. The maximum speed m can be doubled by doubling the amplitude, A. 5. The maximum speed of a simple harmonic oscillator is given by v = A

Transcription

1 CHAPTER 4: Oscillations Responses to Questions. Exaples are: a child s swing (SHM, for sall oscillations), stereo speaers (coplicated otion, the addition of any SHMs), the blade on a jigsaw (approxiately SHM), the string on a guitar (coplicated otion, the addition of any SHMs).. The acceleration of a siple haronic oscillator is oentarily zero as the ass passes through the equilibriu point. At this point, there is no force on the ass and therefore no acceleration.. When the engine is running at constant speed, the piston will have a constant period. The piston has zero velocity at the top and botto of its path. Both of these properties are also properties of SHM. In addition, there is a large force exerted on the piston at one extree of its otion, fro the cobustion of the fuel air ixture, and in SHM the largest forces occur at the extrees of the otion. 4. The true period will be larger and the true frequency will be saller. The spring needs to accelerate not only the ass attached to its end, but also its own ass. As a ass on a spring oscillates, potential energy is converted into inetic energy. The axiu potential energy depends on the displaceent of the ass. This axiu potential energy is converted into the axiu inetic energy, but if the ass being accelerated is larger then the velocity will be saller for the sae aount of energy. A saller velocity translates into a longer period and a saller frequency. 5. The axiu speed of a siple haronic oscillator is given by v = A. The axiu speed can be doubled by doubling the aplitude, A. 6. Before the trout is released, the scale reading is zero. When the trout is released, it will fall downward, stretching the spring to beyond its equilibriu point so that the scale reads soething over 5 g. Then the spring force will pull the trout bac up, again to a point beyond the equilibriu point, so that the scale will read soething less than 5 g. The spring will undergo daped oscillations about equilibriu and eventually coe to rest at equilibriu. The corresponding scale readings will oscillate about the 5-g ar, and eventually coe to rest at 5 g. 7. At high altitude, g is slightly saller than it is at sea level. If g is saller, then the period T of the pendulu cloc will be longer, and the cloc will run slow (or lose tie). 8. The tire swing is a good approxiation of a siple pendulu. Pull the tire bac a short distance and release it, so that it oscillates as a pendulu in siple haronic otion with a sall aplitude. Measure the period of the oscillations and calculate the length of the pendulu fro the expression l T = π. The length, l, is the distance fro the center of the tire to the branch. The height of the g branch is l plus the height of the center of the tire above the ground. 9. The displaceent and velocity vectors are in the sae direction while the oscillator is oving away fro its equilibriu position. The displaceent and acceleration vectors are never in the sae direction. 8 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This aterial is protected under all copyright laws as they currently exist. No portion of this aterial ay be reproduced, in any for or by any eans, without perission in writing fro the publisher. 44

2 Physics for Scientists & Engineers with Modern Physics, 4 th Edition Instructor Solutions Manual. The period will be unchanged, so the tie will be (c), two seconds. The period of a siple pendulu oscillating with a sall aplitude does not depend on the ass.. The two asses reach the equilibriu point siultaneously. The angular frequency is independent of aplitude and will be the sae for both systes.. Epty. The period of the oscillation of a spring increases with increasing ass, so when the car is epty the period of the haronic otion of the springs will be shorter, and the car will bounce faster.. When waling at a noral pace, about s (tied). The faster you wal, the shorter the period. The shorter your legs, the shorter the period. 4. When you rise to a standing position, you raise your center of ass and effectively shorten the length of the swing. The period of the swing will decrease. 5. The frequency will decrease. For a physical pendulu, the period is proportional to the square root of the oent of inertia divided by the ass. When the sall sphere is added to the end of the rod, both the oent of inertia and the ass of the pendulu increase. However, the increase in the oent of inertia will be greater because the added ass is located far fro the axis of rotation. Therefore, the period will increase and the frequency will decrease. 6. When the 64-Hz for is set into vibration, the sound waves generated are close 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 will not begin to vibrate. 7. If you shae the pan at a resonant frequency, standing waves will be set up in the water and it will slosh bac and forth. Shaing the pan at other frequencies will not create large waves. The individual water olecules will ove but not in a coherent way. 8. Exaples of resonance are: pushing a child on a swing (if you push at one of the liits of the oscillation), blowing across the top of a bottle, producing a note fro a flute or organ pipe. 9. Yes. Rattles which occur only when driving at certain speeds are ost liely resonance phenoena.. Building with lighter aterials doesn t necessarily ae it easier to set up resonance vibrations, but it does shift the fundaental frequency and decrease the ability of the building to dapen oscillations. Resonance vibrations will be ore noticeable and ore liely to cause daage to the structure. Solutions to Probles. The particle would travel four ties the aplitude: fro x = A to x = to x = A to x = to x = A. So the total distance = 4A = 4(.8 ) =.7.. The spring constant is the ratio of external applied force to displaceent. Fext 8 N 75 N 5 N = = = = 55N 5 N x Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This aterial is protected under all copyright laws as they currently exist. No portion of this aterial ay be reproduced, in any for or by any eans, without perission in writing fro the publisher. 444

