CHAPTER 14: Oscillations. Answers to Questions. l. The length, l, is the distance from the center of the tire to the branch.

Transcription

1 CHAPTER 4: Osciations Answers to Questions. Eapes are: a chid s swing (SHM, for sa osciations), stereo speaers (copicated otion, the addition of any SHMs), the bade on a jigsaw (approiatey SHM), the string on a guitar (copicated otion, the addition of any SHMs).. The acceeration of a sipe haronic osciator is oentariy zero as the ass passes through the equiibriu point. At this point, there is no force on the ass and therefore no acceeration.. When the engine is running at constant speed, the piston wi have a constant period. The piston has zero veocity at the top and botto of its path. Both of these properties are aso properties of SHM. In addition, there is a arge force eerted on the piston at one etree of its otion, fro the cobustion of the fue air iture, and in SHM the argest forces occur at the etrees of the otion. 4. The true period wi be arger and the true frequency wi be saer. The spring needs to acceerate not ony the ass attached to its end, but aso its own ass. As a ass on a spring osciates, potentia energy is converted into inetic energy. The aiu potentia energy depends on the dispaceent of the ass. This aiu potentia energy is converted into the aiu inetic energy, but if the ass being acceerated is arger then the veocity wi be saer for the sae aount of energy. A saer veocity transates into a onger period and a saer frequency. 5. The aiu speed of a sipe haronic osciator is given by v = A. The aiu speed can be doubed by doubing the apitude, A. 6. Before the trout is reeased, the scae reading is zero. When the trout is reeased, it wi fa downward, stretching the spring to beyond its equiibriu point so that the scae reads soething over 5 g. Then the spring force wi pu the trout bac up, again to a point beyond the equiibriu point, so that the scae wi read soething ess than 5 g. The spring wi undergo daped osciations about equiibriu and eventuay coe to rest at equiibriu. The corresponding scae readings wi osciate about the 5-g ar, and eventuay coe to rest at 5 g. 7. At high atitude, g is sighty saer than it is at sea eve. If g is saer, then the period T of the penduu coc wi be onger, and the coc wi run sow (or ose tie). 8. The tire swing is a good approiation of a sipe penduu. Pu the tire bac a short distance and reease it, so that it osciates as a penduu in sipe haronic otion with a sa apitude. Measure the period of the osciations and cacuate the ength of the penduu fro the epression T = π. The ength,, is the distance fro the center of the tire to the branch. The height of the g branch is pus the height of the center of the tire above the ground. 9. The dispaceent and veocity vectors are in the sae direction whie the osciator is oving away fro its equiibriu position. The dispaceent and acceeration vectors are never in the sae direction. 8 Pearson Education, Inc., Upper Sadde River, NJ. A rights reserved. This ateria is protected under a copyright aws as they currenty eist. No portion of this ateria ay be reproduced, in any for or by any eans, without perission in writing fro the 448

2 Giancoi Physics for Scientists & Engineers, 4 th Edition. The period wi be unchanged, so the tie wi be (c), two seconds. The period of a sipe penduu osciating with a sa apitude does not depend on the ass.. The two asses reach the equiibriu point siutaneousy. The anguar frequency is independent of apitude and wi be the sae for both systes.. Epty. The period of the osciation of a spring increases with increasing ass, so when the car is epty the period of the haronic otion of the springs wi be shorter, and the car wi bounce faster.. When waing at a nora pace, about s (tied). The faster you wa, the shorter the period. The shorter your egs, the shorter the period. 4. When you rise to a standing position, you raise your center of ass and effectivey shorten the ength of the swing. The period of the swing wi decrease. 5. The frequency wi decrease. For a physica penduu, the period is proportiona to the square root of the oent of inertia divided by the ass. When the sa sphere is added to the end of the rod, both the oent of inertia and the ass of the penduu increase. However, the increase in the oent of inertia wi be greater because the added ass is ocated far fro the ais of rotation. Therefore, the period wi increase and the frequency wi decrease. 6. When the 64-Hz for is set into vibration, the sound waves generated are cose enough in frequency to the resonance frequency of the 6-Hz for to cause it to vibrate. The 4-Hz for has a resonance frequency far fro 64 Hz and far fro the haronic at 58 Hz, so it wi not begin to vibrate. 7. If you shae the pan at a resonant frequency, standing waves wi be set up in the water and it wi sosh bac and forth. Shaing the pan at other frequencies wi not create arge waves. The individua water oecues wi ove but not in a coherent way. 8. Eapes of resonance are: pushing a chid on a swing (if you push at one of the iits of the osciation), bowing across the top of a botte, producing a note fro a fute or organ pipe. 9. Yes. Rattes which occur ony when driving at certain speeds are ost iey resonance phenoena.. Buiding with ighter aterias doesn t necessariy ae it easier to set up resonance vibrations, but it does shift the fundaenta frequency and decrease the abiity of the buiding to dapen osciations. Resonance vibrations wi be ore noticeabe and ore iey to cause daage to the structure. Soutions to Probes. The partice woud trave four ties the apitude: fro = A to = to = A to = to = A. So the tota distance = 4A = 4(.8) =.7.. The spring constant is the ratio of eterna appied force to dispaceent. 8 Pearson Education, Inc., Upper Sadde River, NJ. A rights reserved. This ateria is protected under a copyright aws as they currenty eist. No portion of this ateria ay be reproduced, in any for or by any eans, without perission in writing fro the 449

3 Chapter 4 Osciations Fet 8 N 75 N 5 N = = = = 55 N 5 N The spring constant is found fro the ratio of appied force to dispaceent. F g ( 68 g)( 9.8 s ) et 5 = = = =. N 5. The frequency of osciation is found fro the tota ass and the spring constant. 5. N.467 Hz.5 Hz f = = = π π 568 g 4. (a) The otion starts at the aiu etension, and so is a cosine. The apitude is the dispaceent at the start of the otion. π π = Acos ( ωt) = Acos t = ( 8.8c) cos t = ( 8.8c) cos ( 9.5t) T.66 ( 8.8c) cos( 9.5t) (b) Evauate the position function at t =.8 s. = 8.8c cos 9.5s.8s =.5 c.c 5. The period is. seconds, and the ass is 5 g. The spring constant can be cacuated fro Eq. 4-7b. 5g T = T = = = = T π 4 π 5N (.s) 6. (a) The spring constant is found fro the ratio of appied force to dispaceent. F g (.4 g)( 9.8 s ) 65N et = = = =.6 65 N (b) The apitude is the distance pued down fro equiibriu, so A =.5c The frequency of osciation is found fro the osciating ass and the spring constant. 65N f = = =.65Hz.6 Hz π π.4g 7. The aiu veocity is given by Eq. 4-9a. π A π (.5) v = ωa= = =. s a T 7.s The aiu acceeration is given by Eq. 4-9b. A (.5) a = ω A= = =.9 s. s a T 7.s a a g.9 s = = = 9.8 s..% 8 Pearson Education, Inc., Upper Sadde River, NJ. A rights reserved. This ateria is protected under a copyright aws as they currenty eist. No portion of this ateria ay be reproduced, in any for or by any eans, without perission in writing fro the 45

