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**Unformatted text preview: **14 OscillationsThis loudspeaker cone generatessound waves by oscillating backand forth at audio frequencies. Looking Ahead The goal of Chapter 14 is to understand systems that oscillate with simple harmonic motion.Simple Harmonic MotionSpringsPendulumsThe most basicoscillation, withsinusoidal motion,is called simpleharmonic motion.Simple harmonic motion occurs whenthere is a linear restoring force. Thesimplest example isa mass on a spring.You will learn how todetermine the periodof oscillation.A mass swinging at the end of a string orrod is a pendulum. Its motion is anotherexample of simple harmonic motion.OscillationThe oscillating cartis an example ofsimple harmonicmotion. Youll learnhow to use themass and the springconstant to determine the frequencyof oscillation.In this chapter you will learn to: Representsimple harmonic motionboth graphically and mathematically. Understand the dynamics of oscillating systems. Recognize the similarities amongmany types of oscillating systems.Simple harmonic motion has a veryclose connection to uniform circularmotion. Youll learn that an edge-onview of uniform circular motion is noneother than simple harmonic motion.Looking BackSection 4.5 Uniform circular motionThe bounce at thebottom of a bungeejump is an exhilaratingexample of a massoscillating on a spring.Looking BackSection 10.4 Restoring forcesDamping and ResonanceEnergy of OscillationsIf theres drag or other dissipation, thenthe oscillation runs down. This iscalled a damped oscillation.If there is no friction or other dissipation, then the mechanical energy of anoscillator is conserved. Conservation ofenergy will be an important tool.All kineticThe system oscillates between allkinetic energy andall potential energyA0All potentialLooking BackSection 10.5 Elastic potential energySection 10.6 Energy diagramsIllustrated Chapter Previewsgive an overview of theupcoming ideas for eachchapter, setting them in context,explaining their utility, and tyingthem to existing knowledge(through Looking Backreferences). These previewsbuild on the cognitivepsychology concept of anadvance organizer.The period of a pendulum is determined bythe length of the string;neither the mass northe amplitude matters.Consequently, the pendulum was the basis oftime keeping for manycenturies.AxThe amplitude ofa damped oscillation undergoesexponentialdecay.Ax0AOscillations can increase in amplitude,sometimes dramatically, when driven attheir natural oscillation frequency. Thisis called resonance.t378 c h a p t e r 14 . OscillationsFiguRE 14.1 Examples of positionversustime graphs for oscillating systems.The oscillation takesplace around anequilibrium position.PositionTt14.1 Simple Harmonic MotionObjects or systems of objects that undergo oscillatory motiona repetitive motionback and forth around an equilibrium positionare called oscillators. FiguRE 14.1shows position-versus-time graphs for three different oscillating systems. Althoughthe shapes of the graphs are different, all these oscillators have two things in common:1. The oscillation takes place about an equilibrium position, and2. The motion is periodic, repeating at regular intervals of time.PositionThe motion is periodic.One cycle takes time T.TtPositionThis oscillationis sinusoidal.TThe time to complete one full cycle, or one oscillation, is called the period of themotion. Period is represented by the symbol T.A closely related piece of information is the number of cycles, or oscillations, com1pleted per second. If the period is 10 s, then the oscillator can complete 10 cycles1in one second. Conversely, an oscillation period of 10 s allows only 10 of a cycle to becompleted per second. In general, T seconds per cycle implies that 1/T cycles will becompleted each second. The number of cycles per second is called the frequency f ofthe oscillation. The relationship between frequency and period istf=1TorT=1f(14.1)The units of frequency are hertz, abbreviated Hz, named in honor of the Germanphysicist Heinrich Hertz, who produced the first artificially generated radio waves in1887. By definition,TABLE 14.1Units of frequencyFrequencyPeriod103 Hz = 1 kilohertz = 1 kHz1 ms106 Hz = 1 megahertz = 1 MHz1 msFigures are carefully1 ns109 Hz = 1 gigahertz = 1 GHzstreamlined in detail andcolor so students focus on thephysicsfor instance, the objectof interest in mechanics.FiguRE 14.2 A prototype simpleExplicit instruction asannotations directly on figureshelps students to interpretfigures and graphs.harmonicmotion experiment.