using this value of xm with the given spring constant leads to a potential energy of U = k xm2 = 0.00296 J. Setting this equal to the kinetic energy the block has as it passes back through the initial position, we have K = 0.00296 J = I ωm2 where ωm is the maximum angular speed of the block (and is not to be confused with the angular frequency ω of the oscillation, though they are related by ωm = θoω if θo is expressed in radians). The rotational inertia of the block is I = Md2 = 0.0018 kg·m2. Thus, we can solve the above relation for the maximum angular speed of the block: 2 2 2(0.00296 J) 1.81 rad/s 0.0018 kg mm K I ω = = = ⋅ . Therefore the angular frequency of the oscillation is ω = ωm/θo = 34.6 rad/s. Using Eq. 15-5, then, the period is T = 0.18 s. 154 51. 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 will oscillate in simple harmonic motion. If τ = –Cθ, where τ is the torque, θ is the angle of rotation, and C is a constant of proportionality, then the angular frequency of oscillation is ω = C I/ and the period is T I C= =2 2π π/ /ω , where I is the rotational inertia of the rod. The plan is to find the torque as a function of θ 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 θ, with the left end moving away from the wall. This end is now (L/2) sin θ further from the wall and has moved a distance (L/2)(1 – cos θ) to the right. The length of the spring is now 2 2 2 0( / 2) (1 cos ) [ ( / 2)sin ]L Lθ θ= − + +l l . If the angle θ is small we may approximate cos θ with 1 and sin θ with θ in radians. Then the length of the spring is given by 0 / 2Lθ≈ +l l and its elongation is ∆x = Lθ/2. The force it exerts on the rod has magnitude F = k∆x = kLθ/2. Since θ is small we may approximate the torque exerted by the spring on the rod by τ = –FL/2, where the pivot point was taken as the origin. Thus τ = –(kL2/4)θ. 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 its center is I = mL2/12, where m is its mass. See Table 10-2. Thus the period of oscillation is T I C mL kL m k = = =2 2 12 4 2 3 2 2π π π / / . With m = 0.600 kg and k = 1850 N/m, we obtain T = 0.0653 s. 52. (a) For the “physical pendulum” we have T = 2 π = 2 π . If we substitute r for h and use item (i) in Table 10-2, we have 2 22 12 a bT r rg π += + In the figure below, we plot T as a function of r, for a = 0.35 m and b = 0.45 m. 155 CHAPTER 15 (b) The minimum of T can be located by setting its derivative to zero, / 0dT dr = . This yields 2 2 2 2(0.35 m) (0.45 m) 0.16 m. 12 12 a br + += = = (c) The direction from the center does not matter, so the locus of points is a circle around the center, of radius [(a2 + b2)/12]1/2. 53. Replacing x and v in Eq. 15-3 and Eq. 15-6 with θ and dθ/dt, respectively, we identify 4.44 rad/s as the angular frequency ω. Then we evaluate the expressions at t = 0 and divide the second by the first: = − ω tanφ . (a) The value of θ at t = 0 is 0.0400 rad, and the value of dθ/dt then is –0.200 rad/s, so we are able to solve for the phase constant: φ = tan−1[0.200/(0.0400 x 4.44)] = 0.845 rad. (b) Once φ is determined we can plug back in to θo = θmcosφ to solve for the angular amplitude. We find θm = 0.0602 rad. 54. We note that the initial angle is θo = 7º = 0.122 rad (though it turns out this value will cancel in later calculations). If we approximate the initial stretch of the spring as the arc- length that the corresponding point on the plate has moved through (x = r θo where r = 0.025 m) then the initial potential energy is approximately kx2 = 0.0093 J. This should equal to the kinetic energy of the plate ( I ωm2 where this ωm is the maximum angular speed of the plate, not the angular frequency ω). Noting that the maximum angular speed of the plate is ωm = ωθo where ω = 2π/T with T = 20 ms = 0.02 s as determined from the graph, then we can find the rotational inertial from I ωm2 = 0.0093 J. Thus, 5 21.3 10 kg mI −= × ⋅ . 