4 Giancoi Physics for Scientists & Engineers, 4 th Edition 8. The tabe of data is shown, aong with the soothed graph. Every quarter of a period, the ass oves fro an etree point to the equiibriu. The tie position - A T/4 T/ A T/4 T - A 5T/4 graph resebes a cosine wave (actuay, the opposite of a cosine wave). position / A tie / T 9. The reationship between frequency, ass, and spring constant is Eq. 4-7a, f (a) f π f π ( 4 ) =. π = = 4 = 4 4. Hz.5 g =.579 N.6 N π.579n (b) f = = = 4.8 Hz π π 5. g. The spring constant is the sae regardess of what ass is attached to the spring. f = = f = constant f = f π (.68 g)(.6 Hz) (.8Hz) (.6 Hz) g.8hz = g +.68 g.6 Hz = =.74 g. We assue that the spring is stretched soe distance y whie the rod is in equiibriu and horizonta. Cacuate the net torque F s about point A whie the object is in equiibriu, with cocwise A θ torques as positive. τ = Mg ( ) F = M g y = s Mg Now consider the rod being dispaced an additiona distance y beow the horizonta, so that the rod aes a sa ange of θ as shown in the free-body diagra. Again write the net torque about point A. If the ange is sa, then there has been no appreciabe horizonta dispaceent of the rod. d θ Mg F Mg y y I M dt τ = ( ) = s ( + ) = α = Incude the equiibriu condition, and the approiation that y = sin θ θ. d θ d θ = = dt dt Mg y y M Mg y Mg M d θ d θ θ + = M θ = dt dt M 8 Pearson Education, Inc., Upper Sadde River, NJ. A rights reserved. This ateria is protected under a copyright aws as they currenty eist. No portion of this ateria ay be reproduced, in any for or by any eans, without perission in writing fro the 45

5 Chapter 4 Osciations This is the equation for sipe haronic otion, corresponding to Eq. 4-, with ω =. M ω = 4 π f = f = M π M. (a) We find the effective spring constant fro the ass and the frequency of osciation. f = π = π f = π = g.Hz 9.54N N sig fig (b) Since the objects are the sae size and shape, we anticipate that the spring constant is the sae. 9.54N f = = =.4 Hz π π.5g. (a) For A, the apitude is A =.5. For B, the apitude is A =.5. A B (b) For A, the frequency is cyce every 4. seconds, so f =.5Hz A. For B, the frequency is cyce every. seconds, so f =.5 Hz B. (c) For C, the period is T = 4.s. For B, the period is T =.s A B (d) Object A has a dispaceent of when t =, so it is a sine function. ( π ) ( π ) = A sin f t =.5 sin t A A A A Object B has a aiu dispaceent when t =, so it is a cosine function. ( π ) ( π ) = A cos f t =.5 cos t B B B B 4. Eq. 4-4 is = Acos( ωt+ φ). (a) If = A, then A= Acos φ φ = cos φ = π. (b) If =, then (c) If = Acos φ φ = cos φ =± π. = A, then A= Acos φ φ = cos () φ =. (d) If = A, then A= Acos φ φ = cos φ =± π. (e) If = A, then (f) If A A= Acos φ φ = cos φ =± π. =, then A = Acos φ φ = cos φ =± π. 4 The abiguity in the answers is due to not nowing the direction of otion at t =. 5. We assue that downward is the positive direction of otion. For this otion, we have = 5 N, A.8,.6 g, = = and ω = = 5 N.6 g = 4.5 rad s. (a) Since the ass has a zero dispaceent and a positive veocity at t =, the equation is a sine function. () = (.8 ) sin [( 4.rad s) t] y t 8 Pearson Education, Inc., Upper Sadde River, NJ. A rights reserved. This ateria is protected under a copyright aws as they currenty eist. No portion of this ateria ay be reproduced, in any for or by any eans, without perission in writing fro the 45

6 Giancoi Physics for Scientists & Engineers, 4 th Edition π π (b) The period of osciation is given by T = = =.845s. The spring wi have ω 4.5 rad s its aiu etension at ties given by the foowing. T t = + nt = 4.59 s + n a (.8 s ), n =,,, 4 The spring wi have its iniu etension at ties given by the foowing. T t = + nt =.8 s + n in (.8 s ), n =,,, 4 6. (a) Fro the graph, the period is.69 s. The period and the ass can be used to find the spring constant..95g T = = = = T π.7877n.79n (.69s) (b) Fro the graph, the apitude is.8 c. The phase constant can be found fro the initia conditions. π π = Acos t+ φ = (.8c) cos t+ φ T.69.4 = (.8c) cosφ =.4c φ = cos = ±.rad.8 Because the graph is shifted to the RIGHT fro the -phase cosine, the phase constant ust be subtracted. π = t t.69 (.8 c) cos. or (.8 c) cos( 9..) 7. (a) The period and frequency are found fro the anguar frequency. 5π 5 ω = π f f = ω Hz T.6 s π = π 4 = 8 = f = (b) The veocity is the derivative of the position. 5π π d 5π 5π π = (.8 ) cos t+ v = = (.8 ) sin t+ 4 6 dt π 5π π = (.8) cos =. v = (.8) sin = 7.5 s (c) The acceeration is the derivative of the veocity. 5π 5π π dv 5π 5π π v = (.8 ) sin t+ a = = (.8 ) cos t dt π 5π π v (.) = (.8 ) sin (.) + = s a (.) 5π 5π π =.8 cos. + = 9 s Pearson Education, Inc., Upper Sadde River, NJ. A rights reserved. This ateria is protected under a copyright aws as they currenty eist. No portion of this ateria ay be reproduced, in any for or by any eans, without perission in writing fro the 45

