OSCILLAIONS 4 Q4.. Reason: here are any exaples in daily life, such as a ass hanging fro a spring, a tennis ball being volleyed back and forth, washboard road bups, a beating heart, AC electric voltage, or a pendulu swinging. Assess: he study of oscillations is iportant precisely because they occur so often in nature. Q4.. Reason: Frequency is the rate at which an event is occurring. Since the heartbeat is given as beats per inute, this is a frequency. Assess: Any description of an event that tells you its rate of occurrence is a frequency. Q4.3. Reason: We are given the graph of x versus t. However, we want to think about the slope of this graph to answer velocity questions. (a) When the x versus t graph is increasing, the particle is oving to the right. It has iu speed when the positive slope of the x versus t graph is greatest. his occurs at 0 s, 4 s, and 8 s. (b) When the x versus t graph is decreasing, the particle is oving to the left. It has iu speed when the negative slope of the x versus t graph is greatest. his occurs at s and 6 s. (c) he particle is instantaneously at rest when the slope of the x versus t graph is zero. his occurs at s, 3 s, 5 s, and 7 s. Assess: his is reiniscent of aterial studied in Chapter ; what is new is that the otion is oscillatory and the graph periodic. Q4.4. Reason: Synthesis 4. shows that both the iu speed and iu acceleration are proportional to the aplitude. If the aplitude doubles, then both the iu speed and iu acceleration increase by a factor of two. Assess: he iu acceleration depends ore stongly on the frequency. Q4.5. Reason: Synthesis 4. shows that the iu speed is proportional to the aplitude. For sall angles doubling the angle corresponds to doubling the aplitude, so this increases her iu speed by a factor of two. Assess: he iu acceleration also doubles. Q4.6. Reason: Since the energy and aplitude of an oscillator are related by E ka /, we see that the aplitude is proportional to the square root of E( A E/ k). If E is doubled, A will be increased by a factor of. hat is ANew AOld. What we have done aounts to a qualitative analysis. We inspect the relationship and write down the answer. Soe students find this difficult to do. If that is the case, you should attept to solve the question in this anner and then use a ore quantitative approach to convince yourself that your qualitative answer is correct. For exaple, write the relationship for the old and the new case as follows: E ka / and E ka / Old Old New New Copyright 05 Pearson Education, Inc. All rights reserved. his aterial is protected under all copyright laws as they currently exist. 4-
4- Chapter 4 Divide the new expression by the old one, solve for A, and insert New ENew EOld to obtain he new aplitude is (0 c) 8 c. A A E / E A E / E A New Old New Old Old Old Old Old Assess: he qualitative ethod above requires ore insight and the quantitative ethod requires ore tie. Both skills are needed and copleent each other. Q4.7. Reason: he iu kinetic energy is the sae as the total echanical energy. he total energy and aplitude of an oscillator are related by E ka /, we see that the energy is proportional to the square of the aplitude. If A is doubled, E will increase by a factor of four. hat is ENew 4EOld 4(J) 8J. Assess: his question ay also be answered using a ore quantitative approach as outlined in Question 4.6. Q4.8. Reason: elling us that the initial elongation is doubled is a way of telling us that the aplitude is doubled. Since energy is conserved, the iu potential energy is equal to the iu kinetic energy. ka / v/ Solving this for the iu speed we obtain the following: v A k/ he above expression infors us that the iu speed is proportional to the aplitude: If the aplitude doubles, the speed will double, hence ( v ) ( v ) (0.3/s) 0.6/s New Old Assess: Notice that we are solving the question by inspecting the relationship. his question ay also be answered using a ore quantitative approach as outlined in Question 4.6. Q4.9. Reason: Fro the graph the strategy is to deterine the period, then use f /. As is done in Figure 4.4, one can easure the period between two crests; in this case it appears to be s. f 0.5 Hz s he aplitude is the iu distance fro the equilibriu position. On this graph it appears that A 0 c. Assess: he aplitude is not the distance fro the iu to the iniu that would be A. See Figure 4.6. Q4.0. Reason: Fro the graph we read the period as s and the iu speed as 0.0 /s. Knowing the period we can deterine the frequency as follows: f / / s 0.50 Hz Knowing the iu speed and the frequency we can deterine the aplitude. v ωa ( π/ ) A or A v / π (0.0 /s)(.0 s)/ π 0.03 Assess: hese are fairly typical values for the frequency and aplitude of an oscillator. Q4.. Reason: he period of a block oscillating on a spring is given in Equation 4.6, π / k. We are told that.0 s. (a) In this case the ass is doubled,. So (.0s).8s. π / k π k / Copyright 05 Pearson Education, Inc. All rights reserved. his aterial is protected under all copyright laws as they currently exist.
Oscillations 4-3 (b) In this case the spring constant is quadrupled, k k. So (.0 s).8 s. π k / k k π k / k k (c) he forula for the period does not contain the aplitude; that is, the period is independent of the aplitude. Changing (in particular, doubling) the aplitude does not affect the period, so the new period is still.0 s. Assess: It is equally iportant to understand what doesn t appear in a forula. It is quite startling, really, the first tie you realize it, that the aplitude doesn t affect the period. But this is crucial to the idea of siple haronic otion. Of course, if the spring is stretched too far, out of its linear region, then the aplitude would atter. Q4.. Reason: he period of a siple pendulu is given in Equation 4.7, π Lg /. We are told that.0 s. (a) In this case the ass is doubled,. However, the ass does not appear in the forula for the period of a pendulu; that is, the period does not depend on the ass. herefore the period is still.0 s. (b) In this case the length is doubled, L L. (.0s).8s. So π L / g L L π Lg / L L (c) he forula for the period of a siple sall-angle pendulu does not contain the aplitude; that is, the period is independent of the aplitude. Changing (in particular, doubling) the aplitude, as long as it is still sall, does not affect the period, so the new period is still.0 s. Assess: It is equally iportant to understand what doesn t appear in a forula. It is quite startling, really, the first tie you realize it, that the aplitude ( θ ) doesn t affect the period. But this is crucial to the idea of siple haronic otion. Of course, if the pendulu is swung too far, out of its linear region, then the aplitude would atter. he aplitude does appear in the forula for a pendulu not restricted to sall angles because the sallangle approxiation is not valid; but then the otion is not siple haronic otion. Q4.3. Reason: If it behaves like a ass on a spring, then triing the wings will reduce the ass, and this will k increase the frequency because f. π Assess: It would be easier to beat a wing quickly if it had less ass. Q4.4. Reason: he relationship between the length of a pendulu, its period and the value of the acceleration due to gravity at that site is given by, π L/ g. Inspecting this relationship, we see that the period is inversely proportional to the square root of the acceleration due to gravity. Going fro Miai to Denver results in a saller value of g and hence a larger value of the period of the clock. If the period of the clock is increased, it takes longer for an oscillation and the clock will run slower. he clock will run slower in Denver copared to Miai. Assess: Notice that we solved the question by inspecting the relationship between the period and the acceleration due to gravity. his question ay also be answered using a ore quantitative approach as outlined in Question 4.6. k Q4.5. Reason: Reducing the ass increases the frequency because f So you would reove water. Assess: ry it! Copyright 05 Pearson Education, Inc. All rights reserved. his aterial is protected under all copyright laws as they currently exist.. π L 4π L Q4.6. Reason: Solve the period equation for a siple pendulu for g. π g. So to easure g local g one should easure the length and period of a sall angle siple pendulu very accurately, and plug into this forula for g. Assess: here are solid state devices that do this now.