3 Chapter 4 Oscillations. The spring constant is found fro the ratio of applied force to displaceent. F ext g 68 g 9.8 s 5 = = = =. N x x 5. The frequency of oscillation is found fro the total ass and the spring constant. 5. N.467 Hz.5 Hz f = = = π π 568 g 4. (a) The otion starts at the axiu extension, and so is a cosine. The aplitude is the displaceent at the start of the otion. π π x = Acos ( ωt) = Acos t = ( 8.8c) cos t = ( 8.8c) cos ( 9.5t) T c cos 9.5t (b) Evaluate the position function at t =.8 s. x = 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 calculated fro Eq. 4-7b. 5g T = T = = = = T π 4 π 5N (.s) 6. (a) The spring constant is found fro the ratio of applied force to displaceent. F g (.4 g)( 9.8 s ) ext = = = = 65N 65 N x x.6 (b) The aplitude is the distance pulled down fro equilibriu, so A =.5c The frequency of oscillation is found fro the oscillating ass and the spring constant. 65N f = = =.65Hz.6 Hz π π.4g 7. The axiu velocity is given by Eq. 4-9a. π A π (.5) v = ωa= = =. s ax T 7.s The axiu acceleration is given by Eq. 4-9b. A (.5) a = ω A= = =.9 s. s ax T 7.s a g ax.9 s = = = 9.8 s..% 8 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This aterial is protected under all copyright laws as they currently exist. No portion of this aterial ay be reproduced, in any for or by any eans, without perission in writing fro the publisher. 445

4 Physics for Scientists & Engineers with Modern Physics, 4 th Edition Instructor Solutions Manual 8. The table of data is shown, along with the soothed graph. Every quarter of a period, the ass oves fro an extree point to the equilibriu. The tie position - A T/4 T/ A T/4 T - A 5T/4 graph resebles a cosine wave (actually, the opposite of a cosine wave). position / A tie / T 9. The relationship 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.579 N (b) f = = = 4.8 Hz π π 5. g. The spring constant is the sae regardless of what ass is attached to the spring. f = = f = constant f = f π (.68 g)(.6 Hz) ( g)(.8hz ) = ( g +.68 g)(.6 Hz ) = =.74 g.8hz.6 Hz. We assue that the spring is stretched soe distance y while the rod is in equilibriu and horizontal. Calculate the net torque F s about point A while the object is in equilibriu, with clocwise A θ torques as positive. τ = Mg ( l ) F l = M gl y l = s Mg Now consider the rod being displaced an additional distance y below the horizontal, so that the rod aes a sall angle of θ as shown in the free-body diagra. Again write the net torque about point A. If the angle is sall, then there has been no appreciable horizontal displaceent of the rod. d θ Mg F Mg y y I M dt τ = ( l ) l = l s ( + ) l = α = l Include the equilibriu condition, and the approxiation that y = l sin θ l θ. d θ d θ = = dt dt Mgl yl y l Ml Mgl yl Mgl Ml d θ d θ θ l + l = M θ = dt dt M This is the equation for siple haronic otion, corresponding to Eq. 4-, with ω =. M 8 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This aterial is protected under all copyright laws as they currently exist. No portion of this aterial ay be reproduced, in any for or by any eans, without perission in writing fro the publisher. 446

5 Chapter 4 Oscillations ω = 4 π f = f = M π M. (a) We find the effective spring constant fro the ass and the frequency of oscillation. f = π = π f = π = g. Hz 9.54 N 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 aplitude is A =.5. For B, the aplitude is A =.5. A B (b) For A, the frequency is cycle every 4. seconds, so f =.5Hz A. For B, the frequency is cycle 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 displaceent of when t =, so it is a sine function. ( π ) ( π ) x = A sin f t x =.5 sin t A A A A Object B has a axiu displaceent when t =, so it is a cosine function. ( π ) ( π ) x = A cos f t x =.5 cos t B B B B 4. Eq. 4-4 is x = Acos( ωt+ φ). (a) If x = A, then (b) If x =, then (c) If x = A, then () (d) If x = A, then (e) If x = A, then (f) If x A A= Acos φ φ = cos φ = π. = Acos φ φ = cos φ = ± π. A= Acos φ φ = cos φ =. A= Acos φ φ = cos φ = ± π. 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, ω = = 5 N.6 g = 4.5 rad s. (a) Since the ass has a zero displaceent and a positive velocity at t =, the equation is a sine function. y t =.8 sin 4.rad s t = = and () [ ] 8 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This aterial is protected under all copyright laws as they currently exist. No portion of this aterial ay be reproduced, in any for or by any eans, without perission in writing fro the publisher. 447

6 Physics for Scientists & Engineers with Modern Physics, 4 th Edition Instructor Solutions Manual π π (b) The period of oscillation is given by T = = =.845s. The spring will have ω 4.5 rad s its axiu extension at ties given by the following. T t = + nt = 4.59 s + n ax (.8 s ), n=,,, 4 The spring will have its iniu extension at ties given by the following. 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 = π = = =.7877 N.79 N T (.69s) (b) Fro the graph, the aplitude is.8 c. The phase constant can be found fro the initial conditions. π π x = Acos t+ φ = (.8c) cos t+ φ T.69.4 x = (.8 c) 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 x = (.8 c) cos t. or (.8 c) cos ( 9.t.) (a) The period and frequency are found fro the angular frequency. 5π 5 ω = π f f = ω Hz T.6 s π = π 4 = 8 = f = (b) The velocity is the derivative of the position. 5π π dx 5π 5π π x = (.8 ) cos t+ v = = (.8 ) sin t+ 4 6 dt π 5π π x = (.8 ) cos =. v = (.8 ) sin = 7.5 s (c) The acceleration is the derivative of the velocity. 5π 5π π dv 5π 5π π v = (.8 ) sin t+ a = = (.8 ) cos t dt π 5π π v (.) = (.8 ) sin (.) + = s π 5π π a (. ) = (.8) cos (.) + = 9 s Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This aterial is protected under all copyright laws as they currently exist. No portion of this aterial ay be reproduced, in any for or by any eans, without perission in writing fro the publisher. 448