(a)OscillationThe point on theobject that ismeasuredAir track(b)0ATurningpointAAAtxx is measured fromthe equilibriumposition wherethe object wouldbe at rest.The motionis sinusoidal,indicatingSHM.The motion is symmetrical about theequilibrium position. Maximum distanceto the left and to the right is A.1 Hz K 1 cycle per second = 1 s -1We will frequently deal with very rapid oscillations and make use of the units shownin Table 14.1.Uppercase and lowercase letters are important. 1 MHz is 1 megahertz =106 Hz, but 1 mHz is 1 millihertz = 10-3 Hz! NOTE ExAMPLE 14.1Frequency and period of a loudspeaker coneWhat is the oscillation period of a loudspeaker cone that vibrates back and forth 5000 timesper second?SOLvE The oscillation frequency is f = 5000 cycles/s = 5000 Hz = 5.0 kHz. The periodis the inverse of the frequency; henceT=11== 2.0 * 10-4 s = 200 msf5000 HzA system can oscillate in many ways, but we will be especially interested inthe smooth sinusoidal oscillation (i.e., like a sine or cosine) of the third graph inFigure 14.1. This sinusoidal oscillation, the most basic of all oscillatory motions, iscalled simple harmonic motion, often abbreviated SHM. Lets look at a graphicaldescription before we dive into the mathematics of simple harmonic motion.FiguRE 14.2a shows an air-track glider attached to a spring. If the glider is pulled outa few centimeters and released, it will oscillate back and forth on the nearly frictionless air track. FiguRE 14.2b shows actual results from an experiment in which a computer was used to measure the gliders position 20 times every second. This is aposition-versus-time graph that has been rotated 90 from its usual orientation in orderfor the x-axis to match the motion of the glider.The objects maximum displacement from equilibrium is called the amplitude Aof the motion. The objects position oscillates between x = - A and x = + A. Whenusing a graph, notice that the amplitude is the distance from the axis to the maximum,not the distance from the minimum to the maximum.14.1 . Simple Harmonic Motionshows the data with the graph axes in their normal positions. Youcan see that the amplitude in this experiment was A = 0.17 m, or 17 cm. You canalso measure the period to be T = 1.60 s. Thus the oscillation frequency wasf = 1/T = 0.625 Hz.FiguRE 14.3b is a velocity-versus-time graph that the computer produced by usingx/ t to find the slope of the position graph at each point. The velocity graph is alsosinusoidal, oscillating between - vmax (maximum speed to the left) and + vmax (maximum speed to the right). As the figure shows,FiguRE 14.3aThe instantaneous velocity is zero at the points where x = { A. These are the turning points in the motion.The maximum speed vmax is reached as the object passes through the equilibriumposition at x = 0 m. The velocity is positive as the object moves to the right butnegative as it moves to the left.Position and velocity graphsof the experimental data.FiguRE 14.3(a) The speed is zerowhen xA.00.10.2x012vvmaxvt (s)6.0vmax002.04.0t (s)6.0The positionversustimegraph for simple harmonic motion.FiguRE 14.4(14.2)where the notation x(t) indicates that the position x is a function of time t. Becausecos(2p) = cos(0), its easy to see that the position at time t = T is the same as the position at t = 0. In other words, this is a cosine function with period T. Be sure to convinceyourself that this function agrees with the five special points shown in Figure 14.4.NOTE The argument of the cosine function is in radians. That will be true throughout this chapter. Its especially important to remember to set your calculator toradian mode before working oscillation problems. Leaving it in degree mode willlead to errors. We can write Equation 14.2 in two alternative forms. Because the oscillation frequency is f = 1/T, we can write1. Starts at x A2. Passes throughxx 0 at t 1 T45. Returns tox A at t(14.5)Equations 14.2, 14.3, and 14.5 are equivalent ways to write the position of an objectmoving in simple harmonic motion.TA02TTA4. Passes through x3. Reaches xA at t0 at t12T(14.3)Recall from Chapter 4 that a particle in circular motion has an angular velocity v thatis related to the period by v = 2p/T, where v is in