4-4 Chapter 4 Q4.7. Reason: he leg acts soewhat like a pendulu as it swings forward. By bending their knees to bring their feet up closer to the body, sprinters are shortening the pendulus, which akes the swing faster. Assess: See if you can notice this effect by first running as fast as you can, then again without bending your knees any higher than necessary to clear the ground. Q4.8. Reason: As the gibbons ove through the trees they are essentially swinging pendulus. By bending their knees and bringing their feet closer to their body, they are decreasing the length of the pendulu, which in turn decreases the period of the pendulu. Decreasing the period of the pendulu results in a shorter swing tie and they can ove through the trees faster. Assess: You do the sae thing when you run. As you run, you don t keep your ars extended to their full length, instead you bend the at the elbow. If you didn t the swing of your ars could not keep up with the swing of your legs and you would run in a very awkward anner try it! Q4.9. Reason: Natural frequency is the frequency that an oscillator will oscillate at on its own. You ay drive an oscillator at a frequency other than its natural frequency. Assess: If you are pushing a child in a swing, you build up the aplitude of the oscillation by driving the oscillator at its natural frequency. You can achieve resonance by driving an oscillator at its natural frequency. Q4.0. Reason: Saller auditory systes eans less ass to ove, so they would naturally oscillate at higher frequencies. Assess: A Web search will show that the range of hearing for ice is about khz to 70 khz. Q4.. Reason: he truck is a driven oscillator. At the interediate speed where the vertical otion is large and unpleasant, the driving frequency due to hitting the bups of the washboard is close to the natural (resonance) frequency (which is deterined by the suspension syste and the ass of the truck); the resulting large-aplitude otion is resonance. At speeds either significantly above or below that interediate speed, the driving frequency of the bups in the road is either saller or greater than the resonance frequency and the response of the driven oscillator is sall. Assess: It should be noted that a driven oscillator does oscillate at the driving frequency, whether that happens to be close to the resonance frequency or not. So the frequency of the up-and-down otion of the truck is the frequency with which it hits the regular bups in the washboard road. You could try an experient by varying the natural frequency of your truck by loading it ore or less and see if the speed that produces resonance on the sae washboard road changes siilarly. Q4.. Prepare: When the kangaroo increases its speed the tendons stretch ore which stores ore energy in the, so they spend ore tie in the air propelled by greater spring energy. When in contact with the ground it is like a spring in siple haronic otion. Hopping faster would increase the aplitude of the oscillation, but that doesn t change the period, or tie in contact with the ground. Assess: Kangaroo hopping can be quite efficient at high speeds. Q4.3. Reason: (a) he unstretched equilibriu position is 0 c. When we load the spring with 00 g we establish a new equilibriu position at 30 c. When we pull the oscillator down to 40 c (i.e., an additional 0 c) and release it, it will oscillate with an aplitude of A 0 c. he correct choice is B. (b) Knowing that 00 g stretched the spring 0 c, we can deterine the spring constant. k g/ x (0.kg)(9.8 /s )/(0.0 ) 9.8 N/ Knowing the spring constant and the ass on the spring, we can deterine the oscillation frequency as follows: f k/ / π (9.8 N/)/(0.0 kg)/π.6 Hz he correct choice is C. (c) he frequency of oscillation depends on k and, neither of which would change on the oon. he correct choice is C. Assess: hese are rather sall but acceptable values for the spring constant and frequency. Copyright 05 Pearson Education, Inc. All rights reserved. his aterial is protected under all copyright laws as they currently exist.
Oscillations 4-5 Q4.4. Reason: his is siple haronic otion where the equilibriu position is at x 0. (a) One can read the period off the graph by seeing how uch tie elapses fro peak to peak or trough to trough. Reading peak to peak gives about 38 s 4 s 4 s. So the correct choice is B. (b) he aplitude is the iu displaceent fro the equilibriu position. his is easily read at a nuber of places on the graph, but the first positive peak occurs at about t 4 s; the displaceent there is 4.5 c.he correct choice is C. (c) At the tie t 30 s the graph crosses a grid line and allows us to read the answer, x c. he correct choice is B. (d) he velocity of the object is given by the slope of the x versus t graph. We are looking for the first tie that slope is zero (the tangent line is horizontal); this occurs at t s. he correct choice is B. (e) Kinetic energy is iu when the speed is greatest; this occurs as the object oves through the equilibriu position, not at the end points of its otion where it is instantaneously at rest. Of the choices given, 8 s is the tie where the graph crosses the t-axis and where x 0 and the object is in equilibriu (but oving quickly). he correct choice is B. Assess: It would be very instructive to construct a v x versus t graph for this sae situation (it would be the slope of this x versus t graph) and think about the sae questions in relation to the new graph. Q4.5. Reason: Soe of the question ay be answered by coparing the expression given with the general equation for siple haronic otion. (a) he expression given for the displaceent is x (0.350 ) cos(5.0 t). he general expression for the displaceent is x Acos ωt. Coparing these two expressions for the displaceent we see that the aplitude of oscillation is A 0.350, choice B. (b) he frequency of oscillation ay be deterined by f ω/ π (5.0/s)/ π.39 Hz, choice B. (c) he ass attached to the spring ay be deterined by kω / (00 N/)/(5/s) 0.89 kg, choice B. (d) he total echanical energy of the oscillator is equal to its iu potential energy, which ay be deterined by Eotal U ka/ (00 N/)(0.350 ) /. J, choice E. (e) he iu speed ay be deterined by v ωa (5.0/s)(0.350 ) 5.5 /s, choice E. Assess: Working this proble brings our attention to two things. First, there is a lot of inforation tucked away in the function given for the displaceent of the oscillator. Second, there are a lot of details associated with an oscillation. It is iportant to know these details and their interconnection. Q4.6. Prepare: his is a ass on a spring syste. Increasing the ass increases the period and decreases the frequency. he answer is C. Assess: Even though the ass of the people inside is five ties greater that is only a part of the ass of the whole car. Q4.7. Prepare: Your ars act like siple pendulus and carrying the weights in your hands erely increases g the ass of the bob, but that doesn t affect the frequency of oscillation, f. he answer is B. π L Assess: he ass of the bob doesn t appear in the equation for the frequency of a siple pendulu. Q4.8. Reason: he relationship between the period, the length of the pendulu, and the acceleration due to gravity at the site of the pendulu is π L/ g. his expression ay be solved for the length L to obtain: L t g ρ π /4 (5.5 s) (9.80 /s )/4 7.5 he correct choice is C. Assess: While this is a rather long pendulu, it is in the acceptable range. Q4.9. Reason: We see in Figure 4.6 that the cells on the basilar ebrane close to the stapes correspond to higher frequencies. he correct choice is B. Assess: People often lose hearing sensitivity in the higher frequencies with age. Copyright 05 Pearson Education, Inc. All rights reserved. his aterial is protected under all copyright laws as they currently exist.