7 Chapter 4 Oscillations 8. (a) The axiu speed is given by Eq. 4-9a. v = π f A= π ax ( 44Hz)(.5 ) = 4. s. (b) The axiu acceleration is given by Eq. 4-9b. a Hz.5. s ax 4 = π f A= π =. 9. When the object is at rest, the agnitude of the spring force is equal to the force of gravity. This deterines the spring constant. The period can then be found. g F = x g = vertical x T = π x.4 = π = π = π g g 9.8 s =.75s x. The spring constant can be found fro the stretch distance corresponding to the weight suspended on the spring. F (.6 g)( 9.8 s ) ext g = = = = 7.84 N x x.5 After being stretched further and released, the ass will oscillate. It taes one-quarter of a period for the ass to ove fro the axiu displaceent to the equilibriu position. π.6 g T = π = =.s N. Each object will pass through the origin at the ties when the arguent of its sine function is a ultiple 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 axiu displaceent in the positive direction, and so will be represented by a cosine function. The ass, period, and aplitude are given. A =.6 ; ω = π = π =.4 rad s y = (.6) cos ( t) T.55s (b) The tie to reach the equilibriu is one-quarter of a period, so (.55s 4 ) =.4s. (c) The axiu speed is given by Eq. 4-9a. v = ωa= ax (.4 rad s)(.6 ) =.8 s (d) The axiu acceleration is given by Eq. 4-9b. a = ω A= ax (.4 rad s) (.6 ) =. s The axiu acceleration occurs at the endpoints of the otion, and is first attained at the release point. 8 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This aterial is protected under all copyright laws as they currently exist. No portion of this aterial ay be reproduced, in any for or by any eans, without perission in writing fro the publisher. 449

8 Physics for Scientists & Engineers with Modern Physics, 4 th Edition Instructor Solutions Manual 4. s. The period of the juper s otion is T = = 5.75 s. The spring constant can then be found 8 cycles fro the period and the juper s ass. ( 5.75s) 65.g T = π = = = 88.8N 88.8 N T The stretch of the bungee cord needs to provide a force equal to the weight of the juper when he is at the equilibriu point. ( 65. g)( 9.8 s ) g Δ x = g Δ x = = = N Thus the unstretched bungee cord ust be = Consider the first free-body diagra for the bloc while it is at equilibriu, so that the net force is zero. Newton s second law for vertical 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 bloc is displaced a distance x fro the equilibriu point. Each upward force will have increased by an aount x, since x <. Again write Newton s second law for vertical forces. F = F = F + F g = F x+ F x g = x+ F + F g = x y net A B A B A B This is the general for of a restoring force that produces SHM, with an effective spring constant of. Thus the frequency of vibration is as follows. f = = effective π π 5. (a) If the bloc is displaced a distance x to the right in Figure 4-a, then the length of spring # will be increased by a distance x and the length of spring # will be increased by a distance x, where x = x + x. The force on the bloc can be written F = x. Because the springs eff are assless, they act siilar to a rope under tension, and the sae force F is exerted by each spring. Thus F = x = x = x. eff F F F x = x + x = = F + = = + eff eff T = π = π + eff (b) The bloc will be in equilibriu when it is stationary, and so the net force at that location is zero. Then, if the bloc is displaced a distance x to the right in the diagra, then spring # will exert an additional force of F = x, in the opposite direction to x. Liewise, spring # will exert an additional force F = x, in the sae direction as F. Thus the net force on the F A F F B F A B g g x 8 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This aterial is protected under all copyright laws as they currently exist. No portion of this aterial ay be reproduced, in any for or by any eans, without perission in writing fro the publisher. 45

9 Chapter 4 Oscillations F = F + F = x x= + x. The effective spring constant is thus displaced bloc is = +, and the period is given by T = π = π The ipulse, which acts for a very short tie, changes the oentu of the ass, giving it an initial velocity v. Because this occurs at the equilibriu position, this is the axiu velocity of the ass. Since the otion starts at the equilibriu position, we represent the otion by a sine function. J J =Δ p = Δ v = v = v v = = v = Aω = A ax J J J = A A= x = Asinωt = sin t 7. The various values can be found fro the equation of otion, x = Acosωt =.65cos7.4 t. (a) The aplitude is the axiu value of x, and so A =.65. ω 7.4 rad s (b) The frequency is f = = =.8 Hz. π π rad (c) The total energy can be found fro the axiu potential energy. ω.5g 7.4 rad s.65.j.j ax E = U = A = A = = (d) The potential energy can be found fro U = x, and the inetic energy fro E = U + K. = = ω = = U x x.5g 7.4 rad s.6.j K = E U =.J.J =. J 8. (a) The total energy is the axiu potential energy. U = E x = A x = A A.77 (b) Now we are given that x = A U x x = = = E A A 9. 8 Thus the energy is divided up into potential and inetic The total energy can be found fro the spring constant and the aplitude. E = A = 95 N. =.9 J That is represented by the horizontal line on the graph. (a) Fro the graph, at x =.5c, we have U.J. (b) Fro energy conservation, at x =.5c, we have K = E U =.8J. 8 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This aterial is protected under all copyright laws as they currently exist. No portion of this aterial ay be reproduced, in any for or by any eans, without perission in writing fro the publisher. 45