7 Chapter 4 Osciations 8. (a) The aiu speed is given by Eq. 4-9a. v = π f A= π 44Hz.5 = 4. s. a (b) The aiu acceeration is given by Eq. 4-9b. a Hz.5. s a 4 = π f A= π =. 9. When the object is at rest, the agnitude of the spring force is equa to the force of gravity. This deterines the spring constant. The period can then be found. g F = g = vertica T =.4 = = = g g 9.8 s = π π π π.75s. The spring constant can be found fro the stretch distance corresponding to the weight suspended on the spring. F g (.6 g)( 9.8 s ) et = = = = N After being stretched further and reeased, the ass wi osciate. It taes one-quarter of a period for the ass to ove fro the aiu dispaceent to the equiibriu position. π.6 g T = π = =.s 7.84N 4 4. Each object wi pass through the origin at the ties when the arguent of its sine function is a utipe of π. 5 7 A:. t = n π t = n π, n =,,, so t = π, π, π, π, π, π, π,4 π, A A A A A A B:. t = n π t = n π, n =,,, so t = π, π, π, π, π, π, π, π, π, B B B B B B Thus we see the first three ties are πs, πs, π s or.s, 6.s, 9.4s.. (a) The object starts at the aiu dispaceent in the positive direction, and so wi be represented by a cosine function. The ass, period, and apitude are given. A =.6 ; ω = π = π =.4 rad s y = (.6) cos ( t) T.55s (b) The tie to reach the equiibriu is one-quarter of a period, so (.55s 4 ) =.4s. (c) The aiu speed is given by Eq. 4-9a. v = ωa=.4 rad s.6 =.8 s a (d) The aiu acceeration is given by Eq. 4-9b. a = ω A=.4 rad s.6 =. s a The aiu acceeration occurs at the endpoints of the otion, and is first attained at the reease point. 8 Pearson Education, Inc., Upper Sadde River, NJ. A rights reserved. This ateria is protected under a copyright aws as they currenty eist. No portion of this ateria ay be reproduced, in any for or by any eans, without perission in writing fro the 454

8 Giancoi Physics for Scientists & Engineers, 4 th Edition 4. s. The period of the juper s otion is T = = 5.75 s. The spring constant can then be found 8 cyces fro the period and the juper s ass. ( 5.75s) 65.g T = π = = = 88.8N 88.8N T The stretch of the bungee cord needs to provide a force equa to the weight of the juper when he is at the equiibriu point. ( 65. g)( 9.8 s ) g = g = = = N Thus the unstretched bungee cord ust be = Consider the first free-body diagra for the boc whie it is at equiibriu, so that the net force is zero. Newton s second aw for vertica forces, with up as positive, gives this. F = F + F g = F + F = g y A B A B Now consider the second free-body diagra, in which the g boc is dispaced a distance fro the equiibriu point. Each upward force wi have increased by an aount, since <. Again write Newton s second aw for vertica forces. F = F = F + F g = F + F g = + F + F g = y net A B A B A B This is the genera for of a restoring force that produces SHM, with an effective spring constant of. Thus the frequency of vibration is as foows. f = = effective π π 5. (a) If the boc is dispaced a distance to the right in Figure 4-a, then the ength of spring # wi be increased by a distance and the ength of spring # wi be increased by a distance, where = +. The force on the boc can be written F =. Because the springs eff are assess, they act siiar to a rope under tension, and the sae force F is eerted by each spring. Thus F = = =. eff F F F = + = = F + = = + eff eff T = π = π + eff (b) The boc wi be in equiibriu when it is stationary, and so the net force at that ocation is zero. Then, if the boc is dispaced a distance to the right in the diagra, then spring # wi eert an additiona force of F =, in the opposite direction to. Liewise, spring # wi eert an additiona force F =, in the sae direction as F. Thus the net force on the F A F F B F A B g 8 Pearson Education, Inc., Upper Sadde River, NJ. A rights reserved. This ateria is protected under a copyright aws as they currenty eist. No portion of this ateria ay be reproduced, in any for or by any eans, without perission in writing fro the 455

9 Chapter 4 Osciations F = F + F = = +. The effective spring constant is thus dispaced boc is = +, and the period is given by T = π = π The ipuse, which acts for a very short tie, changes the oentu of the ass, giving it an initia veocity v. Because this occurs at the equiibriu position, this is the aiu veocity of the ass. Since the otion starts at the equiibriu position, we represent the otion by a sine function. J J = p = v = v = v v = = v = Aω = A a J J J = A A= = Asinωt = sin t 7. The various vaues can be found fro the equation of otion, = Acosωt =.65cos7.4 t. (a) The apitude is the aiu vaue of, and so A =.65. ω 7.4 rad s (b) The frequency is f = = =.8 Hz. π π rad (c) The tota energy can be found fro the aiu potentia energy. ω.5g 7.4 rad s.65.j.j a E = U = A = A = = (d) The potentia energy can be found fro U =, and the inetic energy fro E = U + K. = = ω = = U.5g 7.4 rad s.6.j K = E U =.J.J =. J 8. (a) The tota energy is the aiu potentia energy. U = E = A = A A.77 (b) Now we are given that = A U = = = E A A 9. 8 Thus the energy is divided up into potentia and inetic The tota energy can be found fro the spring constant and the apitude. E = A = 95 N. =.9 J That is represented by the horizonta ine on the graph. (a) Fro the graph, at =.5c, we have U.J. (b) Fro energy conservation, at =.5c, we have K = E U =.8J. 8 Pearson Education, Inc., Upper Sadde River, NJ. A rights reserved. This ateria is protected under a copyright aws as they currenty eist. No portion of this ateria ay be reproduced, in any for or by any eans, without perission in writing fro the 456

10 Giancoi Physics for Scientists & Engineers, 4 th Edition (c) Find the speed fro the estiated inetic energy. K = v K v = = =.5 s.8j.55 g The spreadsheet used for this probe (c) can be found on the Media Manager, with fienae PSE4_ISM_CH4.XLS, on tab Probe (a) At equiibriu, the veocity is its aiu. Use Eq. 4-9a, and reaize that the object can be oving in either direction. a v = ωa= π fa= π.5hz.5 =.56 s v ±.4 s (b) Fro Eq. 4-b, we find the veocity at any position. (. ) (.5) v =± v =± a (.56 s) =±.756 s ±.8 s A (c) E tota = = = v.5g.56 s.974 J.97 J a (d) Since the object has a aiu dispaceent at t =, the position wi be described by the cosine function. (.5) cos ( π(.5 Hz ) ) (.5) cos ( 5.π ) = t = t. The spring constant is found fro the ratio of appied force to dispaceent. F 95. N = = = 54.9 N.75 Assuing that there are no dissipative forces acting on the ba, the eastic potentia energy in the oaded position wi becoe inetic energy of the ba N E = E = v v = = a a a a (.75) =. s i f.6 g. The energy of the osciator wi be conserved after the coision. E = A = + M v v = A + M a a This speed is the speed that the boc and buet have iediatey after the coision. Linear oentu in one diension wi have been conserved during the (assued short tie) coision, and so the initia speed of the buet can be found. p = p v = + M v v before after o a o U (J) M.55g 5 N = A = (.4 ) = 6 s + M.5 g.55 g equib 8 Pearson Education, Inc., Upper Sadde River, NJ. A rights reserved. This ateria is protected under a copyright aws as they currenty eist. No portion of this ateria ay be reproduced, in any for or by any eans, without perission in writing fro the 457