rad/s. Now that weve defined thefrequency f, you can see that v and f are related by2pv (in rad/s) == 2pf (in Hz)(14.4)TIn this context, v is called the angular frequency. The position can be written in termsof v asx ( t ) = A cos vt4.00.7redraws the position-versus-time graph of Figure 14.3a as a smooth curve.Although these are empirical data (we dont yet have any theory of oscillation) thegraph for this particular motion is clearly a cosine function. The objects position isx ( t ) = A cos (2pft )0.17 mA2.0(b)vx (m/s)Kinematics of Simple Harmonic Motion2ptT0.17 m1.60 s0.10.7x ( t ) = A cosAT0.2A mass oscillating on a spring is the prototype of simple harmonic motion. Ouranalysis, in which we answer these questions, will be of a spring-mass system. Evenso, most of what we learn will be applicable to other types of SHM.FiguRE 14.4The speed is maximumas the object passesthrough x 0.xx (m)We can ask three important questions about this oscillating system:1. How is the maximum speed vmax related to the amplitude A?2. How are the period and frequency related to the objects mass m, the springconstant k, and the amplitude A?3. Is the sinusoidal oscillation a consequence of Newtons laws?379Derivatives of sine andcosine functionsTABLE 14.2d1 a sin ( bt + c )2 = + ab cos ( bt + c )dtd1 a cos ( bt + c )2 = - ab sin ( bt + c )dtt34TNOTE paragraphs throughoutguide students away fromknown preconceptions andaround common stickingpoints and highlight manymath- and vocabulary-relatedissues that have been provento cause difficulties.380 c h a p t e r 14 . OscillationsPosition and velocity graphsfor simple harmonic motion.FiguRE 14.5Position xTAvx ( t ) = - vmax sint02TTAVelocity vxTvmaxNOTE vmaxvx ( t ) =t0vmax2TTv (t)2pt= - vmax sin(2pft ) = - vmax sin vtT(14.6)is the maximum speed and thus is a positive number. 12We deduced Equation 14.6 from the experimental results, but we could equally wellfind it from the position function of Equation 14.2. After all, velocity is the time derivative of position. Table 14.2 on the previous page reminds you of the derivatives of thesine and cosine functions. Using the derivative of the position function, we findA cos vtx (t)12Just as the position graph was clearly a cosine function, the velocity graph shownin FiguRE 14.5 is clearly an upside-down sine function with the same period T. Thevelocity vx, which is a function of time, can be writtendx2pA2pt=sin= - 2pfA sin(2pft ) = - vA sin vt (14.7)dtTTComparing Equation 14.7, the mathematical definition of velocity, to Equation 14.6,the empirical description, we see that the maximum speed of an oscillation isvmax sin vtvmax =2pA= 2pfA = vAT(14.8)Equation 14.8 answers the first question we posed above, which was how the maximumspeed vmax is related to the amplitude A. Not surprisingly, the object has a greater maximum speed if you stretch the spring farther and give the oscillation a larger amplitude.ExAMPLE 14.2Worked examples follow aconsistent problem-solvingstrategy and include carefulexplanations of the underlying,and often unstated, reasoning.A system in simple harmonic motionAn air-track glider is attached to a spring, pulled 20.0 cm to theright, and released at t = 0 s. It makes 15 oscillations in 10.0 s.a. What is the period of oscillation?b. What is the objects maximum speed?c. What are the position and velocity at t = 0.800 s?MODELSOLvE15 oscillations= 1.50 oscillations/s = 1.50 Hzf=10.0 sThus the period is T = 1/f = 0.667 s.b. The oscillation amplitude is A = 0.200 m. Thusvmax =ExAMPLE 14.32pA 2p(0.200 m)== 1.88 m/sT0.667 sFinding the timeA mass oscillating in simple harmonic motion starts at x = A andhas period T. At what time, as a fraction of T, does the object firstpass through x = 1 A?2Figure 14.4 showed that the object passes through theequilibrium position x = 0 at t = 1 T. This is one-quarter of the4total distance in one-quarter of a period. You might expect it totake 1 T to reach 1 A, but this is not the case because the SHM82graph is not linear between x = A and x = 0. We need to usex ( t ) = A cos (2pt/T ). First, we write the equation with x = 1 A:2SOLvEx = A cosAn object oscillating on a spring is in SHM.a. The oscillation frequency is1212c. The object starts at x = + A at t = 0 s. This is exactly theoscillation described by Equations 14.2 and 14.6. The positionat t = 0.800 s is2p (0.800 s)2pt= (0.200 m) cosT0.667 s1212= (0.200 m)cos (7.54 