10 Physics for Scientists & Engineers with Modern Physics, 4 th Edition Instructor Solutions Manual (c) Find the speed fro the estiated inetic energy. K = v K v = = =.5 s.8j.55 g The spreadsheet used for this proble x (c) can be found on the Media Manager, with filenae PSE4_ISM_CH4.XLS, on tab Proble (a) At equilibriu, the velocity is its axiu. Use Eq. 4-9a, and realize that the object can be oving in either direction. v = ωa= π fa= π.5hz.5 =.56 s v ±.4 s ax equib (b) Fro Eq. 4-b, we find the velocity at any position. (. ) (.5) x v =± v =± ax (.56 s) =±.756 s ±.8 s A (c) E total = = = v.5g.56 s.974 J.97 J ax (d) Since the object has a axiu displaceent at t =, the position will be described by the cosine function. x =.5 cos π.5 Hz t x =.5 cos 5.π t ( ). The spring constant is found fro the ratio of applied force to displaceent. F 95. N = = = 54.9 N x.75 Assuing that there are no dissipative forces acting on the ball, the elastic potential energy in the loaded position will becoe inetic energy of the ball N E = E x = v v = x = ax ax ax ax (.75) =. s i f.6 g. The energy of the oscillator will be conserved after the collision. E = A = + M v v = A + M ax ax This speed is the speed that the bloc and bullet have iediately after the collision. Linear oentu in one diension will have been conserved during the (assued short tie) collision, and so the initial speed of the bullet can be found. p = p v = + M v v before after o ax o U (J) M.55g 5 N = A = (.4 ) = 6 s + M.5 g.55 g 8 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This aterial is protected under all copyright laws as they currently exist. No portion of this aterial ay be reproduced, in any for or by any eans, without perission in writing fro the publisher. 45

11 Chapter 4 Oscillations. To copare the total energies, we can copare the axiu potential energies. Since the frequencies and the asses are the sae, the spring constants are the sae. E A A A E = A = A = 5 A = 5 high high high high energy energy energy energy low low low low energy energy energy energy 4. (a) The spring constant can be found fro the ass and the frequency of oscillation. ω = = π = π = π = (b) The energy can be found fro the axiu potential energy. E = A = 85.7 N.45 = 8.64 J.86 J f 4 f 4. Hz.4 g 85.7 N 85 N 5. (a) The wor done in copressing the spring is stored as potential energy. The copressed location corresponds to the axiu potential energy and the aplitude of the ensuing otion. W (.6J) W = A = = = 46 N 4 N A (.) (b) The axiu acceleration occurs at the copressed location, where the spring is exerting the axiu force. If the copression distance is positive, then the acceleration is negative. x ( 46 N )(.) F = x = a = = =.7 g a 5 s 6. (a) The total energy of an object in SHM is constant. When the position is at the aplitude, the speed is zero. Use that relationship to find the aplitude. E = v + x = A tot.7 g A= v + x = (.55 s) + (. ) = N (b) Again use conservation of energy. The energy is all inetic energy when the object has its axiu velocity. E = v + x = A = v tot ax 8 N v = A = ax ( ) =.5865 s.59 s.7 g 7. We assue that the collision of the bullet and bloc is so quic that there is no significant otion of the large ass or spring during the collision. Linear oentu is conserved in this collision. The speed that the cobination has right after the collision is the axiu speed of the oscillating syste. Then, the inetic energy that the cobination has right after the collision is stored in the spring when it is fully copressed, at the aplitude of its otion. p = p v = before after ( + M) v v = v ax ax + M ( + M) v = A ax ( + M) v = A + M 8 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This aterial is protected under all copyright laws as they currently exist. No portion of this aterial ay be reproduced, in any for or by any eans, without perission in writing fro the publisher. 45

12 Physics for Scientists & Engineers with Modern Physics, 4 th Edition Instructor Solutions Manual 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 initial position of the oscillator be x. x t dx dx dx x cos = cos + cos = ± x ( A x ) x ( A x ) v =± A x = =± dt =± dt dt x x x t A A A Mae these definitions: ω x A φ ; cos. Then we have the following. x x x cos + cos = ± t cos + t x Acos ( t ) A A A φ = ± ω = ± ω + φ The phase angle definition could be changed so that the function is a sine instead of a cosine. And the ± sign can be resolved if the initial velocity 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 initial speed is the axiu speed, and that can be used to find the aplitude. v ax = A A= v =.6 s.785g 84 N = ax (c) The axiu acceleration can be found fro the ass, spring constant, and aplitude a = A = ax (.476)( 84 N ) (.785g) = 4.6 s (d) Because the ass started at the equilibriu position of x =, the position function will be proportional to the sine function. x = (.48 ) sin[ π(.47 Hz ) t] x = (.48 ) sin ( 4.87π t) (e) The axiu energy is the inetic energy that the object has when at the equilibriu position. (f) E = v =.785g.6 s =.J ax Use the conservation of echanical energy for the oscillator. E = x + v = A.4 A + K = A A.4.J.84.68J K = = = 8 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This aterial is protected under all copyright laws as they currently exist. No portion of this aterial ay be reproduced, in any for or by any eans, without perission in writing fro the publisher. 454

13 Chapter 4 Oscillations 4. We solve this using conservation of energy, equating the energy at the copressed point with the energy as the ball leaves the launcher. Tae the location for gravitational potential energy to be at the level where the ball is on the copressed spring. The location for elastic potential energy is the uncopressed position of the spring. Initially, the ball has only elastic potential energy. At the point where the spring is uncopressed and the ball just leaves the spring, there will be gravitational potential energy, translational inetic energy, and rotational inetic energy. The ball is rolling without slipping. v E = E x = gh + v + Iω = gx sin θ + v + i f ( 5) r r 7.5 g 7 = ( gx sinθ + v ) = 9.8 s.6 sin5 x +. s (.6 ) = 89.6N 9 N sig. fig. 4. The period of a pendulu is given by T = π L g. The length is assued to be the sae for the pendulu 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. cycles cycles (b) The frequency is given by f = =.64 Hz. 5s 4. We consider this a siple pendulu. Since the otion starts at the aplitude position at t =, we ay describe it by a cosine function with no phase angle, θ = θ cosωt. The angular velocity can ax be written as a function of the length, cos g θ = θ. ax t l 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 Saddle River, NJ. All rights reserved. This aterial is protected under all copyright laws as they currently exist. No portion of this aterial ay be reproduced, in any for or by any eans, without perission in writing fro the publisher. 455