11 Chapter 4 Osciations. To copare the tota energies, we can copare the aiu potentia energies. Since the frequencies and the asses are the sae, the spring constants are the sae. E A A A high high high high energy energy energy energy = = = 5 = 5 E A A A ow ow ow ow energy energy energy energy 4. (a) The spring constant can be found fro the ass and the frequency of osciation. ω = = π = π = π = (b) The energy can be found fro the aiu potentia energy. E f 4 f 4.Hz.4g 85.7N 85N = A = = 85.7 N J.86 J 5. (a) The wor done in copressing the spring is stored as potentia energy. The copressed ocation corresponds to the aiu potentia energy and the apitude of the ensuing otion. W (.6J) W = A = = = 46 N 4 N A. (b) The aiu acceeration occurs at the copressed ocation, where the spring is eerting the aiu force. If the copression distance is positive, then the acceeration is negative. ( 46 N )(.) F = = a = = =.7 g a 5 s 6. (a) The tota energy of an object in SHM is constant. When the position is at the apitude, the speed is zero. Use that reationship to find the apitude. E = v + = A tot.7 g A= v + = (.55 s) + (. ) = N (b) Again use conservation of energy. The energy is a inetic energy when the object has its aiu veocity. E = v + = A = v tot a 8 N v = A = a ( ) =.5865 s.59 s.7 g 7. We assue that the coision of the buet and boc is so quic that there is no significant otion of the arge ass or spring during the coision. Linear oentu is conserved in this coision. The speed that the cobination has right after the coision is the aiu speed of the osciating syste. Then, the inetic energy that the cobination has right after the coision is stored in the spring when it is fuy copressed, at the apitude of its otion. p = p v = before after ( + M) v v = v a a + M + M v = A + M v = A + M a 8 Pearson Education, Inc., Upper Sadde River, NJ. A rights reserved. This ateria is protected under a copyright aws as they currenty eist. No portion of this ateria ay be reproduced, in any for or by any eans, without perission in writing fro the 458

12 Giancoi Physics for Scientists & Engineers, 4 th Edition 9.8 s ( 9.46 ) ( 7.87 g) A v = ( + M) = + = ( 4.7 N )( 7.87 g g) 8. The hint says to integrate Eq. 4-a, which coes fro the conservation of energy. Let the initia position of the osciator be. t d d d cos = cos + cos = ± ( A ) ( A ) v =± A = =± dt =± dt dt t A A A Mae these definitions: ω ; cos. Then we have the foowing. A φ cos + cos = ± t cos + t Acos ( t ) A A A φ = ± ω = ± ω + φ The phase ange definition coud be changed so that the function is a sine instead of a cosine. And the ± sign can be resoved if the initia veocity is nown. 9. (a) Find the period and frequency fro the ass and the spring constant..785g T = π = π =.44s.4s 84 N 84N f = = = =.47 Hz.44 Hz T π π.785g (b) The initia speed is the aiu speed, and that can be used to find the apitude. v a = A A= v =.6 s.785g 84 N = a (c) The aiu acceeration can be found fro the ass, spring constant, and apitude a = A = N.785g = 4.6 s a (d) Because the ass started at the equiibriu position of =, the position function wi be proportiona to the sine function. (.48 ) sin[ π(.47 Hz ) ] (.48 ) sin ( 4.87π ) = t = t (e) The aiu energy is the inetic energy that the object has when at the equiibriu position. (f) E = v =.785g.6 s =.J a Use the conservation of echanica energy for the osciator. (.4 ) E = + v = A A + K = A K = A = =.4.J.84.68J 8 Pearson Education, Inc., Upper Sadde River, NJ. A rights reserved. This ateria is protected under a copyright aws as they currenty eist. No portion of this ateria ay be reproduced, in any for or by any eans, without perission in writing fro the 459

13 Chapter 4 Osciations 4. We sove this using conservation of energy, equating the energy at the copressed point with the energy as the ba eaves the auncher. Tae the ocation for gravitationa potentia energy to be at the eve where the ba is on the copressed spring. The ocation for eastic potentia energy is the uncopressed position of the spring. Initiay, the ba has ony eastic potentia energy. At the point where the spring is uncopressed and the ba just eaves the spring, there wi be gravitationa potentia energy, transationa inetic energy, and rotationa inetic energy. The ba is roing without sipping. v E = E = gh+ v + Iω = gsin θ + v + i f ( 5) r r 7.5 g 7 = ( gsinθ + v ) = 9.8 s.6 sin5 +. s (.6 ) = 89.6N 9 N sig. fig. 4. The period of a penduu is given by T = π L g. The ength is assued to be the sae for the penduu both on Mars and on Earth. T T Mars Mars Earth = = = π L g g Earth Earth = T = = Mars Earth g Mars T T π Lg π Lg Earth.5s.s.7 5s 4. (a) The period is given by T = =.6s. cyces cyces (b) The frequency is given by f = =.64 Hz. 5s 4. We consider this a sipe penduu. Since the otion starts at the apitude position at t =, we ay describe it by a cosine function with no phase ange, θ = θ cosωt. The anguar veocity can a be written as a function of the ength, cos g θ = θ. a t 9.8 s (a) θ ( t =.5s) = cos (.5s) = 5.4. (b) ( t ) g g Mars 9.8 s θ =.45s = cos.45s = s θ = 6.s = cos 6.s =. (c) ( t ) 8 Pearson Education, Inc., Upper Sadde River, NJ. A rights reserved. This ateria is protected under a copyright aws as they currenty eist. No portion of this ateria ay be reproduced, in any for or by any eans, without perission in writing fro the 46

14 Giancoi Physics for Scientists & Engineers, 4 th Edition 44. The period of a penduu is given by T = π L g..5 (a) T = π L g = π =.5s 9.8 s (b) If the penduu is in free fa, there is no tension in the string supporting the penduu bob, and so no restoring force to cause osciations. Thus there wi be no period the penduu wi not osciate and so no period can be defined. 45. If we consider the penduu as starting fro its aiu dispaceent, then the equation of otion π t can be written as θ = θ cosωt = θ cos. Sove for the tie for the position to decrease to haf the T apitude. πt πt / / π θ = θ = θ cos = cos = t = T / / 6 T T It taes T for the position to change fro + to +. 5 It taes T for the position to change fro to. Thus it taes T T = T for the position to change fro + to 5. Due to the 4 6 syetric nature of the cosine function, it wi aso tae T for the position to change fro to 5, and so fro + 5 to 5 taes T. The second haf of the cyce wi be identica to the first, 6 and so the tota tie spent between + 5 and 5 is T. So the penduu spends one-third of its tie between + 5 and There are ( 4 h)( 6in h)( 6s in ) = 86,4s in a day. The coc shoud ae one cyce in eacty two seconds (a tic and a toc ), and so the coc shoud ae 4, cyces per day. After one day, the coc in question is 6 seconds sow, which eans that it has ade ess cyces than required for precise tieeeping. Thus the coc is ony aing 4,87 cyces in a day. Accordingy, the period of the coc ust be decreased by a factor of 4,87 4,. 4,87 4,87 T = T π g = π g new od new od 4, 4, 4,87 4,87 = new = od (.99 ) =.994 4, 4, Thus the penduu shoud be shortened by Use energy conservation to reate the potentia energy at the aiu height of the penduu to the inetic energy at the owest point of the swing. Tae the owest point to be the zero ocation for gravitationa potentia energy. See the diagra. E = E K + U = K + U top botto top top botto botto a a + gh = v v = gh = g cosθ h = cosθ θ cosθ 8 Pearson Education, Inc., Upper Sadde River, NJ. A rights reserved. This ateria is protected under a copyright aws as they currenty eist. No portion of this ateria ay be reproduced, in any for or by any eans, without perission in writing fro the 46