rad) = 0.0625 m = 6.25 cmThe velocity at this instant of time isvx = - vmax sin2p(0.800 s)2pt= - (1.88 m/s) sinT0.667 s= - (1.88 m/s) sin (7.54 rad) = - 1.79 m/s = - 179 cm/sAt t = 0.800 s, which is slightly more than one period, the object is 6.25 cm to the right of equilibrium and moving to the leftat 179 cm/s. Notice the use of radians in the calculations.1212A2pt= A cos2TThen we solve for the time at which this position is reached:T1Tp 1t=cos -1==T2p22p 36x=The motion is slow at the beginning and then speeds up,so it takes longer to move from x = A to x = 1 A than it does to2move from x = 1 A to x = 0. Notice that the answer is indepen2dent of the amplitude A.ASSESS14.2 . Simple Harmonic Motion and Circular Motion381An object moves with simple harmonic motion. If the amplitudeand the period are both doubled, the objects maximum speed isStop to think 14.1a. Quadrupled.d. Halved.b. Doubled.e. Quartered.c. Unchanged.14.2 Simple Harmonic Motionand Circular MotionThe graphs of Figure 14.5 and the position function x ( t ) = A cos vt are for an oscillation in which the object just happened to be at x0 = A at t = 0. But you will recall thatt = 0 is an arbitrary choice, the instant of time when you or someone else starts a stopwatch. What if you had started the stopwatch when the object was at x0 = - A, or whenthe object was somewhere in the middle of an oscillation? In other words, what if theoscillator had different initial conditions. The position graph would still show an oscillation, but neither Figure 14.5 nor x ( t ) = A cos vt would describe the motion correctly.To learn how to describe the oscillation for other initial conditions it will help toturn to a topic you studied in Chapter 4circular motion. Theres a very close connection between simple harmonic motion and circular motion.Imagine you have a turntable with a small ball glued to the edge. FiguRE 14.6a showshow to make a shadow movie of the ball by projecting a light past the ball and ontoa screen. The balls shadow oscillates back and forth as the turntable rotates. This iscertainly periodic motion, with the same period as the turntable, but is it simple harmonic motion?To find out, you could place a real object on a real spring directly below the shadow, as shown in FiguRE 14.6b. If you did so, and if you adjusted the turntable to have thesame period as the spring, you would find that the shadows motion exactly matchesthe simple harmonic motion of the object on the spring. Uniform circular motionprojected onto one dimension is simple harmonic motion.To understand this, consider the particle in FiguRE 14.7. It is in uniform circularmotion, moving counterclockwise in a circle with radius A. As in Chapter 4, we canlocate the particle by the angle f measured ccw from the x-axis. Projecting the ballsshadow onto a screen in Figure 14.6 is equivalent to observing just the x-componentof the particles motion. Figure 14.7 shows that the x-component, when the particle isat angle f, isx = A cos f(14.9)Recall that the particles angular velocity, in rad/s, isv=dfdtA projection of the circularmotion of a rotating ball matches thesimple harmonic motion of an object ona spring.FiguRE 14.6(a)Light from projectorBallShadow(b)Simple harmonic motion of blockA particle in uniformcircular motion with radius A andangular velocity v.FiguRE 14.7y(14.10)Particle in uniformcircular motion(14.11)vAfA0A cos fAs f increases, the particles x-component isx ( t ) = A cos vtScreenOscillation of balls shadowThis is the rate at which the angle f is increasing. If the particle starts from f0 = 0 att = 0, its angle at a later time t is simplyf = vtTurntableCircularmotionof ballxv(14.12)This is identical to Equation 14.5 for the position of a mass on a spring! Thus thex-component of a particle in uniform circular motion is simple harmonic motion.When used to describe oscillatory motion, v is called the angular frequency rather than the angular velocity. The angular frequency of an oscillator hasthe same numerical value, in rad/s, as the angular velocity of the correspondingparticle in circular motion. AThe x-component ofthe particles positiondescribes the positionof the balls shadow.NOTE A0xA cos fANew concepts are introducedthrough observations aboutthe real world and theories,grounded by making sense ofobservations. This inductiveapproach illustrates howscience operates, and hasbeen shown to improvestudent