14 Physics for Scientists & Engineers with Modern Physics, 4 th Edition Instructor Solutions Manual 44. The period of a pendulu is given by T = π L g..5 (a) T = π L g = π =.5s 9.8 s (b) If the pendulu is in free fall, there is no tension in the string supporting the pendulu bob, and so no restoring force to cause oscillations. Thus there will be no period the pendulu will not oscillate and so no period can be defined. 45. If we consider the pendulu as starting fro its axiu displaceent, then the equation of otion π t can be written as θ = θ cosωt = θ cos. Solve for the tie for the position to decrease to half the T aplitude. 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 will also tae T for the position to change fro to, 5 and so fro + 5 to taes 5 T. The second half of the cycle will be identical to the first, 6 and so the total tie spent between + 5 and is 5 T. So the pendulu spends one-third of its tie between + 5 and There are ( 4h)( 6in h)( 6s in) = 86,4s in a day. The cloc should ae one cycle in exactly two seconds (a tic and a toc ), and so the cloc should ae 4, cycles per day. After one day, the cloc in question is 6 seconds slow, which eans that it has ade less cycles than required for precise tieeeping. Thus the cloc is only aing 4,87 cycles in a day. Accordingly, the period of the cloc ust be decreased by a factor of 4,87 4,. 4,87 4,87 T = T π l g = π g new old new l old 4, 4, 4,87 4,87 l old l = =.99 =.994 new 4, 4, Thus the pendulu should be shortened by Use energy conservation to relate the potential energy at the axiu height of the pendulu to the inetic energy at the lowest point of the swing. Tae the lowest point to be the zero location for gravitational potential energy. See the diagra. E = E K + U = K + U top botto top top botto botto ax ax + gh = v v = gh = gl cosθ h l = l l cosθ θ l cosθ 8 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This aterial is protected under all copyright laws as they currently exist. No portion of this aterial ay be reproduced, in any for or by any eans, without perission in writing fro the publisher. 456

15 Chapter 4 Oscillations 48. (a) For a physical pendulu with the sall angle approxiation, we ay apply Eq We need the oent of inertia and the distance fro the suspension point to the center of ass. We approxiate the cord as a rod, and find the center of ass relative to the stationary end of the cord. I = I + I = Ml + l = bob cord ( M + ) l Ml + ( l ) M + h = x = = CM l M + M + M ( M + ) g I M + l M + l T = π = π = π gh + total M + g l M + (b) If we use the expression for a siple pendulu we would have T = π l g. Find the siple fractional error. ( M + ) l l ( M + ) π π T T siple ( M + ) g g ( M + ) ( M + ) error = = = = T ( M + ) l ( M + ) ( M + ) π M + g M + Note that this is negative, indicating that the siple pendulu approxiation is too large. 49. The balance wheel of the watch is a torsion pendulu, described by τ = Kθ. A specific torque and angular displaceent are given, and so the torsional constant can be deterined. The angular frequency is given by ω = K I. Use these relationships to find the ass. τ =Kθ 5 θ. in K = = τ π 4rad ω = π f = K K = I r 5. in K π 4rad 4 = = = 4. g =.4g f r. Hz.95 M L 5. (a) We call the upper ass M and the lower ass. Both asses have length l. The period of the physical pendulu 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 parallel axis theore is used to find the oent of inertia. 7 I = I + I = Ml + l + l = M + l ( l ) ( l ) upper lower M + M + h = x = = CM l M + M + M ( M + ) g 7 7 I M + l M + l T = π = π = π gh + total M + g l M + M l l 8 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This aterial is protected under all copyright laws as they currently exist. No portion of this aterial ay be reproduced, in any for or by any eans, without perission in writing fro the publisher. 457

16 Physics for Scientists & Engineers with Modern Physics, 4 th Edition Instructor Solutions Manual 7 [ ( 7.g) + ( 4.g) ](.55 ) ( 7.g) + ( 4.g) 9.8 s = π =.6495s.6s [ ] (b) It too 7. seconds for 5 swings, which gives a period of.4 seconds. That is reasonable qualitative agreeent. 5. (a) In the text, we are given that τ = Kθ. Newton s second law for rotation, d θ Eq. -4, says that τ = Iα = I. We assue that the torque applied by the twisting of dt the wire is the only 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 differential equation for siple haronic oscillation. We exchange variables with Eq. 4-4, and write the equation for the angular otion. K x = Acos ( ωt+ φ) θ = θ cos ( ωt+ φ), ω = I (b) The period of the otion is found fro the angular velocity ω. K K π I ω = ω = = T = π I I T K 5. The eter stic used as a pendulu is a physical pendulu. The period is given by Eq. 4-4, I T = π. Use the parallel axis theore to find the oent of inertia about the pin. Express gh the distances fro the center of ass. / I l + h π l I = I + h = l + h T = π = π = h CM gh gh + h g / dt l l = π ( ) + h + = h = =.887 dh h h l x = l h = fro the end Use the distance for h to calculate the period. (.) / π l π T = + h.887.5s = + = g h 9.8 s.887 / 5. This is a torsion pendulu. The angular frequency is given in the text as ω = K I, where K is the torsion constant (a property of the wire, and so a constant in this proble). The rotational inertia of a rod about its center is Ml. K π I ω = = T = π I T K 8 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This aterial is protected under all copyright laws as they currently exist. No portion of this aterial ay be reproduced, in any for or by any eans, without perission in writing fro the publisher. 458