15 Chapter 4 Osciations 48. (a) For a physica penduu with the sa ange approiation, we ay appy Eq We need the oent of inertia and the distance fro the suspension point to the center of ass. We approiate the cord as a rod, and find the center of ass reative to the stationary end of the cord. I = I + I = M + = M + ( ) bob cord M + M + h = = = CM M + M + I M + M + T = π = π = π M gh tota + M + g ( M + ) g M + (b) If we use the epression for a sipe penduu we woud have T = π. sipe g fractiona error. π Find the M + M + π π T T M + g g M + sipe M + error = = = = T M + M + M M + g M + ( + ) Note that this is negative, indicating that the sipe penduu approiation is too arge. 49. The baance whee of the watch is a torsion penduu, described by τ = Kθ. A specific torque and anguar dispaceent are given, and so the torsiona constant can be deterined. The anguar K frequency is given by ω =. Use these reationships to find the ass. I θ τ = Kθ K = = τ K K ω = π f = = I r 5. in π 4rad 5. in K π 4rad = = = = π ( ) 4 f r.hz g.4g 5. (a) We ca the upper ass M and the ower ass. Both asses have ength. The period of the physica penduu is given by Eq Note that we ust find both the oent of inertia of the syste about the upperost point, and the center of ass of the syste. The parae ais theore is used to find the oent of inertia. ( ) ( ) 7 I = I + I = M + + = M + upper ower M + M + h = = = CM M + M + M L 8 Pearson Education, Inc., Upper Sadde River, NJ. A rights reserved. This ateria is protected under a copyright aws as they currenty eist. No portion of this ateria ay be reproduced, in any for or by any eans, without perission in writing fro the 46

16 Giancoi Physics for Scientists & Engineers, 4 th Edition 7 [ ( 7.g) + ( 4.g) ](.55 ) ( 7.g) + ( 4.g) 9.8 s [ ] 7 7 I M + M + T = π = π = π M gh tota + M + g ( M + ) g M + = π =.6495s.6s (b) It too 7. seconds for 5 swings, which gives a period of.4 seconds. That is reasonabe quaitative agreeent. 5. (a) In the tet, we are given that τ = Kθ. Newton s second aw for rotation, d θ Eq. -4, says that τ = Iα = I. We assue that the torque appied dt by the twisting of the wire is the ony torque. θ d d K τ = Iα = I =Kθ = θ = ω θ θ dt dt I This is the sae for as Eq. 4-, which is the differentia equation for sipe haronic osciation. We echange variabes with Eq. 4-4, and write the equation for the anguar otion. K = Acos ( ωt+ φ) θ = θ cos ( ωt+ φ), ω = I (b) The period of the otion is found fro the anguar veocity ω. K K π I ω = ω = = T = π I I T K M 5. The eter stic used as a penduu is a physica penduu. The period is given by Eq. 4-4, I T = π. Use the parae ais theore to find the oent of inertia about the pin. Epress gh the distances fro the center of ass. I + h π I = I + h = + h T = π = π = + h CM gh gh g h / dt = π ( ) + h + = h = =.887 dh h h = h = fro the end Use the distance for h to cacuate the period. (.) / π π T = + h.887.5s = + = g h 9.8 s.887 / / 8 Pearson Education, Inc., Upper Sadde River, NJ. A rights reserved. This ateria is protected under a copyright aws as they currenty eist. No portion of this ateria ay be reproduced, in any for or by any eans, without perission in writing fro the 46

17 Chapter 4 Osciations K 5. This is a torsion penduu. The anguar frequency is given in the tet as ω =, where K is the I torsion constant (a property of the wire, and so a constant in this probe). The rotationa inertia of a rod about its center is M. K π I ω = = T = π I T K I π T K I M (.7M )(.7 ) = = = = = T I M M I π K T T = = s =.9s 54. The torsiona constant is reated to the period through the reationship given in probe 5. The rotationa inertia of a dis in this configuration is I = MR. I I MR T = K = = = MR f = K T T π π π (.75g)(.65) (.Hz).7 N rad = i 55. This is a physica penduu. Use the parae ais theore to find the oent of inertia about the pin at point A, and then use Eq. 4-4 to find the period. ( + + ) I = I + Mh = MR + Mh = M R + h pin CM I M R h R h T = π = π = π Mgh Mgh gh R h A M (. ) + (.8 ) ( 9.8 s )(.8) = π =.8s 56. (a) The period of the otion can be found fro Eq. 4-8, giving the anguar frequency for the daped otion. ( i ) b 4. N.66 N s ω = = = rad s 4.85g 4.85g π π T = = =.898s ω 6.996rad s (b) If the apitude at soe tie is A, then one cyce ater, the apitude wi be to find the fractiona change. 8 Pearson Education, Inc., Upper Sadde River, NJ. A rights reserved. This ateria is protected under a copyright aws as they currenty eist. No portion of this ateria ay be reproduced, in any for or by any eans, without perission in writing fro the 464 Ae γt γt b (.66Nis ) T (.898s ) γt.85g fractiona change. Ae A = = e = e = e = A And so the apitude decreases by % fro the previous apitude, every cyce.. Use this