4-6 Chapter 4 Proble P4.. Prepare: he period of the vibration is the inverse of the frequency, so we will use Equation 4.. he frequency generated by a guitar string is 440 Hz., hence 3.3 0 s.3 s f 440 Hz Assess: he units of frequency are Hz, that is, cycles per second, or s, so the period is in seconds. P4.. Prepare: Knowing the relationship between period and frequency, we can deterine the frequency fro the period. 4 he period is deterined by f / /(54 in)( in/60 s) 3. 0 Hz. Assess: We expect the frequency to be sall for such a large period. P4.3. Prepare: Your pulse or heartbeat is 75 beats per inute or 75 beats/60 s.5 beats/s. he period is the inverse of the frequency, so we will use Equation 4.. he frequency of your heart s oscillations is 75 beats f.5 beats/s.3 Hz 60 s he period is the inverse of the frequency, hence 0.80 s f.3 Hz Assess: A heartbeat of.3 beats per second eans that one beat takes a little less than second, which is what we obtained above. P4.4. Prepare: Use ratios. Call x the new stretch. Solve Hooke s law for x. F x k g.5 kg 3 x F g.0 kg k 3 3 herefore x x (6.4 c) 9.6 c. Assess: We expected a greater weight to stretch the spring ore. P4.5. Prepare: For a sall angle pendulu the restoring force is proportional to the displaceent because sin θ θ. If we odel this as a Hooke s law situation, then doubling the distance will double the restoring force fro 0 N to 40 N. Assess: his would not work for angles uch larger than 0. P4.6. Prepare: he air-track glider oscillating on a spring is in siple haronic otion. he glider copletes 0 oscillations in 33 s, and it oscillates between the 0 c ark and the 60 c ark. We will use Equations 4. and 4.3. (a) 33s 3.3s/oscillation 3.3s 0 oscillations (b) f 0.30 Hz 3.3s Copyright 05 Pearson Education, Inc. All rights reserved. his aterial is protected under all copyright laws as they currently exist.
Oscillations 4-7 (c) he oscillation fro one side to the other is equal to 60 c 0 c 50 c 0.50. hus, the aplitude is A (0.50 ) 0.5. (d) he iu speed is π v A ( π /3.3s)(0.5 ) 0.48 /s Assess: he glider takes 3.3 seconds to coplete on oscillation. Its iu speed of 0.48 /s, or ph, is reasonable. P4.7. Model: he air-track glider attached to a spring is in siple haronic otion. he glider is pulled to the right and released fro rest at t 0 s. It then oscillates with a period.0 s and a iu speed 4v 0 c/s 0.40 /s. While the aplitude of the oscillation can be obtained fro Equation 4.3, the position π t of the glider can be obtained fro Equation 4.0, xt () Acos( ). (a) v (0.40 /s)(.0 s) v π A A π π (b) he glider s position at t 0.5 s is ( / ) 0.7 0.3 π (0.5 s) x0.5 s (0.7 ) cos 0.090 9.0 c.0 s Assess: At t 0.5 s, which is less than one quarter of the tie period, the object has not reached the equilibriu position and is still oving toward the left. P4.8. Prepare: he oscillation is the result of siple haronic otion. As the graph shows, the tie to coplete one cycle (or the period) is.0 s. We will use Equation 4. to find frequency. (a) he aplitude A 0 c. (b) f 0.50 Hz.0 s Assess: It is iportant to know how to find inforation fro graphs. P4.9. Prepare: he oscillation is the result of siple haronic otion. As the graph shows, the tie to coplete one cycle (or the period) is 4.0 s. We will use Equation 4. to find frequency. (a) he aplitude (b) he period 4.0 s, thus A 0 c. f 0.5 Hz 4.0 s Assess: It is iportant to know how to find inforation fro a graph. P4.0. Prepare: reating the building as an oscillator, the agnitude of the iu displaceent is the aplitude, the agnitude of the iu velocity is deterined by v π fa, and the agnitude of the iu acceleration is deterined by a π f A. (a) he agnitude of the iu displaceent is x A 0.30. (b) he agnitude of the iu velocity is v π fa π(. Hz)(0.30 ).3 /s. (c) he agnitude of the iu acceleration is a ( π f) A 4 π (. Hz) (0.30 ) 7 /s. Assess: hese are reasonable values for the agnitude of the iu displaceent, velocity, and acceleration. Copyright 05 Pearson Education, Inc. All rights reserved. his aterial is protected under all copyright laws as they currently exist.