156 ω ′ ≈ ≈ 49000 500 , N / m kg 9.9 rad / s so that T ≈ 0.63 s. Taking the (natural) log of both sides of the above equation, and rearranging, we find ( ) ( ) 32 500 kg2 ln2 0.69 1.1 10 kg/s. 0.63 s mb T = ≈ = × Note: if one worries about the ω´ ≈ ω approximation, it is quite possible (though messy) to use Eq. 15-43 in its full form and solve for b. The result would be (quoting more figures than are significant) = 2 2 ( 2) + 4 = 1086 2 2 b mkln ln kg / s π which is in good agreement with the value gotten “the easy way” above. 61. (a) We set ω = ωd and find that the given expression reduces to xm = Fm/bω at resonance. (b) In the discussion immediately after Eq. 15-6, the book introduces the velocity amplitude vm = ωxm. Thus, at resonance, we have vm = ωFm/bω = Fm/b. 62. With ω = 2π/T then Eq. 15-28 can be used to calculate the angular frequencies for the given pendulums. For the given range of 2.00 < ω < 4.00 (in rad/s), we find only two of the given pendulums have appropriate values of ω: pendulum (d) with length of 0.80 m (for which ω = 3.5 rad/s) and pendulum (e) with length of 1.2 m (for which ω = 2.86 rad/s). 63. With M = 1000 kg and m = 82 kg, we adapt Eq. 15-12 to this situation by writing 2 4 k T M m πω = = + . If d = 4.0 m is the distance traveled (at constant car speed v) between impulses, then we may write T = v/d, in which case the above equation may be solved for the spring constant: ( ) 22 2= 4 . 4 v k vk M m d M m d π π ⇒ = +  +   159 CHAPTER 15 Before the people got out, the equilibrium compression is xi = (M + 4m)g/k, and afterward it is xf = Mg/k. Therefore, with v = 16000/3600 = 4.44 m/s, we find the rise of the car body on its suspension is = 4 = 4 + 4 2 = 0.050 . 2 x x mg k mg M m d vi f − FH IKπ m 64. Its total mechanical energy is equal to its maximum potential energy kxm2, and its potential energy at t = 0 is kxo2 where xo = xmcos(π/5) in this problem. The ratio is therefore cos2(π/5) = 0.655 = 65.5%. 65. (a) From the graph, we find xm = 7.0 cm = 0.070 m, and T = 40 ms = 0.040 s. Thus, the angular frequency is ω = 2π/T = 157 rad/s. Using m = 0.020 kg, the maximum kinetic energy is then mv2 = m ω2 xm2 = 1.2 J. (b) Using Eq. 15-5, we have f = ω/2π = 50 oscillations per second. Of course, Eq. 15-2 can also be used for this. 66. (a) From the graph we see that xm = 7.0 cm = 0.070 m and T = 40 ms = 0.040 s. The maximum speed is xmω = xm2π/T = 11 m/s. (b) The maximum acceleration is xmω2 = xm(2π/T)2 = 1.7 × 103 m/s2. 67. Setting 15 mJ (0.015 J) equal to the maximum kinetic energy leads to vmax = 0.387 m/s. Then one can use either an “exact” approach using vmax = or the “SHM” approach where vmax = Lωmax = Lωθmax = Lθmax to find L. Both approaches lead to L = 1.53 m. 68. Since ω = 2πf where f = 2.2 Hz, we find that the angular frequency is ω = 13.8 rad/s. Thus, with x = 0.010 m, the acceleration amplitude is am = xm ω 2 = 1.91 m/s2. We set up a ratio: = = 1.91 9.8 = 0.19 . a a g g g gm m F HG I KJ FH IK 69. (a) Assume the bullet becomes embedded and moves with the block before the block moves a significant 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 final speed of the block and bullet. Conservation of momentum yields mv0 = (m + M)v, so 160 = + = 0.050 150 0.050 + 4.0 = 1.85 . 