learning by reconcilingnew ideas with what theyalready know.382 c h a p t e r 14 . OscillationsAnalogy is used throughout thetext and figures to consolidatestudent understanding bycomparing with a more familiarconcept or situation.A cup on the turntable in a microwaveoven moves in a circle. But from theoutside, you see the cup sliding back andforthin simple harmonic motion!The names and units can be a bit confusing until you get used to them. It may helpto notice that cycle and oscillation are not true units. Unlike the standard meter or thestandard kilogram, to which you could compare a length or a mass, there is no standard cycle to which you can compare an oscillation. Cycles and oscillations are simplycounted events. Thus the frequency f has units of hertz, where 1 Hz = 1 s -1. We maysay cycles per second just to be clear, but the actual units are only per second.The radian is the SI unit of angle. However, the radian is a defined unit. Further, itsdefinition as a ratio of two lengths (u = s/r) makes it a pure number without dimensions. As we noted in Chapter 4, the unit of angle, be it radians or degrees, is really justa name to remind us that were dealing with an angle. The 2p in the equation v = 2pf(and in similar situations), which is stated without units, means 2p rad/cycle. Whenmultiplied by the frequency f in cycles/s, it gives the frequency in rad/s. That is why,in this context, v is called the angular frequency.NOTE Hertz is specifically cycles per second or oscillations per second. It isused for f but not for v. Well always be careful to use rad/s for v, but you shouldbe aware that many books give the units of v as simply s -1. The Phase ConstantA particle in uniformcircular motion with initial angle f0.FiguRE 14.8Angle at time t isf vt f0.yInitial position ofparticle at t 0vAf00 A cos ff = vt + f0(14.13)In this case, the particles projection onto the x-axis at time t isAfNow were ready to consider the issue of other initial conditions. The particle inFigure 14.7 started at f0 = 0. This was equivalent to an oscillator starting at the farright edge, x0 = A. FiguRE 14.8 shows a more general situation in which the initial anglef0 can have any value. The angle at a later time t is thenx0xAA cos f0vThe initial x-component of theparticles position can be anywherebetween A and A, depending on f0.x ( t ) = A cos ( vt + f0 )(14.14)If Equation 14.14 describes the particles projection, then it must also be the position of an oscillator in simple harmonic motion. The oscillators velocity vx is foundby taking the derivative dx/dt. The resulting equations,x ( t ) = A cos ( vt + f0 )vx ( t ) = - vA sin( vt + f0 ) = - vmax sin( vt + f0 )(14.15)are the two primary kinematic equations of simple harmonic motion.The quantity f = vt + f0, which steadily increases with time, is called the phaseof the oscillation. The phase is simply the angle of the circular-motion particle whoseshadow matches the oscillator. The constant f0 is called the phase constant. It specifies the initial conditions of the oscillator.To see what the phase constant means, set t = 0 in Equations 14.15:x0 = A cos f0v0x = - vA sin f0(14.16)The position x0 and velocity v0x at t = 0 are the initial conditions. Different values ofthe phase constant correspond to different starting points on the circle and thusto different initial conditions.The perfect cosine function of Figure 14.5 and the equation x ( t ) = A cos vt are foran oscillation with f0 = 0 rad. You can see from Equations 14.16 that f0 = 0 rad implies x0 = A and v0 = 0. That is, the particle starts from rest at the point of maximumdisplacement.FiguRE 14.9 illustrates these ideas by looking at three values of the phase constant:f0 = p/3 rad (60 ), - p/3 rad ( - 60 ), and p rad (180 ). Notice that f0 = p/3 rad andf0 = - p/3 rad have the same starting position, x0 = 1 A. This is a property of the co2sine function in Equation 14.16. But these are not the same initial conditions. In one casethe oscillator starts at 1 A while moving to the right, in the other case it starts at 1 A while22moving to the left. You can distinguish between the two by visualizing the motion.14.2 . Simple Harmonic Motion and Circular MotionFiguRE 14.9383Oscillations described by the phase constants f0 = p/3 rad, - p/3 rad, and p rad.f0p/3 radyf012Ap3A12Axp3f0tThe starting point of the oscillation isshown on the circle and on the