4-8 Chapter 4 P4.. Prepare: We will assue that ship and passengers are approxiately in siple haronic otion. Synthesis box 4. gives the iu acceleration for an object in siple haronic otion, a ( π f) A. A and f / /5 s 0.067 Hz. (a) a ( π f) A (π 0.067 Hz) () 0. /s (b) o one significant figure, g 0 /s, so the passenger s acceleration is about g. 50 Assess: his is not a large acceleration, but it can play havoc with soe people s stoachs. P4.. Prepare: Solve a ( π f) A for A. a 0. 0 / s (a) A 30. ( π f ) ( π(. 3 Hz)) (b) v π fa π(. 3 Hz)(0. 0030 ) 0. 04 /s Assess: hese both see like sall but possible answers. P4.3. Prepare: he total side-to-side otion is A. Solve Assess: he height of the building is not needed. a π f A ( ) for A. a (0. 00)(9. 80 / s ) A 34 c ( π f ) ( π(0. 7 Hz)) P4.4. Prepare: he distance between upper and lower liits of otion is actually A, so v π fa π(40 Hz)(0. 0050 ).3 /s Assess: his sees reasonable. P4.5. Prepare: Use the equation for v. Solve the equation for A. v.5 /s v π fa A π f π(50 Hz) Assess: his sees about the right size for a bublebee..6 a (0. 00)(9. 80 / s ). A 0.50 c. P4.6. Prepare: he figure shows the aplitude is 40. It also shows the period is 40 s, so the frequency is f 5 Hz. 40 s Use the equation for the iu acceleration. a ( π f) A (π 5Hz) (40 ) 987/s 990 /s Dividing by 9.8 /s to get in units of g we get a 00 g. Assess: his is a very large acceleration for a biological syste. P4.7. Prepare: he spring undergoes siple haronic otion. he elastic potential energy in a spring stretched by a distance x fro its equilibriu position is given by Equation 4.9, and the total echanical energy of the object is the su of kinetic and potential energies as in Equation 4.0. At iu displaceent, the total energy is siply E KA, Equation 4.. (a) When the displaceent is x A the potential energy is, 3 U kx k A ka E K E U E 4 4 4 Copyright 05 Pearson Education, Inc. All rights reserved. his aterial is protected under all copyright laws as they currently exist.
Oscillations 4-9 hus, one quarter of the energy is potential and three-quarters is kinetic. (b) o have U E requires A U kx E ka x P4.8. Prepare: he block attached to the spring is in siple haronic otion. At iu displaceent position, x A, and the kinetic energy is zero. We can use the energy conservation Equation 4.0 to find the aplitude. (a) he conservation of echanical energy equation Kf + U f Ki + Ui is v + k( Δ x) v0 + 0J 0J+ ka v0 + 0J.0 kg A v0 (0.40/s) 0.0 0c k 6 N/ (b) We have to find the velocity at a point where x A/. he conservation of echanical energy equation K + U K + U is i i A 3 v + k v0 + 0J v v0 ka v0 v0 v0 4 4 4 3 3 v v0 (0.40 /s) 0.346 /s 35 c/s 4 4 Assess: Speed decreases as an object oves away fro the equilibriu position, so a decreased speed of 35 /s copared to 40 /s at the equilibriu position is reasonable. P4.9. Prepare: he block attached to the spring is in siple haronic otion. he period of an oscillating ass on a spring is given by Equation 4.6. he period of an object attached to a spring is π 0.00s k where is the ass and k is the spring constant. (a) For ass, π ( ) 0.83s k Copyright 05 Pearson Education, Inc. All rights reserved. his aterial is protected under all copyright laws as they currently exist.
4-0 Chapter 4 (b) For ass, π 0/.4s k (c) he period is independent of aplitude. hus 0.00 s. (d) For a spring constant k, π 0/.4s k Assess: As would have been expected, increase in ass leads to slower siple haronic otion. P4.0. Prepare: he air-track glider attached to a spring is in siple haronic otion. Experientally, the period of otion is (.0 s)/(0 oscillations).0 s. Equation 4.6 relates the period to the glider s ass and the spring constant. Using Equation 4.6 for the period, π π π k (0.00 kg) 5.5 N/ k.0 s Assess: k 5.5 N/ eans that a force of 5.5 N or a ass of 0.56 kg (a little over a pound) will stretch/copress it by 00 c. his is a soft spring. P4.. Prepare: he oscillating ass is in siple haronic otion. he position of the oscillating ass is given by x( t) (.0 c)cos(0 t), where t is in seconds. We will copare this with Equation 4.0. (a) he aplitude A.0 c. (b) he period is calculated as follows: π π 0 rad/s 0.63s 0 rad/s (c) he spring constant is calculated fro Equation 4.6 as follows: π k π k (0.050 kg)(0 rad/s) 5.0 N/ (d) he iu speed fro Equation 4.6 is π v π fa A (0 rad/s)(.0 c) 0 c/s (e) he total energy fro Equation 4. is (5.0 N/)(0.0 ).0 0 3 E ka J (f) At t 0.40 s, the velocity fro Equation 4. is v (0.0 c/s) sin[(0 rad/s)(0.40 s)] 5 c/s x Assess: Velocity at t 0.40 s, is less than the iu velocity, as would be expected. P4.. Prepare: he oscillating ass is at rest (instantaneously) at the top and at the botto. Between these is a distance of A 0c. he equilibriu position is halfway between those points (when the spring is stretched 0 c) and the net force (the su of the upward spring force and the downward gravitational force) is zero. We are given 0.5 kg. (a) Apply Newton s second law to the equilibriu position 0 c below the release point. g (0.5 kg)(9.8 /s ) Σ F kx g 0 k 4.5 N/ 5 N/ x 0.0 (b) he aplitude of the oscillation is half the total travel distance, or 0 c. Copyright 05 Pearson Education, Inc. All rights reserved. his aterial is protected under all copyright laws as they currently exist.