0v mv m M kg m / s kg kg m / s a fa f When the block is in its initial position the spring and gravitational forces balance, so the spring is elongated by Mg/k. After the collision, however, the block oscillates with simple harmonic motion about the point where the spring and gravitational forces balance with the bullet embedded. At this point the spring is elongated a distance  = +M m g ka f / , somewhat different from the initial elongation. Mechanical energy is conserved during the oscillation. At the initial position, just after the bullet is embedded, the kinetic energy is 12 2( )M m v+ and the elastic potential energy is 12 2k Mg k( / ) . We take the gravitational potential energy to be zero at this point. When the block and bullet reach the highest point in their motion the kinetic energy is zero. The block is then a distance ym above the position where the spring and gravitational forces balance. Note that ym is the amplitude of the motion. The spring is compressed by ym −  , so the elastic potential energy is 1 2 2k ym( )−  . The gravitational potential energy is (M + m)gym. Conservation of mechanical energy yields 1 2 + + 1 2 = 1 2 + + . 2 2 2M m v k Mg k k y M m gym ma f b g a fFH IK −  We substitute  = +M m g ka f / . Algebraic manipulation leads to y m M v k mg k M mm = + − + = + − + = a f a f a fa f a fc h a f 2 2 2 2 2 2 2 0 050 4 0 185 500 0 050 9 8 500 2 4 0 0 050 0 166 . . . . . ( ) . . . . kg kg m / s N / m kg m / s N / m kg kg m 2 (b) The original energy of the bullet is E mv0 12 0 2 1 2 20 050 150 563= = =( . )( )kg m / s J . The kinetic energy of the bullet-block system just after the collision is = 1 2 + = 1 2 0.050 + 4.0 1.85 = 6.94 . 2 2E m M va f a fa f kg kg m / s J Since the block does not move significantly during the collision, the elastic and gravitational potential energies do not change. Thus, E is the energy that is transferred. The ratio is E/E0 = (6.94 J)/(563 J) = 0.0123 or 1.23%. 70. (a) We note that ω = = = 165.1 rad/s. 161 CHAPTER 15 r R R R R R= ± − = 3 3 8 4 2 2 2a f or . Thus, our result is r = 0.126/2 = 0.0630 m. 75. (a) The frequency for small amplitude oscillations is f g L= 1 2/ /πa f , where L is the length of the pendulum. This gives f = =1 2 9 80 2 0 0 352/ ( . / ) / ( . ) . .πa f m s m Hz (b) The forces acting on the pendulum are the tension force  T of 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 a ae= + ′ , where ae is the acceleration of the elevator and  ′a is the acceleration of the pendulum relative to the elevator. Newton’s second law can then be written ( )em g a T− + = rr r ma′ r . 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 find the frequency for small amplitude oscillations by replacing g with g + ae in the expression f g L= ( / ) /1 2π . Thus f g a L e= + = + =1 2 1 2 9 8 2 0 2 0 0 39 π π . . . . .m / s m / s m Hz 2 2 (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. To find the frequency for small amplitude oscillations, replace g with zero in f g L= ( / ) /1 2π . The result is zero. The pendulum does not oscillate. 76. Since the particle has zero speed (momentarily) at x ≠ 0, then it must be at its turning point; thus, xo = xm = 0.37 cm. It is straightforward to infer from this that the phase constant φ in Eq. 15-2 is zero. Also, f = 0.25 Hz is given, so we have ω = 2πf = π/2 rad/s. The variable t is understood to take values in seconds. (a) The period is T = 1/f = 4.0 s. (b) As noted above, ω = π/2 rad/s. (c) The amplitude, as observed above, is 0.37 cm. (d) Eq. 15-3 becomes x = (0.37 cm) cos(πt/2). (e) The derivative of x is v = –(0.37 cm/s)(π/2) sin(πt/2) ≈ (–0.58 cm/s) sin(πt/2). 164 (f) From the previous part, we conclude vm = 0.58 cm/s. (g) The acceleration-amplitude is am = ω2xm = 0.91 cm/s2. (h) Making sure our calculator is in radians mode, we find x = (0.37) cos(π(3.0)/2) = 0. It is important to avoid rounding off the value of π in order to get precisely zero, here. (i) With our calculator still in radians mode, we obtain v = –(0.58 cm/s)sin(π(3.0)/2) = 0.58 cm/s. 77. Since T = 0.500 s, we note that ω = 2π/T = 4π rad/s. We work with SI units, so m = 0.0500 kg and vm = 0.150 m/s. (a) Since ω = k m/ , the spring constant is ( ) ( )22 4 rad/s 0.0500 kg 7.90 N/m.k mω π= = = (b) We use the relation vm = xmω and obtain = = 0.150 4 = 0.0119 . x vm mω π m (c) The frequency is f = ω/2π = 2.00 Hz (which is equivalent to f = 1/T). 