18 Giancoi Physics for Scientists & Engineers, 4 th Edition (c) Since the object is at the origin at t =, we wi use a sine function to epress the equation of otion. ( ω ) (.66Nis ).s.85g sin ( rad) (.96s ) t =.7 e sin 7. rad s t [ ] (.66Nis ).s.85g b (.66 Nis ) γ (.85g) γ t = Ae sin t. = Ae sin 6.996rad. A = =.7 ; = = =.96s e 57. We assue that initiay, the syste is criticay daped, so b = 4. Then, after aging, we critica assue that after cyces, the car s osciatory apitude has dropped to 5% of its origina bt. apitude. That is epressed by A= Ae bt b( T) b f b.5 n (.5 ) A= Ae A = Ae = Ae = f n.5 b b π 6πb = = = b b b b b critica critica π / b 6π = + =.6 bcritica n(.5) And so b has decreased to about 6% of its origina vaue, or decreased by a factor of 6. If we used % instead of 5%, we woud have found that b decreased to about % of its origina vaue. And if we used % instead of 5%, we woud have found that b decreased to about 6% of its origina vaue. γ t 58. (a) Since the anguar dispaceent is given as θ Ae ( ω t) = cos, we see that the dispaceent at t = is the initia apitude, so A = 5. We evauate the apitude 8. seconds ater. ( 8.s) e γ = γ = n =.54s.s 8.s 5 (b) The approiate period can be found fro the daped anguar frequency. The undaped anguar frequency is aso needed for the cacuation. ( ) gh g ω = = = I g g 9.8s ω = ω γ = γ = (.54s ) = 4.57 rad s.85 π π rad T = = =.5s ω 4.57 rad s (c) We sove the equation of otion for the tie when the apitude is haf the origina apitude. γ t n n 7.5 = 5 e t = = = 5.5s / γ.54s 8 Pearson Education, Inc., Upper Sadde River, NJ. A rights reserved. This ateria is protected under a copyright aws as they currenty eist. No portion of this ateria ay be reproduced, in any for or by any eans, without perission in writing fro the 465

19 Chapter 4 Osciations 59. (a) The energy of the osciator is a potentia energy when the cosine (or sine) factor is, and so bt. E A A e = = The osciator is osing 6.% of its energy per cyce. Use this to find the actua frequency, and then copare to the natura frequency. bt ( + T) bt bt E( t+ T) =.94 E( t) Ae = A e e = b ω n = (.94 ) = n (.94 ) T b ω ω f f [ n(.94) ] [ n(.94) ] 4 b = π π = = f ω 4ω 6π 6π π [ n(.94) ] 5 f f = =. % diff = =. % 6π f bt. (b) The apitude s decrease in tie is given by A= Ae Find the decrease at a tie of nt, and sove for n. The vaue of b was found in part (a). bnt n (.94 ) bt b A= Ae Ae = Ae = nt = nt T n = =. periods n.94 bt. γ t 6. The apitude of a daped osciator decreases according to A= Ae = Ae The data can be used to find the daping constant. bt.75g 5. A A= Ae b= = = t A.5s. n n.9 g s b 6. (a) For the ighty daped haronic osciator, we have b 4 ω ω. 4 We aso assue that the object starts to ove fro aiu dispaceent, and so bt bt bt bt d b = Ae cosω t and v = = Ae cosω tω Ae sinω t ω Ae sin ω t. dt bt bt cos sin E = + v = A e ω t + ω A e ω t bt bt bt bt A e cos ω t A e sin ω t A e E e = + = = (b) The fractiona oss of energy during one period is as foows. Note that we use the b π bt bt approiation that ω = 4 π. T 8 Pearson Education, Inc., Upper Sadde River, NJ. A rights reserved. This ateria is protected under a copyright aws as they currenty eist. No portion of this ateria ay be reproduced, in any for or by any eans, without perission in writing fro the 466

20 Giancoi Physics for Scientists & Engineers, 4 th Edition bt bt ( + T) bt bt E = E() t E( t+ T) = E e E e = E e e bt bt Ee e bt E bt bt bπ π = e bt = = = = E ω Q Ee 6. (a) Fro probe 5 (b), we can cacuate the frequency of the undaped otion. T = π = π + 5N s f = = = = 5.4Hz π π π.5g γ t (b) Eq. 4-6 says = Ae cos ω t, which says the apitude foows the reationship = Ae γ t. a Use the fact that = A after 55 periods have eapsed, and assue that the daping is ight a enough that the daped frequency is the sae as the natura frequency. γ ( T ) n f 5.4Hz A= Ae γ = = n = n =.684s.684s 55T (c) Again use a = Ae γ t. 55 γt n4 n 4 γt = Ae A= Ae t = = =.s a 4 γ.684s This is the tie for osciations, since 55 osciations corresponds to a haf-ife. 6. (a) Eq. 4-4 is used to cacuate. φ ω ω ω ω φ = tan if ω = ω, φ = tan = ω ( b) ω( b) (b) With ω = ω, we have F = F cosω t and = A sin ω t. The dispaceent and the driving et π rad or 9 out of phase with each other. The dispaceent is when the driving force is a aiu, and the dispaceent is a aiu (+A or A) when the driving force is. force are one-quarter cyce (c) As entioned above, the phase difference is Eq. 4- gives the apitude A as a function of driving frequency ω. To find the frequency for da aiu apitude, we set = and sove for. dω ω / F F A = = ( ω ω ) + b ω ω ω + b ω ( ) / da F = ( ) ( ω ω) + b ω ( ω ω) ω+ b ω = dω 8 Pearson Education, Inc., Upper Sadde River, NJ. A rights reserved. This ateria is protected under a copyright aws as they currenty eist. No portion of this ateria ay be reproduced, in any for or by any eans, without perission in writing fro the 467

21 Chapter 4 Osciations ( ) F ω ω ω+ b ω = / ( ω ω ) ω+ b ω = ω ω + b ω b b ω ω ω ω = = 65. We approiate that each spring of the car wi effectivey support one-fourth of the ass. The rotation of the ipropery-baanced car tire wi force the spring into osciation. The shaing wi be ost prevaent at resonance, where the frequency of the tire atches the frequency of the spring. At v resonance, the anguar veocity of the car tire, ω =, wi be the sae as the anguar frequency of r the spring syste, ω =. v 6, N ω = = v = r = (.4 ) =. s r 4 ( 5 g) 66. First, we put Eq. 4- into a for that epicity shows A as a function of Q and has the ratio ωω. F F A = = ω ω b ω ω b ω ω + ω ω + ω ω F ( ω ω ) ω ω ( ω ω) ( ω ) = = = 4 ω 4 b ω ω ω ω ω ω ω ω ω + ω ω ω + + Q ω ω Q ω F A = = F + Q For a vaue of Q = 6., the foowing graph is obtained. F + ( ωω) F Q The spreadsheet used for this probe can be found on the Media Manager, with fienae PSE4_ISM_CH4.XLS, on tab Probe F A ωω 8 Pearson Education, Inc., Upper Sadde River, NJ. A rights reserved. This ateria is protected under a copyright aws as they currenty eist. No portion of this ateria ay be reproduced, in any for or by any eans, without perission in writing fro the 468