17 Chapter 4 Oscillations I π T K I Ml (.7M )(.7l ) = = = = = T I M M I l l π K T T = = s =.9s 54. The torsional constant is related to the period through the relationship given in proble 5. The rotational 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 ) = i.7 N rad 55. This is a physical pendulu. Use the parallel axis 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 angular 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 aplitude at soe tie is A, then one cycle later, the aplitude will be to find the fractional change. Ae γt γt b (.66Nis ) T (.898s ) γt.85g fractional change.. Use this Ae A = = e = e = e = A And so the aplitude decreases by % fro the previous aplitude, every cycle. (c) Since the object is at the origin at t =, we will use a sine function to express the equation of otion. ( ω ) (.66Nis ).s.85g sin( rad) (.66Nis ).s.85g b (.66 Ni s ) γ (.85g) γ t x = Ae sin t. = Ae sin rad. A = =.7 ; = = =.96s e 8 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This aterial is protected under all copyright laws as they currently exist. No portion of this aterial ay be reproduced, in any for or by any eans, without perission in writing fro the publisher. 459

18 Physics for Scientists & Engineers with Modern Physics, 4 th Edition Instructor Solutions Manual (.96s ) t x =.7 e sin 7. rad s t [ ] 57. We assue that initially, the syste is critically daped, so b = 4. Then, after aging, we critical assue that after cycles, the car s oscillatory aplitude has dropped to 5% of its original bt. aplitude. That is expressed by A= Ae bt b( T) b f b.5 ln (.5 ) A = A e A = Ae = Ae = f b b π 6πb ln (.5 ) = = = b b b b b critical critical π / b 6π = + =.6 bcritical ln(.5) And so b has decreased to about 6% of its original value, or decreased by a factor of 6. If we used % instead of 5%, we would have found that b decreased to about % of its original value. And if we used % instead of 5%, we would have found that b decreased to about 6% of its original value. γ t 58. (a) Since the angular displaceent is given as θ Ae ( ω t) = cos, we see that the displaceent at t = is the initial aplitude, so A = 5. We evaluate the aplitude 8. seconds later. ( 8.s) e γ = γ = ln =.54s.s 8.s 5 (b) The approxiate period can be found fro the daped angular frequency. The undaped angular frequency is also needed for the calculation. gh g ( l ) g ω = = = I l l g 9.8s ω = ω γ = γ = (.54s ) = 4.57 rad s l (.85) π π rad T = = =.5s ω 4.57 rad s (c) We solve the equation of otion for the tie when the aplitude is half the original aplitude. γ t ln ln 7.5 = 5 e t = = = 5.5s / γ.54s 59. (a) The energy of the oscillator is all potential energy when the cosine (or sine) factor is, and so bt. E A A e = = The oscillator is losing 6.% of its energy per cycle. Use this to find the actual frequency, and then copare to the natural frequency. bt ( + T) bt bt E( t+ T) =.94 E( t) Ae =.94 A e e =.94 8 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This aterial is protected under all copyright laws as they currently exist. No portion of this aterial ay be reproduced, in any for or by any eans, without perission in writing fro the publisher. 46

19 Chapter 4 Oscillations b ω = ln (.94 ) = ln (.94 ) T b ω ω f f [ ln (.94) ] [ ln (.94) ] 4 b = π π = = f ω 4ω 6π 6π π [ ln (.94) ] f f = = = = 6π 5. % diff. % f bt. (b) The aplitude s decrease in tie is given by A= Ae Find the decrease at a tie of nt, and b solve for n. The value of was found in part (a). bnt ln (.94 ) bt b A= Ae Ae = Ae = nt = nt T n = =. periods ln.94 bt. γ t 6. The aplitude of a daped oscillator 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. ln ln.9 g s b 6. (a) For the lightly daped haronic oscillator, we have b 4 ω ω. 4 We also assue that the object starts to ove fro axiu displaceent, and so bt bt bt bt dx b x = Ae cosω t and v = = Ae cosω tω Ae sinω t ω Ae sin ω t. dt bt bt cos sin E = x + 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 fractional loss of energy during one period is as follows. Note that we use the b π bt bt approxiation that ω = 4 π. T bt bt ( + T) bt bt Δ E= E() t E( t+ T) = Ee Ee = Ee e bt bt Ee e bt Δ E bt bt bπ π = e bt = = = = E ω Q Ee 8 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This aterial is protected under all copyright laws as they currently exist. No portion of this aterial ay be reproduced, in any for or by any eans, without perission in writing fro the publisher. 46

20 Physics for Scientists & Engineers with Modern Physics, 4 th Edition Instructor Solutions Manual 6. (a) Fro proble 5 (b), we can calculate 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 x = Ae cos ω t, which says the aplitude follows the relationship x = Ae γ t. ax Use the fact that x = A after 55 periods have elapsed, and assue that the daping is light ax enough that the daped frequency is the sae as the natural frequency. γ ( 55T ) ln f 5.4Hz A= Ae γ = = ln = ln =.684s.684s 55T (c) Again use x ax = Ae γ t. γt ln4 ln 4 γt x = Ae A= Ae t = = =.s ax 4 γ.684s This is the tie for oscillations, since 55 oscillations corresponds to a half-life. 6. (a) Eq. 4-4 is used to calculate. 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 x = A sin ω t. The displaceent and the driving ext force are one-quarter cycle ( π rad or 9 ) out of phase with each other. The displaceent is when the driving force is a axiu, and the displaceent is a axiu (+A or A) when the driving force is. (c) As entioned above, the phase difference is Eq. 4- gives the aplitude A as a function of driving frequency ω. To find the frequency for da axiu aplitude, we set = and solve 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 F ( ω ω ) ω+ b ω = / ( ω ω ) ω+ b ω = ( ω ω ) + b ω b b ω ω ω ω = = 8 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This aterial is protected under all copyright laws as they currently exist. No portion of this aterial ay be reproduced, in any for or by any eans, without perission in writing fro the publisher. 46