78. (a) Hooke’s law readily yields (0.300 kg)(9.8 m/s2)/(0.0200 m) = 147 N/m. (b) With m = 2.00 kg, the period is = 2 0 733T m k π = . s . 79. Using ∆m = 2.0 kg, T1 = 2.0 s and T2 = 3.0 s, we write T m k T m m k1 2 2 2= = +π πand ∆ . Dividing one relation by the other, we obtain =2 1 T T m m m + ∆ which (after squaring both sides) simplifies to 2 2 1 1.6kg. ( / ) 1 mm T T ∆= = − 165 CHAPTER 15 80. (a) Comparing with Eq. 15-3, we see ω = 10 rad/s in this problem. Thus, f = ω/2π = 1.6 Hz. (b) Since vm = ωxm and xm = 10 cm (see Eq. 15-3), then vm = (10 rad/s)(10 cm) = 100 cm/s or 1.0 m/s. (c) The maximum occurs at t = 0. (d) Since am = ω2xm then vm = (10 rad/s)2(10 cm) = 1000 cm/s2 or 10 m/s2. (e) The acceleration extremes occur at the displacement extremes: x = ±xm or x = ±10 cm. (f) Using Eq. 15-12, we find = 0 10 10 102ω k m k⇒ = =. .kg rad / s N / ma fa f Thus, Hooke’s law gives F = –kx = –10x in SI units. 81. (a) We require U E= 12 at some value of x. Using Eq. 15-21, this becomes 1 2 = 1 2 1 2 = 2 . 2 2kx kx x xm m FHG IKJ⇒ We compare the given expression x as a function of t with Eq. 15-3 and find xm = 5.0 m. Thus, the value of x we seek is x = ≈5 0 2 3 5. / . m. (b) We solve the given expression (with x = 5 0 2. / ), making sure our calculator is in radians mode: = 4 + 3 1 2 = 1.54 . 1t π π cos s− FHG IKJ Since we are asked for the interval teq – t where teq specifies the instant the particle passes through the equilibrium position, then we set x = 0 and find = 4 + 3 0 = 2.29 . 1teq cos s π π − bg Consequently, the time interval is teq – t = 0.75 s. 82. The distance from the relaxed position of the bottom end of the spring to its equilibrium position when the body is attached is given by Hooke’s law: ∆x = F/k = (0.20 kg)(9.8 m/s2)/(19 N/m) = 0.103 m. 166 (d) To find the frequency when spring 2 is attached to the mass, we replace k with k(n + 1) to obtain 2 2 1 ( 1)= 1 0.70 1.0(200 Hz) 2.6 10 Hz. 2 n kf n f mπ + = + = + = × 87. The magnitude of the downhill component of the gravitational force acting on each ore car is = 10000 9.8 2wx kg m / s sinb gc h θ where θ = 30° (and it is important to have the calculator in degrees mode during this problem). We are told that a downhill pull of 3ωx causes the cable to stretch x = 0.15 m. Since the cable is expected to obey Hooke’s law, its spring constant is = 3 = 9.8 10 . 5k w x x × N / m (a) Noting that the oscillating mass is that of two of the cars, we apply Eq. 15-12 (divided by 2π). 51 1 9.8 10 N / m 1.1 Hz. 2 2 2 20000 kg kf m ω π π π ×= = = = (b) The difference between the equilibrium positions of the end of the cable when supporting two as opposed to three cars is = 3 2 = 0.050 . ∆ −x w w k x x m 88. Since the centripetal acceleration is horizontal and Earth’s gravitational g is downward, we can define the magnitude of an “effective” gravitational acceleration using the Pythagorean theorem: 2 2 2( / ) . eff g g v R= + Then, since frequency is the reciprocal of the period, Eq. 15-28 leads to f g L g v R L eff= = +1 2 1 2 2 4 2 π π . With v = 70 m/s, R = 50m, and L = 0.20 m, we have 13.5 s 3.5 Hz.f −≈ = 89. (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 = 0.096 m is the 169 CHAPTER 15 elongation of the spring at equilibrium, k is the spring constant, and m = 1.3 kg is the mass of the block. Thus k = mg/y = (1.3 kg)(9.8 m/s2)/(0.096 m) = 1.33 ×102 N/m. (b) The period is given by 1 2 1.3 kg2 2 0.62 s. 133 N / m mT f k π π π ω = = = = = (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. At the initial position, the block is not moving but it has potential energy ( ) ( ) ( ) ( ) ( ) 22 21 11.3 kg 9.8 m/s 0.146 m 133 N / m 0.146 m 0.44 J. 2 2i i i U mgy ky= − + = − + = − When