Oscillations 4- (c) he frequency is Assess: hese are typical nubers. f k 4.5 N/.6 Hz π π 0.5 kg P4.3. Prepare: he ass attached to the spring oscillates in siple haronic otion. he ass oscillates at a frequency of.0 Hz. We will need the spring constant k which we will deterine using Equation 4.6. (a) he period using Equation 4. is / f /.0 Hz 0.50 s. (b) Using energy conservation ka kx ( ). 0 + vx 0 Using x 0 5.0 c, ( v x) 0 30c/s, and k ( π f) 0. kg[ π(.0 Hz)] 3.58 N/, we get A 5.54 c, which is to be reported as 5.5 c. (c) he iu speed fro Equation 4.6 is v π fa π(.0 Hz)(5.54 c) 69.6 c/s, which will be reported as 70 c/s. (d) he total energy is E v (0.00 kg)(0.696 /s) 0.049 J. P4.4. Prepare: he ass attached to the spring is in siple haronic otion. (a) he period using Equation 4.6 is (0.507 kg) π π.00s k (0 N/) (b) Using Equation 4.6, the iu speed v π fa ( π/ ) A ( π/.00 s)(0.0 ) 0.68 /s. (c) he total energy fro Equation 4. is E v (0.507 kg)(0.68 /s) 0.00 J. P4.5. Prepare: Assue a sall angle of oscillation so there is siple haronic otion. We will use Equation 4.7 for the pendulu s tie period. he period of the pendulu is L π g 0 0 4.00s (a) he period is independent of the ass and depends only on the length. hus 0 4.00 s. (b) For a new length L L0, (c) For a new length L L /, 0 L π g 0 0 5.66 s L / π g 0 0.83s (d) he period is independent of the aplitude as long as there is siple haronic otion. hus 4.00 s. P4.6. Prepare: Assue a sall angle of oscillation so there is siple haronic otion. We will use Equation 4.7 for the pendulu s tie period. A coplete swing is back and forth, so.00 s. Solve for the length. L (.00 s) π L g (9.80 /s ) 0.48 g 4π 4π Assess: A pendulul a eter long has a period of s, which fits with this anwer. P4.7. Prepare: Because the angle of displaceent is less than 0, the sall-angle approxiation holds and the pendulu exhibits siple haronic otion. We will use Equation 4.7 and g 9.80 /s. Copyright 05 Pearson Education, Inc. All rights reserved. his aterial is protected under all copyright laws as they currently exist.
4- Chapter 4 he period is.0 s/0 oscillations.0 s and is given by the forula L.0 s π L g (9.80 /s) 35.7 c g π π Assess: A length of 35.7 c for the siple pendulu is reasonable. P4.8. Prepare: Assue the pendulu to have sall-angle oscillations. In this case, the pendulu undergoes siple haronic otion. We will use Equation 4.7. he periods of the pendulus on the oon and on the earth are Because, earth oon earth L π earth π and oon gearth Learth Loon goon.6 /s π π Loon Learth gearth g oon gearth 9.8 /s (.00) 33.c Assess: Because of saller g on the oon, a saller pendulu length to have the sae period as on the earth was indeed expected. P4.9. Prepare: Assue a sall angle of oscillation so that the pendulu has siple haronic otion. We will use Equation 4.7. he tie periods of the pendulus on the earth and on Mars are L L earth π and Mars π gearth g Mars Dividing these two equations, earth g Mars earth.50 s gmars gearth (9.80 /s ) 3.67 /s Mars gearth Mars.45 s Assess: Because, Mars > earth the sae length of the pendulu would iply saller g on Mars, as obtained above. P4.30. Prepare: o coplete a whole period, the wrecking ball will have to swing down, up to the other side, back down, and up again to the original position. So the tie it takes to swing fro iu height down to lowest height once is one-quarter of a period. We will assue that the wrecking ball is a siple sall-angle pendulu and so it s period is given by π L/ g. L g L π 0 π.6s 4 4 g 9.80/s Assess: his is enough tie to dive out of the way, but it is still wiser to not stand in the way of wrecking balls. P4.3. Prepare: reating the lower leg as a physical pendulu we can deterine the oent of inertia by cobining π I/ gl and / f. Cobining the above expressions and solving for the oent of inertia we obtain I gl/( π f) (5.0 kg)(9.80 /s )(0.8 )/[ π(.6 Hz)] 8.7 0 kg oon oon Assess: NASA deterines the oent of inertia of the shuttle in a siilar anner. It is suspended fro a heavy cable, allowed to oscillate about its vertical axis of syetry with a very sall aplitude, and fro the period of oscillation one ay deterine the oent of inertia. his arrangeent is called a torsion pendulu. P4.3. Prepare: We will odel the rope as a siple sall-angle pendulu. We want to hang on to the rope for a quarter of a period to get as far out as possible. Copyright 05 Pearson Education, Inc. All rights reserved. his aterial is protected under all copyright laws as they currently exist.
Oscillations 4-3 t π L π 5 9. s 4 g 9. 80/s hang Assess: his sees reasonable for such a long pendulu. P4.33. Prepare: reating the hoop suspended fro the nail as a physical pendulu we can deterine the period of oscillation using π I/ gl. he period of oscillation for the hoop suspended fro its ri on a nail ay be deterined by I gl R gr R g π / π / π / π (0. )/(9.80 /s ).3s Assess: his is a reasonable period for this situation. P4.34. Prepare: Model the elephant legs as physical pendulus with I L. he distance fro the pivot 3 point to the center of gravity is d L For a leg to swing forward requires half a period. (a). L π π π π. 4 s. s gd g 3 g 3(9 8 s ) I 3 L (. 3) L. / (b) he right leg takes.4 s to swing forward, then stay planted while the left leg takes.4 s to swing forward, so it takes.48 s for a whole period of the process. hat is, each leg hits the ground /.48 ties per second. here are 4 legs. 60s (4 legs) 97 steps/in 48s. in Assess: his sees to jibe with the nature shows we ve seen. P4.35. Prepare: he otion is a daped oscillation. he iu displaceent or aplitude of the oscillation / at tie t is given by Equation 4.9, () t x, t Ae τ where τ is the tie constant. Using x 0.368 A and t 0.0 s, we can calculate the tie constant. Fro Equation 4.9. 0.0 s/ τ 0.0 s 0.0 s 0.368A Ae ln (0.368) τ 0.0 s τ ln (0.368) Assess: he above result says that the oscillation decreases to about 37% of its initial value after one tie constant. P4.36. Prepare: he otion is a daped oscillation. he iu displaceent or aplitude of the oscillation / at tie t is given by Equation 4.9, () t x, t Ae τ where τ is the tie constant. Using x 0.50 A and t in 30 s, we can calculate the tie constant. Fro Equation 4.9 30 s/ τ 30 s 30 s 0.50A Ae ln (0.50) τ 900 s 3 in τ ln (0.50) he tie constant concept is such that if the aplitude decreased by half in in, then it will take another in to decrease by half again. So it will take an additional in to decrease to 5% of its initial value. Assess: We converted the tie to SI units, but would have gotten the sae answer if we had done the whole proble in inutes, without converting to seconds. he pronunciation of Foucault is foo-koh. P4.37. Prepare: Assue A in arbitrary units. he object continues to oscillate but x decreases due to the daping. / t (4. 0 s) he equation for the graph is x() t e cos( π (0. Hz)). t Copyright 05 Pearson Education, Inc. All rights reserved. his aterial is protected under all copyright laws as they currently exist.