22 Giancoi Physics for Scientists & Engineers, 4 th Edition 67. Appy the resonance condition, ω = ω, to Eq. 4-, aong with the given condition of F A =.7. Note that for this condition to be true, the vaue of.7 ust have units of s. F A = ω ω + b ω A ( ω ω ) ( ) F F F F F Q F = = = = = = =.7 b ω Q ω bω bω ω ω ( ω ) ( π f ) π 8 Q =.7 =.7 4 =.7 s 4 8 Hz =.7 d d = + is a soution of + b + = F cosωt by direct dt dt substitution. d d = A sin ( ωt + φ) ; = ωa cos ( ωt + φ) ; = ω A sin ( ωt+ φ) dt dt 68. We are to show that A sin ( ωt φ ) d d + b + = F cos ωt dt dt [ ] [ ] ω A sin ωt + φ b A cos t A sin t F cos t + ω ω + φ + ω + φ = ω Epand the trig functions. [ ] [ ] A ω A sinωt cosφ + cosωt sinφ + bωa cosωt cosφ sinωt sinφ = F cosωt Group by function of tie. ( A ω A ) cosφ bω A sinφ sin ωt + ( A ω A ) sinφ + bω A cosφ cosωt = F cosωt The equation has to be vaid for a ties, which eans that the coefficients of the functions of tie ust be the sae on both sides of the equation. Since there is no sinωt on the right side of the equation, the coefficient of sinωt ust be. A ω A cosφ bωa sinφ = sinφ A ω A ω ω ω ω ω ω ω 8 Pearson Education, Inc., Upper Sadde River, NJ. A rights reserved. This ateria is protected under a copyright aws as they currenty eist. No portion of this ateria ay be reproduced, in any for or by any eans, without perission in writing fro the 469 = = = = = tan φ φ = tan cosφ bωa bω bω ωb ωb Thus we see that Eq. 4-4 is necessary for A sin ( ωt φ ) = + to be the soution. This can be iustrated with the diagra shown. Equate the coefficients of cos ω t. A ω A sinφ + bωa cos φ = F ( ω ω ) ωb ( ω ω ) A ( ω ) + bω = F ω b ω b ( ω ω ) + ( ω ω ) + ω b + φ ωb ω ω

23 Chapter 4 Osciations A ( ω ω ) A = ω b + = F ω b ω b + + ( ω ω ) ( ω ω ) F ω b + ( ω ω ) Thus we see that Eq. 4- is aso necessary for A sin ( ωt φ ) = + to be the soution. bt. 69. (a) For the daped osciator, the apitude decays according to A= Ae We are aso given the ω Q vaue, and Q =. We use these reationships to find the tie for the apitude to b decrease to one-third of its origina vaue. bt/ ω b ω g Q = = = ; A= Ae = A b Q Q t / Q Q 5 = n = n = n = n = 7.7s 7s b ω g ( 9.8 s ) (.5) (b) The energy is a potentia energy when the dispaceent is at its aiu vaue, which is the apitude. We assue that the actua anguar frequency is very neary the sae as the natura anguar frequency. bt bt bt g de b g ω ; E = A = Ae = A e = A e dt / de g g A.7 g. 9.8 s g = = = dt t= Q Q 5.5 = (c) Use Eq. 4-6 to find the frequency spread. ω π f f = Q = = ω π f f Q ( 9.8 s ) (.5) / 5. W f ω g f = = = = =. Hz Q πq πq π( 5) Since this is the tota spread about the resonance frequency, the driving frequency ust be within. Hz on either side of the resonance frequency. 7. Consider the conservation of energy for the person. Ca the unstretched position of the fire net the zero ocation for both eastic potentia energy and gravitationa potentia energy. The aount of stretch of the fire net is given by, easured positivey in the downward direction. The vertica dispaceent for gravitationa potentia energy is given by the variabe y, easured positivey for the upward direction. Cacuate the spring constant by conserving energy between the window height and the owest ocation of the person. The person has no inetic energy at either ocation. 8 Pearson Education, Inc., Upper Sadde River, NJ. A rights reserved. This ateria is protected under a copyright aws as they currenty eist. No portion of this ateria ay be reproduced, in any for or by any eans, without perission in writing fro the 47

24 Giancoi Physics for Scientists & Engineers, 4 th Edition E = E gy = gy + top botto top botto botto ( y y top ) [. (.) ] = = = botto 4 g 6g 9.8 s.9 N botto (.) (a) If the person were to ie on the fire net, they woud stretch the net an aount such that the upward force of the net woud be equa to their weight. ( 6 g)( 9.8 s ) g Fet = = g = = = N (b) To find the aount of stretch given a starting height of 8, again use conservation of energy. Note that y = botto, and there is no inetic energy at the top or botto positions. E E gy gy y top botto top botto top g g = = + = ( 6 g)( 9.8 s ) ( 6 g)( 9.8 s ) 8 = N.98 N = =.549,.4476 This is a quadratic equation. The soution is the positive root, since the net ust be beow the unstretched position. The resut is Appy the conservation of echanica energy to the car, caing condition # to be before the coision and condition # to be after the coision. Assue that a of the inetic energy of the car is converted to potentia energy stored in the buper. We now that = and v =. E = E v + = v + v = g = v = (. s) =. 4 N 7. (a) The frequency can be found fro the ength of the penduu, and the acceeration due to gravity. g 9.8 s f = = =.677 Hz.6Hz π π.6 (b) To find the speed at the owest point, use the conservation of energy reating the owest point to the reease point of the penduu. Tae the owest point to be the zero eve of gravitationa potentia energy. E = E KE + PE = KE + PE top botto top top botto botto ( θ ) + g L Lcos = v + botto v = gl cosθ = 9.8 s.6 cos5 =.6487 s.65 s botto (c) The tota energy can be found fro the inetic energy at the botto of the otion. E tota = v =.95g.6487 s = 6. J botto h = cosθ θ cosθ 8 Pearson Education, Inc., Upper Sadde River, NJ. A rights reserved. This ateria is protected under a copyright aws as they currenty eist. No portion of this ateria ay be reproduced, in any for or by any eans, without perission in writing fro the 47

25 Chapter 4 Osciations g 7. The frequency of a sipe penduu is given by f =. The penduu is acceerating π L verticay which is equivaent to increasing (or decreasing) the acceeration due to gravity by the acceeration of the penduu. (a) (b) g+ a.5g g f = = =.5 =.5 f =. f new π L π L π L g+ a.5g g f = = =.5 =.5 f =.7 f new π L π L π L 74. The equation of otion is =.5sin 5.5t = Asinωt. (a) The apitude is A= =.5. a 5.5s (b) The frequency is found by ω = π f = 5.5s f = =.875Hz π (c) The period is the reciproca of the frequency. T = π f = 5.5s =.4s. (d) The tota energy is given by ( ω ) E = v = A =.65 g 5.5s tota a.5 =.645J.6J. (e) The potentia energy is given by ω.65 g 5.5s.5. J. J potentia E = = = =. The inetic energy is given by E = E E =.645J. J =.9J.9 J. inetic tota potentia 75. (a) The car on the end of the cabe produces tension in the cabe, and stretches the cabe according F to Equation (-4), =, where E is Young s oduus. Rearrange this equation to o E A EA see that the tension force is proportiona to the aount of stretch, F =, and so the EA effective spring constant is =. The period of the bouncing can be found fro the spring constant and the ass on the end of the cabe. o ( 5 g)(. ) o T = π = π = π =.47s.4s EA 9 ( N ) π (. ) (b) The cabe wi stretch soe due to the oad of the car, and then the apitude of the bouncing wi ae it stretch even farther. The tota stretch is to be used in finding the aiu apitude. The tensie strength is found in Tabe -. F ( + static apitude ) = = tensie strength ( abbrev T.S. ) A π r o 8 Pearson Education, Inc., Upper Sadde River, NJ. A rights reserved. This ateria is protected under a copyright aws as they currenty eist. No portion of this ateria ay be reproduced, in any for or by any eans, without perission in writing fro the 47