graph.A012AxAA012AAtTxA0ttvmaxt0TTAvvmaxTxAxvvmaxxA0vp0AAAf0Extensive use is made ofmultiple representationsplacing different representationsside by side to help studentsdevelop the key skill oftranslating between words,math, and figures. Essential togood problem-solving, this skillis overlooked in most physicstextbooks.0AA0p rady0AtxxxThe graphs each have the sameamplitude and period. They areshifted relative to the f0 0 radgraphs of Figure 14.5 because theyhave different initial conditions.f00tf0p/3 rady0vmaxvmaxtt0TTvmaxAll values of the phase constant f0 between 0 and p rad correspond to a particle inthe upper half of the circle and moving to the left. Thus v0x is negative. All values of thephase constant f0 between p and 2p rad (or, as they are usually stated, between - pand 0 rad) have the particle in the lower half of the circle and moving to the right. Thusv0x is positive. If youre told that the oscillator is at x = 1 A and moving to the right at2t = 0, then the phase constant must be f0 = - p/3 rad, not + p/3 rad.ExAMPLE 14.4using the initial conditionsAn object on a spring oscillates with a period of 0.80 s and anamplitude of 10 cm. At t = 0 s, it is 5.0 cm to the left of equilibrium and moving to the left. What are its position and direction ofmotion at t = 2.0 s?MODELAn object oscillating on a spring is in simple harmonicmotion.1212SOLvE We can find the phase constant f0 from the initial condition x0 = - 5.0 cm = A cos f0. This condition givesf0 = cos -1x012= cos -1 = { p rad = { 120A23Because the oscillator is moving to the left at t = 0, it is in theupper half of the circular-motion diagram and must have a phaseconstant between 0 and p rad. Thus f0 is 2 p rad. The angular3frequency isv=2p2p== 7.85 rad/sT0.80 s1Thus the objects position at time t = 2.0 s isx ( t ) = A cos ( vt + f0 )= (10 cm) cos (7.85 rad/s) (2.0 s) += (10 cm) cos (17.8 rad) = 5.0 cm2p32The object is now 5.0 cm to the right of equilibrium. But whichway is it moving? There are two ways to find out. The direct wayis to calculate the velocity at t = 2.0 s:vx = - vA sin ( vt + f0 ) = + 68 cm/sThe velocity is positive, so the motion is to the right. Alternatively, we could note that the phase at t = 2.0 s is f = 17.8 rad.Dividing by p, you can see thatf = 17.8 rad = 5.67p rad = (4p + 1.67p ) radThe 4p rad represents two complete revolutions. The extraphase of 1.67p rad falls between p and 2p rad, so the particle inthe circular-motion diagram is in the lower half of the circle andmoving to the right.384 c h a p t e r 14 . Oscillationscos -1 is a twovalued function. Your calculatorreturns a single value, an angle between 0 rad and p rad. But the negative of thisangle is also a solution. As Example 14.4 demonstrates, you must use additionalinformation to choose between them. NOTE The inverse-cosine functionStop to think 14.2The figure showsfour oscillators at t = 0. Which onehas the phase constant f0 = p/4 rad?Stop to Think questions atthe end of a section allowstudents to quickly check theirunderstanding. Using powerfulranking-task and graphicaltechniques, they are designedto efficiently probe keymisconceptions and encourageactive reading. (Answers areprovided at the end of thechapter.)AA(a)(b)(c)(d)10071071 100x (mm)14.3 Energy in Simple Harmonic MotionThe energy is transformedbetween kinetic energy and potentialenergy as the object oscillates, but themechanical energy E = K + U doesntchange.FiguRE 14.10Energy is transformed betweenkinetic and potential, but the totalmechanical energy E doesnt change.(a)kA122 mvE122 kxmrvx0xA(b)PotentialenergycurveEnergyETotalenergylineTurningpointA0TurningpointxAEnergy here ispurely kinetic.Energy here is purely potential.Weve begun to develop the mathematical language of simple harmonic motion, butthus far we havent included any physics. Weve made no mention of the mass ofthe object or the spring constant of the spring. An energy analysis, using the tools ofChapters 10 and 11, is a good starting place.FiguRE 14.10a shows an object oscillating on a spring, our prototype of simpleharmonic motion. Now well specify that the object has mass m, the spring hasspring constant k, and the motion takes place on a frictionless surface. You learnedin Chapter 10 that the elastic potential energy when the object is at position x isUs = 1 k( x)2, where x = x - xe is the displacement from the equilibrium position2xe. In this chapter well