21 Chapter 4 Oscillations 65. We approxiate that each spring of the car will effectively support one-fourth of the ass. The rotation of the iproperly-balanced car tire will force the spring into oscillation. The shaing will be ost prevalent at resonance, where the frequency of the tire atches the frequency of the spring. At v resonance, the angular velocity of the car tire, ω =, will be the sae as the angular 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 explicitly 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 F = = = ( ω ω ) ω ω ( ω ω) ( ω ) 4 ω 4 b ω ω ω ω ω ω + ω ω ω + + Q ω ω Q ω ω ω ω F A = = F + Q For a value of Q = 6., the following graph is obtained. + ( ωω) F Q The spreadsheet used for this proble can be found on the Media Manager, with filenae PSE4_ISM_CH4.XLS, on tab Proble F A ωω 67. Apply the resonance condition, ω = ω, to Eq. 4-, along with the given condition of F A =.7. Note that for this condition to be true, the value of.7 ust have units of F A = ω ω + b ω ( ) s. 8 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This aterial is protected under all copyright laws as they currently exist. No portion of this aterial ay be reproduced, in any for or by any eans, without perission in writing fro the publisher. 46

22 Physics for Scientists & Engineers with Modern Physics, 4 th Edition Instructor Solutions Manual F F F F F F A( ω = ω) = = = = = Q =.7 Q =.7 b ω bω bω ω ω Q 68. We are to show that x A sin ( ωt φ ) substitution. dx dx = + is a solution of + b + x = F cosωt by direct dt dt dx d x sin ; cos ; sin ( ω φ ) ω ( ω φ ) ω ( ω φ ) x = A t+ = A t+ = A t+ dt dt dx dx + b + x = F cos ωt dt dt ω A sin ( ωt + φ) + b[ ωa cos ( ωt+ φ) ] + [ A sin ( ωt+ φ) ] = F cosωt Expand 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 bωa sinφ sin ωt + ( A ω A ) sinφ + bωa cosφ cosωt = F cosωt The equation has to be valid for all 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 = = = = = tan φ φ = tan b A b b b b cosφ ω ω ω ω ω = + to be the solution. This can be illustrated with the diagra shown. Thus we see that Eq. 4-4 is necessary for x A sin ( ωt φ ) Equate the coefficients of cos ω t. A ω A sinφ + bωa cos φ = F ωb ( ω ω ) A ( ω ) + bω = F ω b ω b ( ω ω ) + ( ω ω ) + ω b ( ω ω ) F A + = F A = ω b ω b ω b ( ω ω ) + ( ω ω ) + ( ω ω ) + Thus we see that Eq. 4- is also necessary for x A sin ( ωt φ ) = + to be the solution. + ( ω ω ) φ ω b ωb ω ω 8 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This aterial is protected under all copyright laws as they currently exist. No portion of this aterial ay be reproduced, in any for or by any eans, without perission in writing fro the publisher. 464

23 Chapter 4 Oscillations bt. 69. (a) For the daped oscillator, the aplitude decays according to A= Ae We are also given the ω Q value, and Q =. We use these relationships to find the tie for the aplitude to b decrease to one-third of its original value. bt/ ω b ω g l Q = = = ; A= Ae = A b Q Q Q Q ( 5) t = ln= ln= ln= ln= 7.7s 7s / b ω g l 9.8 s.5 (b) The energy is all potential energy when the displaceent is at its axiu value, which is the aplitude. We assue that the actual angular frequency is very nearly the sae as the natural angular frequency. bt bt bt g de b g ω ; E= A = Ae = Ae = Ae l dt l / / de g l g A g (.7 g)(.) 9.8 s = = = = dt t= Q l Q ( 5 ).5 l (c) Use Eq. 4-6 to find the frequency spread. Δω Δπ f Δf = Q = = ω π f f Q ( 9.8 s ) (.5) 5. W f ω g l Δ f = = = = =. Hz Q πq πq π( 5) Since this is the total 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. Call the unstretched position of the fire net the zero location for both elastic potential energy and gravitational potential energy. The aount of stretch of the fire net is given by x, easured positively in the downward direction. The vertical displaceent for gravitational potential energy is given by the variable y, easured positively for the upward direction. Calculate the spring constant by conserving energy between the window height and the lowest location of the person. The person has no inetic energy at either location. E = E gy = gy + x top botto top botto botto ( y y top ) [. (.) ] botto 4 g 6 g 9.8 s.9 N xbotto = = = (.) (a) If the person were to lie on the fire net, they would stretch the net an aount such that the upward force of the net would be equal to their weight. g ( 6 g)( 9.8 s ) Fext = x = g x = = = N (b) To find the aount of stretch given a starting height of 8, again use conservation of energy. Note that y = x botto, and there is no inetic energy at the top or botto positions. 8 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This aterial is protected under all copyright laws as they currently exist. No portion of this aterial ay be reproduced, in any for or by any eans, without perission in writing fro the publisher. 465