26 Giancoi Physics for Scientists & Engineers, 4 th Edition ( T.S. ) π ( T.S. ) r πr g g = = = apitude ( T.S. ) Eπ r Eπr E πr (. ) ( 5 g)( 9.8 s ) = 9 = = ( N ) π (. ) 6 5 N The spring constant does not change, but the ass does, and so the frequency wi change. Use Eq. 4-7a to reate the spring constant, the ass, and the frequency. f = f constant f f O O S S π = = = f f O = =.7 Hz =.6 Hz S O S 6.. Tg 77. The period of a penduu is given by T = π g, and so the ength is =. (a) (b) (c) (. s) ( 9.79 s ) TgAustin Austin = = = (. s) ( 9.89 s ) TgParis Paris = = = = =.6.6 Paris Austin (. s) (.6 s ) TgMoon Moon = = = The force of the an s weight causes the raft to sin, and that causes the water to put a arger upward force on the raft. This etra buoyant force is a restoring force, because it is in the opposite direction of the force put on the raft by the an. This is anaogous to puing down on a ass spring syste that is in equiibriu, by appying an etra force. Then when the an steps off, the restoring force pushes upward on the raft, and thus the raft water syste acts ie a spring, with a spring constant found as foows. F ( 75g)( 9.8 s ) et 4 = = =. N.5 (a) The frequency of vibration is deterined by the spring constant and the ass of the raft. f n 4. N.89 Hz.Hz = = = π π g (b) As epained in the tet, for a vertica spring the gravitationa potentia energy can be ignored if the dispaceent is easured fro the osciator s equiibriu position. The tota energy is thus E 4 A. N.5.86 J J tota = = =. 8 Pearson Education, Inc., Upper Sadde River, NJ. A rights reserved. This ateria is protected under a copyright aws as they currenty eist. No portion of this ateria ay be reproduced, in any for or by any eans, without perission in writing fro the 47

27 Chapter 4 Osciations 79. The reationship between the veocity and the position of a SHO is given by Eq. 4-b. Set that epression equa to haf the aiu speed, and sove for the dispaceent. v =± v A = v ± A = A = A = a =± A ±.866A a For the pebbe to ose contact with the board eans that there is no nora force of the board on the pebbe. If there is no nora force on the pebbe, then the ony force on the pebbe is the force of gravity, and the acceeration of the pebbe wi be g downward, the acceeration due to gravity. This is the aiu downward acceeration that the pebbe can have. Thus if the board s downward acceeration eceeds g, then the pebbe wi ose contact. The aiu acceeration and the apitude are reated by a = π f A. a 4 g 9.8 s a = 4 π f A g A 4. a f.5 Hz 8. Assue the boc has a cross-sectiona area of A. In the equiibriu position, the net force on the boc is zero, and so F = g. When the boc is pushed into the water (downward) an additiona buoy distance, there is an increase in the buoyancy force ( F ) equa to the weight of the additiona etra water dispaced. The weight of the etra water dispaced is the density of water ties the voue dispaced. F = g = ρ V g = ρ ga = ρ ga etra add. water add. water water water water This is the net force on the dispaced boc. Note that if the boc is pushed down, the additiona force is upwards. And if the boc were to be dispaced upwards by a distance, the buoyancy force woud actuay be ess than the weight of the boc by the aount F, and so there woud be etra a net force downwards of agnitude F. So in both upward and downward dispaceent, there is etra but opposite to the direction of dispaceent. As a vector, we a net force of agnitude ( ρ ) can write the foowing. F = ρ net water ga water ga This is the equation of sipe haronic otion, with a spring constant of = ρ 8. (a) Fro conservation of energy, the initia inetic energy of the car wi a be changed into eastic potentia energy by copressing the spring. E = E v + = v + v = g.75 N.4 N ( 5 s ) ( 5. ) v = = = (b) The car wi be in contact with the spring for haf a period, as it oves fro the equiibriu ocation to aiu dispaceent and bac to equiibriu. ( 95 g) T = π = π =.6s.75 N 4 water ga 8 Pearson Education, Inc., Upper Sadde River, NJ. A rights reserved. This ateria is protected under a copyright aws as they currenty eist. No portion of this ateria ay be reproduced, in any for or by any eans, without perission in writing fro the 474

28 Giancoi Physics for Scientists & Engineers, 4 th Edition 8. (a) The effective spring constant is found fro the fina dispaceent caused by the additiona ass on the tabe. The weight of the ass wi equa the upward force eerted by the copressed springs. F = F g = y grav springs (.8 g)( 9.8 s ) g = = =.67 N N y (.6 ) (b) We assue the coision taes pace in such a short tie that the springs do not copress a significant aount during the coision. Use oentu conservation to find the speed iediatey after the coision. p = p v = + v v ( + cay tabe ) before after cay cay cay tabe after.8 g (.65 s ).55 s.4 g cay = v = = after cay As discussed in the tet, if we easure dispaceents fro the new equiibriu position, we ay use an energy anaysis of the spring otion without incuding the effects of gravity. The tota eastic and inetic energy iediatey after the coision wi be the aiu eastic energy, at the apitude ocation. E = E v + = A tota after after.4 g = + = (.55 s) (.6 ) c + = =.67 N tota A v after after 84. (a) The graph is shown. The spreadsheet used for this probe can be found on the Media Manager, with fienae PSE4_ISM_CH4.XLS, on tab Probe 4.84a. (b) Equiibriu occurs at the ocation where the force is. Set the force equa to and sove for the separation distance r. C D F( r ) = + = r r C = D Cr = Dr r = D r r C This does atch with the graph, which shows F = at r = D/C. (c) We find the net force at r = r + r. Use the binoia epansion. F r as a fraction of D/C r r F r r C r r D r r Cr Dr + = = r r C r D r C r D r r + = + r r r r r r C r 8 Pearson Education, Inc., Upper Sadde River, NJ. A rights reserved. This ateria is protected under a copyright aws as they currenty eist. No portion of this ateria ay be reproduced, in any for or by any eans, without perission in writing fro the 475