4-4 Chapter 4 Assess: he oscillations dap out at about the rate we expect for τ 40s.. P4.38. Prepare: he initial aplitude is A 65. c. he equation is () e /τ x t t A with x (8. 0 s). 8 c. 80. s/ τ 80. s (a). 8 c (6. 5 c)e τ 6. 3 s 6. s 8. c ln 65. c / t τ (b) Use x () t Ae with t 405. andτ 6.3s x (4. 0 s) /(6. 3 s) (4. 0 s) (6. 5 s)e 3. 4 c Assess: We expect the aplitude at 4.0 s to be between 6. 5 c and.8 c. P4.39. Prepare: Assue daped oscillations. We exaine the graph carefully. It looks fro the graph that the period is 0. 5 s so f Hz. Looking at the peaks, it appears the aplitude has decreased to about 37% of its value after only 0.5 s, so that is our guess for τ. Assess: You want the tie constant for daping on your car to be short. P4.40. Prepare: We will odel the tuned ass daper as a siple sall-angle pendulu. (a) Use the forula for the period of a siple pendulu with a length of 4. L 4 π π 3s g 9.8 /s (b) With an aplitude of 0.75, the iu speed is π π v π fa A (0.75 ) 0.36 /s 36 c/s 3 s Assess: It is ipressive to think of a 660,000 kg ass oving that fast. Note we did not need the ass. P4.4. Prepare: We will odel the child on the swing as a siple sall-angle pendulu. o ake the aplitude grow large quickly we want to drive (push) the oscillator (child) at the natural resonance frequency. In other words, we want to wait the natural period between pushes. L.0 π π.8s g 9.8 /s Assess: You could also increase the aplitude by pushing every other tie (every ), but that would not ake the aplitude grow as quickly as pushing every period. he ass of the child was not needed; the answer is independent of the ass. P4.4. Prepare: We first deterine the frequency of the oscillation, then look at the graph to deterine whether we need to increase or decrease the frequency to decrease the aplitude. Copyright 05 Pearson Education, Inc. All rights reserved. his aterial is protected under all copyright laws as they currently exist.
Oscillations 4-5 he frequency is 0 i/ h 580 ft h 93. Hz 0 ft i 3600 s his is to the right of the peak of the graph, so we need to increase the frequency to decrease the aplitude; you should speed up. Assess: Experience teaches that soeties it helps to speed up on a washboard road. P4.43. Prepare: Given the ass and the resonant frequency, we can deterine the effective spring constant using the relationship ω π f k /. Solving the above expression for the spring constant, obtain 3 k ( π f) [ π(9 Hz)] (7.5 0 kg) 50 N/ Assess: As spring constants go, this is a fairly large value, however the usculature holding the eyeball in the socket is strong and hence will have a large effective spring constant. P4.44. Prepare: For part (a) start fro Hooke s law ( F ) (a) Solve Equation 4.5 for k. sp (0.080 kg)(9.8 /s ) g k 0 N/ L 0.040 Δ (b) he period of a assive object (such as a ball) on a spring is (c) 0.080 kg π π 0.40 s k 0 N/ y kδ y and use the derivation in Equation 4.4. Assess: hese are typical values for springs like those used in college physics labs. P4.45. Prepare: Insert the forula for the period of a ass on a spring into the forula for the iu speed of a haronic oscillator. π π k N/ v π fa A A A (0.0 )./s π 0.40 kg Assess: his proble can also be done with energy conservation considerations. k P4.46. Prepare: he vertical oscillations constitute siple haronic otion given by Equation 4.6. A pictorial representation of the spring and the book is given. Copyright 05 Pearson Education, Inc. All rights reserved. his aterial is protected under all copyright laws as they currently exist.
4-6 Chapter 4 (a) At equilibriu, Newton s first law applied to the physics book is ( Fsp) y g 0 N kδy g 0 N k g/ Δ y (0.500 kg)(9.8 /s )( 0.0 ) 4.5 N/ which we will report as 5 N/ to two significant figures. (b) o calculate the period fro Equation 4.6 π k 4.5N/ π rad 7.0 rad/s 0.898 s 0.90 s 0.50 kg 7.0 rad/s (c) he iu speed fro Equation 4.6 is π v A( π f) A (0.0 )(7.0 rad/s) 0.70 /s Maxiu speed occurs as the book passes through the equilibriu position. P4.47. Prepare: he vertical oscillations constitute siple haronic otion. A pictorial representation of the spring and the ball is shown in the following figure. he period and frequency of oscillations are 0 s 0.6667 s and f.50 Hz 30 oscillations 0.6667 s Since k is known, we can obtain the ass using Equation 4.6. (a) he ass can be found as follows: k k 5.0N/ f 0.69 kg π ( π f) [ π(.50hz)] (b) he iu speed is given by Equation 4.6, v π fa π(.50hz)(0.0600) 0.565/s. Assess: Both the ass of the ball and its iu speed are reasonable. Copyright 05 Pearson Education, Inc. All rights reserved. his aterial is protected under all copyright laws as they currently exist.
Oscillations 4-7 P4.48. Prepare: he vertical oscillations constitute siple haronic otion. o find the oscillation frequency using Equation 4.6, π f k/, we first need to find the spring constant k. In equilibriu, the weight g of the block and the spring force kδ L are equal and opposite. hat is, g kδl k g/ Δ L. As shown in the next figure, the ass of the ug drops out of the calculations, and the stretch length Δ L is.0 c. he frequency of oscillation f is k g/ ΔL g 9.8/s f 3.5 Hz π π π ΔL π 0.00 P4.49. Prepare: he vertical oscillations constitute siple haronic otion. We will use Equation 4.6. At the equilibriu position, the net force on ass on Planet X is the following: k gx F net k Δ L g X 0N Δ L For siple haronic otion, Equation 4.6 yields k / ( π f), thus gx π π X Δ ( π f) g L (0.3 ) 5.86 /s ΔL 4.5 s/0 Assess: his value of g is of the sae order of agnitude as the one for the earth, and would thus see to be reasonable. P4.50. Prepare: he object is undergoing siple haronic otion. We will use Equation 4.8 for velocity of an object undergoing siple haronic otion. Copyright 05 Pearson Education, Inc. All rights reserved. his aterial is protected under all copyright laws as they currently exist.