24 Physics for Scientists & Engineers with Modern Physics, 4 th Edition Instructor Solutions Manual E E gy gy x x x y top botto top botto top x g g = = + = ( 6 g)( 9.8 s ) ( 6 g)( 9.8 s ) x 8 = N.98 N x.576 x.784 = x =.549,.4476 This is a quadratic equation. The solution is the positive root, since the net ust be below the unstretched position. The result is Apply the conservation of echanical energy to the car, calling condition # to be before the collision and condition # to be after the collision. Assue that all of the inetic energy of the car is converted to potential energy stored in the buper. We now that x = and v =. E = E v + x = v + x v = x x g = v = (. s) =. 4 N 7. (a) The frequency can be found fro the length of the pendulu, and the acceleration due to gravity. g 9.8 s f = = =.677 Hz.6Hz π l π.6 (b) To find the speed at the lowest point, use the conservation of energy relating the lowest point to the release point of the pendulu. Tae the lowest point to be the zero level of gravitational potential energy. E = E KE + PE = KE + PE top botto top top botto botto ( θ ) botto + g L Lcos = v + v = gl cosθ = 9.8 s.6 cos5 =.6487 s.65 s botto (c) The total energy can be found fro the inetic energy at the botto of the otion. E total = v =.95g.6487 s = 6. J botto g 7. The frequency of a siple pendulu is given by f =. The pendulu is accelerating π L vertically which is equivalent to increasing (or decreasing) the acceleration due to gravity by the acceleration of the pendulu. (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 x =.5sin 5.5t = Asinωt. (a) The aplitude is A= x =.5. ax 8 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This aterial is protected under all copyright laws as they currently exist. No portion of this aterial ay be reproduced, in any for or by any eans, without perission in writing fro the publisher. 466 h l = l l cosθ θ l cosθ

25 Chapter 4 Oscillations 5.5s (b) The frequency is found by ω = π f = 5.5s f = =.875 Hz π (c) The period is the reciprocal of the frequency. T = π f = 5.5s =.4s. (d) The total energy is given by ( ω ) E = v = A =.65 g 5.5s total ax.5 =.645J.6J. (e) The potential energy is given by ω.65 g 5.5s.5. J. J potential E = x = x = =. The inetic energy is given by E = E E =.645J. J =.9J.9 J. inetic total potential 75. (a) The car on the end of the cable produces tension in the cable, and stretches the cable according F to Equation (-4), Δ l = l, where E is Young s odulus. Rearrange this equation to o E A EA see that the tension force is proportional to the aount of stretch, F = Δl, and so the l o EA effective spring constant is =. The period of the bouncing can be found fro the spring l o constant and the ass on the end of the cable. l ( 5 g)(. o ) T = π = π = π =.47 s.4s 9 EA ( N ) π (. ) (b) The cable will stretch soe due to the load of the car, and then the aplitude of the bouncing will ae it stretch even farther. The total stretch is to be used in finding the axiu aplitude. The tensile strength is found in Table -. F ( x + x static aplitude ) = = tensile strength ( abbrev T.S. ) A π r x ( T.S. ) π ( T.S. ) r πr gl l g = Δ l = = aplitude ( T.S. ) Eπ r Eπr E πr l (. ) ( 5 g)( 9.8 s ) 6 = 9 5 N 9 9 = = ( N ) π (. ) 76. The spring constant does not change, but the ass does, and so the frequency will change. Use Eq. 4-7a to relate the spring constant, the ass, and the frequency. f = f constant f f O O S S π = = = O 6. f = f = S O (.7 Hz) =.6 Hz. S 8 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This aterial is protected under all copyright laws as they currently exist. No portion of this aterial ay be reproduced, in any for or by any eans, without perission in writing fro the publisher. 467

26 Physics for Scientists & Engineers with Modern Physics, 4 th Edition Instructor Solutions Manual Tg 77. The period of a pendulu is given by T = π l g, and so the length is l =. (a) (b) (c) l l (. s) ( 9.79 s ) TgAustin Austin = = = (. s) ( 9.89 s ) TgParis Paris = = = l l = =.6.6 Paris Austin l (. 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 larger upward force on the raft. This extra 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 analogous to pulling down on a ass spring syste that is in equilibriu, by applying an extra force. Then when the an steps off, the restoring force pushes upward on the raft, and thus the raft water syste acts lie a spring, with a spring constant found as follows. F ( 75g)( 9.8 s ) ext 4 = = =. N x.5 (a) The frequency of vibration is deterined by the spring constant and the ass of the raft. 4. N f = = = n.89 Hz.Hz π π g (b) As explained in the text, for a vertical spring the gravitational potential energy can be ignored if the displaceent is easured fro the oscillator s equilibriu position. The total energy is thus E 4 A. N.5.86 J J total = = =. 79. The relationship between the velocity and the position of a SHO is given by Eq. 4-b. Set that expression equal to half the axiu speed, and solve for the displaceent. v =± v x A = v ± x A = x A = x A = ax x =± A ±.866A ax For the pebble to lose contact with the board eans that there is no noral force of the board on the pebble. If there is no noral force on the pebble, then the only force on the pebble is the force of gravity, and the acceleration of the pebble will be g downward, the acceleration due to gravity. This is the axiu downward acceleration that the pebble can have. Thus if the board s downward acceleration exceeds g, then the pebble will lose contact. The axiu acceleration and the aplitude are related by a = π f A. 4 ax g 9.8 s = π ax a 4 f A g A 4. f.5 Hz 8 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This aterial is protected under all copyright laws as they currently exist. No portion of this aterial ay be reproduced, in any for or by any eans, without perission in writing fro the publisher. 468