the block is at the equilibrium point, the elongation of the spring is y = 9.6 cm and the potential energy is ( ) ( ) ( ) ( ) ( ) 22 21 11.3 kg 9.8 m/s 0.096 m 133 N / m 0.096 m 0.61 J. 2 2f U mgy ky= − + = − + = − We write the equation for conservation of energy as U U mvi f= + 12 2 and solve for v: ( ) ( )2 2 0.44J 0.61J 0.51 m/s. 1.3kg i fU Uv m − − + = = = 90. (a) The Hooke’s law force (of magnitude (100)(0.30) = 30 N) is directed upward and the weight (20 N) is downward. Thus, the net force is 10 N upward. (b) The equilibrium position is where the upward Hooke’s law force balances the weight, which corresponds to the spring being stretched (from unstretched length) by 20 N/100 N/m = 0.20 m. Thus, relative to the equilibrium position, the block (at the instant described in part (a)) is at what one might call the bottom turning point (since v = 0) at x = –xm where the amplitude is xm = 0.30 – 0.20 = 0.10 m. (c) Using Eq. 15-13 with m = W/g ≈ 2.0 kg, we have 170 = 2 0 90T m k π = . .s (d) The maximum kinetic energy is equal to the maximum potential energy 12 2kxm . Thus, = = 1 2 100 0.10 = 0.50 . 2K Um m N / m m Ja fa f 91. We note that for a horizontal spring, the relaxed position is the equilibrium position (in a regular simple harmonic motion setting); thus, we infer that the given v = 5.2 m/s at x = 0 is the maximum value vm (which equals ωxm where ω = =k m/ 20 rad / s ). (a) Since ω = 2π f, we find f = 3.2 Hz. (b) We have vm = 5.2 m/s = (20 rad/s)xm, which leads to xm = 0.26 m. (c) With meters, seconds and radians understood, (0.26 m)cos(20 ) (5.2 m/s)sin(20 ). x t v t φ φ = + = − + The requirement that x = 0 at t = 0 implies (from the first equation above) that either φ = +π/2 or φ = –π/2. Only one of these choices meets the further requirement that v > 0 when t = 0; that choice is φ = –π/2. Therefore, ( )(0.26 m)cos 20 (0.26 m)sin 20 . 2 x t tπ = − =   92. (a) Eq. 15-21 leads to 1 2 2(4.0 J)2 0.20 m. 2 200 N / mm m EE kx x k = ⇒ = = = (b) Since 2 / 2 0.80 kg / 200 N / m 0.4 s ,T m kπ π= = ≈ then the block completes 10/0.4 = 25 cycles during the specified interval. (c) The maximum kinetic energy is the total energy, 4.0 J. (d) This can be approached more than one way; we choose to use energy conservation: = + 4.0 = 1 2 + 1 2 . 2 2E K U mv kx⇒ 171 CHAPTER 15 100. We note (from the graph) that am = ω2xm = 4.00 cm/s2. Also the value at t = 0 is ao = 1.00 cm/s2. Then Eq. 15-7 leads to φ = cos−1(–1.00/4.00) = +1.82 rad or – 4.46 rad. The other “root” (+4.46 rad) can be rejected on the grounds that it would lead to a negative slope at t = 0. 101. (a) The graphs suggest that T = 0.40 s and κ = 4/0.2 = 0.02 N·m/rad. With these values, Eq. 15-23 can be used to determine the rotational inertia: I = κT2/4π2 = 8.11 × 10−5 kg.m2. (b) We note (from the graph) that θmax = 0.20 rad. Setting the maximum kinetic energy ( Iωmax2 ) equal to the maximum potential energy (see the hint in the problem) leads to ωmax = θmax= 3.14 rad/s. 102. The angular frequency of the simple harmonic oscillation is given by Eq. 15-13: k m ω = . Thus, for two different masses 1m and 2m , with the same spring constant k, the ratio of the frequencies would be 11 2 2 12 / / k m m mk m ω ω = = . In our case, with 1m m= and 2 2.5m m= , the ratio is 1 2 2 1 2.5 1.58m m ω ω = = = . 103. For simple harmonic motion, Eq. 15-24 must reduce to τ θ θ= − → −L F L Fg gsinc h c h where θ is in radians. We take the percent difference (in absolute value) − − − − = − LF LF LF g g g sin sin sin θ θ θ θ θ d i d i 1 and set this equal to 0.010 (corresponding to 1.0%). In order to solve for θ (since this is not possible “in closed form”), several approaches are available. Some calculators have built-in numerical routines to facilitate this, and most math software packages have this capability. Alternatively, we could expand sinθ ≈ θ – θ 3/6 (valid for small θ) and thereby 174 find an approximate solution (which, in turn, might provide a seed value for a numerical search). Here we show the latter approach: 1 6 0 010 1 1 6 10103 2− − ≈ ⇒ − ≈θ θ θ θ/ . . which leads to 6(0.01/1.01) 0.24 rad 14.0θ ≈ = = ° . A more accurate value (found numerically) for the θ value which results in a 1.0% deviation is 13.986°. 