4-8 Chapter 4 he velocity of an object oscillating on a spring is vx () t ( π f) Asin( π ft) (a) For A A and f f/, we have v x ( t) ( π f/)( A) sin[ π f/) t] ( π f) Asin( π ft) hat is, the iu velocity A ( π f ) reains the sae, but the frequency of oscillation is halved. (b) For 4, k k f f and A A v A( π f ) Aπ f v / π π 4 Quadrupling of the ass halves the frequency or doubles the tie period, and halves the iu velocity. Assess: It is iportant to know how to present inforation on a graph. P4.5. Prepare: he vertical ass/spring systes are in siple haronic otion. (a) For syste A, y is positive for one second as the ass oves downward and reaches iu negative y after two seconds. It then oves upward and reaches the equilibriu position, y 0, at t 3seconds. he iu speed while traveling in the upward direction thus occurs at t 3.0 s. he frequency of oscillation is 0.5 Hz. (b) For syste B, all the echanical energy is potential energy when the position is at iu aplitude, which for the first tie is at t.5 s. he tie period of syste B is thus 6.0 s. (c) Spring/ass A undergoes three oscillations in s, giving it a period A 4.0 s. Spring/ass B undergoes two oscillations in s, giving it a period B 6.0 s. Fro Equation 4.6, we have A B A A k B 4.0 s A π and B π ka kb B B ka 6.0 s 3 If, then A B kb 4 ka 9.5 ka 9 kb 4 Assess: It is iportant to learn how to read a graph. P4.5. Prepare: First we figure the ass of the chair alone, then the ass of the chair plus astronaut, then subtract. (0.90s) ch k (606 N/).46 kg 4π 4π (.09 s) tot k (606 N/) 67.05 kg 4π 4π * tot ch 67.05 kg.46 kg 54.59 kg which we report as 54.6 kg. Assess: his is a reasonable weight for a light astronaut. P4.53. Prepare: he ball attached to the spring is in siple haronic otion. he position and velocity at tie t are x 0 5c and v 0 0 c/s. An exaination of Synthesis 4. shows that x() t Acos( π ft) and vx () t π Asin( π ft). Adding the squares of these equations and using the trigonoetric relationship f v () + we have ( ) x t + cos θ sin θ, (a) he oscillation frequency is A x() t. π f k.5n/ f 0.796 Hz π π 0.0 kg Copyright 05 Pearson Education, Inc. All rights reserved. his aterial is protected under all copyright laws as they currently exist.
Oscillations 4-9 he aplitude of the oscillation is vt () 0c/s A x( t) + ( 5.00 c) + 6.40 c π f π(0.796hz (b) We can use the conservation of energy between x i 5c and x f 3c as follows: k vi + kxi vf + kxf vf vi + ( xi xf ) 0.83 /s 8.3 c/s Assess: Because k is known in SI units of N/, the energy calculation ust be done using SI units of, /s, and kg. Both the aplitude and speed are reasonable. P4.54. Prepare: he transducer undergoes siple haronic otion. he iu restoring force that the disk can withstand is 40 000 N. By applying Newton s second law to the disk of ass 0.0 g, we will first find its iu acceleration and then use Equation 4.7 to find the iu oscillation aplitude. Once we find the aplitude, we will use Equation 4.4 to find the disk s iu speed. (a) Newton s second law for the transducer is 3 8 Frestoring a 40 000 N (0.0 0 kg) a a 4.0 0 /s Because fro Equation 4.7 a ( π f) A, 8 a 4.0 0 /s 5 A.0 0 0 μ 6 ( π f ) [ π(.0 0 Hz] (b) he iu speed is v ω π 6 5 A (.0 0 Hz)(.0 0 ) 64 /s Assess: Apparently, the aplitude is very sall and the iu oscillation speed is large. However, to oscillate at high frequencies such as.0 MHz, you would expect a sall aplitude and a large speed. P4.55. Prepare: he copact car is in siple haronic otion. he ass on each spring for the epty car is (00 kg)/4 300 kg. However, the car carrying four persons eans that each spring has, on the average, an additional ass of 70 kg. For both parts we will use Equation 4.7. First calculate the spring constant as follows: k f k π f π π Now reapply the equation with 370 kg, so 4 ( ) (300 kg)[ (.0 Hz)] 4.74 0 N/ 4 k 4.74 0 N/.8 Hz f π π 370kg Assess: A sall frequency change fro the additional ass is reasonable because frequency is inversely proportional to the square root of the ass. P4.56. Prepare: he frequency is f (6.0 /s)/(5.0 ). Hz. We solve Hooke s law for the distance oved, F kδx Δ x F/ k where we drop the inus sign because we already know the car will rise up. F is the weight of the people and is 300 kg. k f k ( π f) (400 kg)[ π(. Hz)] 80 kn/ π Assess: his sees to be a reasonable aount for the car to rise when four people get out. Copyright 05 Pearson Education, Inc. All rights reserved. his aterial is protected under all copyright laws as they currently exist.
4-0 Chapter 4 P4.57. Prepare: A copletely inelastic collision between the two gliders results in siple haronic otion. Let us denote the 50 g and 500 g asses as and, which have initial velocities ( v ) iand ( v ). i After collides with and sticks to, the two asses ove together with velocity v f. We will first find the final velocity using oentu conservation and then use the echanical energy conservation equation for the two stuck gliders to deterine the aount of copression or the aplitude. he oentu conservation equation pf pi for the copletely inelastic collision is ( + ) v ( v ) + ( v ). Substituting the given values, f i i (0.750 kg) vf (0.5 kg)(.0 /s) + (0.50 kg)(0 /s) vf 0.40 /s We now use the conservation of echanical energy equation, ( K + Us) copressed ( K + Us) equilibriu 0J + KA ( + ) vf + 0J + 0.750 kg A vf (0.40 /s) 0. k 0 N/ he period is π + k π 0.750 kg 0 N/.7 s Assess: he agnitudes of both the aplitude and the tie period are physically reasonable. P4.58. Prepare: A copletely inelastic collision between the bullet and the block results in siple haronic otion. Let us denote the bullet s and block s asses as b and, which have initial velocities B v b and v. After B b collides with and sticks to B, the two ove together with velocity v f and exhibit siple haronic otion. Since the aplitude of the haronic otion is given, we will first find the final velocity of the bullet + block syste using the echanical energy conservation equation. he oentu conservation will then give us the bullet s speed. Copyright 05 Pearson Education, Inc. All rights reserved. his aterial is protected under all copyright laws as they currently exist.
Oscillations 4- (a) he equation for conservation of energy after the collision is k 500 N/ ka ( b B) f f (0.0 ) 4.975 /s + v v A b + B.00 kg he oentu conservation equation for the perfectly inelastic collision pafter pbefore is ( b + B) vf bvb + BvB (.00 kg)(4.975 /s) (0.00 kg) v + (.00 kg)(0 /s) v 50 /s (b) No. he oscillation frequency k/( b + B) depends on the asses but not on the speeds. π f Assess: he bullet s speed is high but not uniaginable. P4.59. Prepare: When the block is displaced fro the equilibriu position one spring is copressed and exerts a restoring force on the block while the other spring is stretched and also exerts a restoring force on the block. hese two forces have the sae agnitude since the springs are identical and a stretch or copression of the sae aount produces the sae restoring force. (a) Use Hooke s law. he restoring force is the su of the restoring forces fro the two springs. F kx (0 N/ + 0 N/)( 0.00 ) 0.40 N) (b) he restoring force would be the sae if the springs were on the sae side of the block. For springs in parallel like that, the total k tot is the su of the two k s. ktot k + k 0 N/ + 0 N/ 40 N/ (c) f two is the frequency for the syste with two springs. b ktot 40N/ 0.64 Hz f two π π.5kg Assess: he answer sees reasonable. If the two k s are the sae k k k (as in this case), you can see the general forula would be + two f fone π π π he frequency with two identical springs is ties the frequency with one spring. An even ore general result can be obtained with siilar reasoning even where the two k s differ, f f b +. two Copyright 05 Pearson Education, Inc. All rights reserved. his aterial is protected under all copyright laws as they currently exist.
4- Chapter 4 P4.60. Prepare: he Super Bungee will stretch until all of the bus s kinetic energy is stored as elastic potential energy in the cord. he bus will stop at one-fourth the period of oscillation. (a) he bus stops when all its kinetic energy is stored as elastic potential energy in the cord. his inforation ay be used to deterine the spring constant of the Super Bungee. 3 3 v / kδ x / or k v / Δ x ( 0 kg)(. /s) /(50 ).6 0 N/ (b) he bus stops in a tie equal to one fourth the period of oscillation t /4 (/4)( πω / ) (/4)( πa/ v) πa/ v π(50 )/(. /s) 3.70s Assess: Many students will be tepted to use the kineatic equations fro the early chapters of the text. However these equations will not work because they are applicable only for the case of constant acceleration. In this case the acceleration is not a constant. P4.6. Prepare: Assue a sall angle oscillation of the pendulu so that it has siple haronic otion. We will use Equation 4.7. (a) At the equator, the period of the pendulu is he tie for 00 oscillations is 00.9 s. (b) At the North Pole, the period is.000 π.009s 9.780 /s equator.000 pole π.004s 9.83 /s he tie for 00 oscillations is 00.4 s. (c) he period on the top of the ountain is.00 s. he acceleration due to gravity can be calculated by rearranging the forula for the period: π π gountain L (.000 ) 9.77 /s ountain.00 s Assess: he difference between the answers at the equator and the North Pole is 0.5 s, and this difference is quite easurable with a hand-operated stopwatch. he acceleration due to gravity at the top of a high ountain at the equator is reasonable because g decreases with altitude and it has decreased by 0.0/s. P4.6. Prepare: Assue the orangutan is a siple sall-angle pendulu. One swing of the ar would be half a period of the oscillatory otion. he horizontal distance traveled in that tie would be (0.90 ) sin(0 ) 0.66 fro analysis of a right triangle. π L π 0.90 0.95 s g 9.8/s dist 0.66 speed 0.65 /s tie 0.95 s Assess: his isn t very fast, but isn t out of the reasonable range. P4.63. Prepare: he iu speed occurs at the equilibriu position. π π v A (0.30 ) 0.377 /s 0.50 s After it detaches it is in free fall. v0 (3.77 /s) H 0.75 73 c g (9.8/s ) Assess: his is a safe and reasonable nuber. Copyright 05 Pearson Education, Inc. All rights reserved. his aterial is protected under all copyright laws as they currently exist.
Oscillations 4-3 P4.64. Prepare: We won t have good accuracy fro reading off the graph. (a) Looking at the graph, the period appears to be about 0.75 s. 0.75 s (b) k (. N/) 7 g π π (c) Fro peak to peak the aplitude appears to decrease by a factor of four in one period. t 0.75 s τ 0.54 s x ln( ) ln( 4 ) A Assess: he answers see to fit with the graph. P4.65. Prepare: he oscillator is in siple haronic otion and is daped, so we will use Equation 4.9. / he iu displaceent, or aplitude, of a daped oscillator decreases as () t x, t Ae τ where τ is the tie constant. We know x / 0.60 A at t 50 s, so we can find τ as follows: t x () 50s t ln τ 97.88 s τ A ln(0.60) Now we can find the tie t 30 at which x /A 0.30 x() t t30 τ ln (97.88 s)ln(0.30) 8 s A he undaped oscillator has a frequency f Hz oscillations per second. hen the nuber of oscillations before the aplitude decays to 30% of its initial aplitude is N f t30 ( oscillations/s) (8 s) 36 oscillations or 40 oscillations to two significant figures. P4.66. Prepare: With no nubers given we ust get all our inforation fro the graph. (a) Fro the graph count three full periods in about 3.4 s, so it appears that.s. (b) Fro the graph after a full 4.0 s it looks like the aplitude is about 0.4 of the original aplitude, this is close to 37%, so we guess that τ 4.0 s. Assess: Part b can also be done by reading off the tie it takes for the aplitude to decrease by half. his appears.8 s to be about.8 s, so τ ln 4.0 s. (0.5) P4.67. Prepare: Model the situation as siple haronic otion with a spring. We are given 6kg and Δ y.5. F g (6kg)(9.8 /s ) 5 k 39 000 N/.4 0 N/ Δy Δy.5 he correct choice is D. Assess: his is a very large spring constant, but the tendon is tough. P4.68. Prepare: Model the situation as siple haronic otion with a spring. he stored energy in a spring is (39 000 N/)(0.00 ) U kx J So the answer is choice D. Assess: his is not a lot of energy, but in a reasonable range. P4.69. Prepare: Model the situation as siple haronic otion with a spring. he period of oscillation for a ass-spring syste is 6kg π π 0.0s k 39 000 N/ Copyright 05 Pearson Education, Inc. All rights reserved. his aterial is protected under all copyright laws as they currently exist.