104. (a) The graph makes it clear that the period is T = 0.20 s. (b) The period of the simple harmonic oscillator is given by Eq. 15-13: = 2T m k π . Thus, using the result from part (a) with k = 200 N/m, we obtain m = 0.203 ≈ 0.20 kg. (c) The graph indicates that the speed is (momentarily) zero at t = 0, which implies that the block is at x0 = ±xm. From the graph we also note that the slope of the velocity curve (hence, the acceleration) is positive at t = 0, which implies (from ma = –kx) that the value of x is negative. Therefore, with xm = 0.20 m, we obtain x0 = –0.20 m. (d) We note from the graph that v = 0 at t = 0.10 s, which implied a = ±am = ±ω2xm. Since acceleration is the instantaneous slope of the velocity graph, then (looking again at the graph) we choose the negative sign. Recalling ω2 = k/m we obtain a = –197 ≈ –2.0 ×102 m/s2. (e) The graph shows vm = 6.28 m/s, so = 1 2 = 4.0 . 2K mvm m J 105. (a) From the graph, it is clear that xm = 0.30 m. (b) With F = –kx, we see k is the (negative) slope of the graph — which is 75/0.30 = 250 N/m. Plugging this into Eq. 15-13 yields = 2 0 28T m k π = . .s (c) As discussed in §15-2, the maximum acceleration is 175 CHAPTER 15 2 2 21.5 10 m/s .m m m ka x x m ω= = = × Alternatively, we could arrive at this result using am = (2π/T)2 xm. (d) Also in §15-2 is vm = ωxm so that the maximum kinetic energy is = 1 2 = 1 2 = 1 2 2 2 2 2K mv m x kxm m m mω which yields 11.3 ≈ 11 J. We note that the above manipulation reproduces the notion of energy conservation for this system (maximum kinetic energy being equal to the maximum potential energy). 106. (a) The potential energy at the turning point is equal (in the absence of friction) to the total kinetic energy (translational plus rotational) as it passes through the equilibrium position: 2 2 2 2 2 2 2 cm cm cm cm 2 2 2 cm cm cm 1 1 1 1 1 1 2 2 2 2 2 2 1 1 3 2 4 4 m vkx Mv I Mv MR R Mv Mv Mv ω   = + = +        = + = which leads to Mv kxmcm 2 22 3= / = 0.125 J. The translational kinetic energy is therefore 1 2 2 2 3 0 0625Mv kxmcm J= =/ . . (b) And the rotational kinetic energy is 2 2 21 cm4 / 6 0.03125J 3.13 10 JmMv kx −= = ≈ × . (c) In this part, we use vcm to denote the speed at any instant (and not just the maximum speed as we had done in the previous parts). Since the energy is constant, then 2 23 1 3 0 4 2 2 dE d dMv kx Mv a kxv dt dt dt    = + = + =      cm cm cm cm which leads to = 2 3 . a k M xcm − FHG IKJ Comparing with Eq. 15-8, we see that ω = 2 3k M/ for this system. Since ω = 2π/T, we obtain the desired result: T M k= 2 3 2π / . 107. (a) From Eq. 16-12, T m k= =2 0π / .45 s. 176 If we expand the plot near the end of that time interval we have This is close enough to a regular sine wave cycle that we can estimate its period (T = 0.18 s, so ω = 35 rad/s) and its amplitude (ym = 0.008 m). (b) Now, with the new driving frequency (ωd = 13.2 rad/s), the x versus t graph (for the first one second of motion) is as shown below: It is a little more difficult in this case to estimate a regular sine-curve-like amplitude and period (for the part of the above graph near the end of that time interval), but we arrive at roughly ym = 0.07 m, T = 0.48 s, and ω = 13 rad/s. 179 CHAPTER 15 (c) Now, with ωd = 20 rad/s, we obtain (for the behavior of the graph, below, near the end of the interval) the estimates: ym = 0.03 m, T = 0.31 s, and ω = 20 rad/s. 180