always use a coordinate system in which xe = 0, makingx = x. Theres no chance for confusion with gravitational potential energy, so wecan omit the subscript s and write the elastic potential energy as1U = kx 2(14.17)2Thus the mechanical energy of an object oscillating on a spring is11E = K + U = mv 2 + kx 2(14.18)2212FiguRE 14.10b is an energy diagram, showing the potential-energy curve U = 2 kx asa parabola. Recall that a particle oscillates between the turning points where the totalenergy line E crosses the potential-energy curve. The left turning point is at x = - A,and the right turning point is at x = + A. To go beyond these points would require anegative kinetic energy, which is physically impossible.You can see that the particle has purely potential energy at xt A and purely kinetic energy as it passes through the equilibrium point at x0. At maximumdisplacement, with x = { A and v = 0, the energy is1E (at x = { A ) = U = kA2(14.19)2At x = 0, where v = { vmax, the energy is1E (at x = 0) = K = m ( vmax)2(14.20)214.3 . Energy in Simple Harmonic Motion385The systems mechanical energy is conserved because the surface is frictionlessand there are no external forces, so the energy at maximum displacement and the energy at maximum speed, Equations 14.19 and 14.20, must be equal. That is11(14.21)m(vmax)2 = kA222Thus the maximum speed is related to the amplitude bykvmax =A(14.22)AmThis is a relationship based on the physics of the situation.Earlier, using kinematics, we found that2pAvmax == 2pfA = vA(14.23)TComparing Equations 14.22 and 14.23, we see that frequency and period of an oscillating spring are determined by the spring constant k and the objects mass m:v=kmAf=1km2p AT = 2pmAk(14.24)These three expressions are really only one equation. They say the same thing, buteach expresses it in slightly different terms.Equations 14.24 are the answer to the second question we posed at the beginning ofthe chapter, where we asked how the period and frequency are related to the objectsmass m, the spring constant k, and the amplitude A. It is perhaps surprising, but theperiod and frequency do not depend on the amplitude A. A small oscillation and alarge oscillation have the same period.Because energy is conserved, we can combine Equations 14.18, 14.19, and 14.20to write1111E = mv 2 + kx 2 = kA2 = m(vmax)2 (conservation of energy)2222Kinetic energy, potentialenergy, and the total mechanical energyfor simple harmonic motion.FiguRE 14.11The total mechanicalenergy E is constant.Potential energyKinetic energyEnergyT(14.25)k2( A - x2 ) = v 2A2 - x2(14.26)BmFiguRE 14.11 shows graphically how the kinetic and potential energy change withtime. They both oscillate but remain positive because x and v are squared. Energy iscontinuously being transformed back and forth between the kinetic energy of the moving block and the stored potential energy of the spring, but their sum remains constant.Notice that K and U both oscillate twice each period; make sure you understand why.Any pair of these expressions may be useful, depending on the known information.For example, you can use the amplitude A to find the speed at any point x by combining the first and second expressions for E. The speed v at position x ist0Positionv=ExAMPLE 14.5using conservation of energyA 500 g block on a spring is pulled a distance of 20 cm and released.The subsequent oscillations are measured to have a period of 0.80 s.a. At what position or positions is the blocks speed 1.0 m/s?b. What is the spring constant?MODELThe motion is SHM. Energy is conserved.a. The block starts from the point of maximum displacement, where E = U = 1 kA2. At a later time, when the position is2x and the speed is v, energy conservation requires121212mv + k x = k A222Solving for x, we findSOLvEx=BA2 -12mv 2v=A2 vkB2t012where we used k/m = v2 from Equation 14.24. The angular frequency is easily found from the period: v = 2p/T = 7.85 rad/s.Thusx=B(0.20 m)2 -1.0 m/s7.85 rad/s2= { 0.15 m = { 15 cmThere are two positions because the block has this speed oneither side of equilibrium.b. Although part a did not require that we know the spring constant, it is straightforward to find from Equation 14.24:mAk4p2 (0.50 kg)4 p2 mk=== 31 N/m2T(0.80 s)2T = 2p386 c h a p t e r 14 . OscillationsStop to think 14.3The four springs shown here have been compressed from their equilibrium position atx = 0 cm. When released, the attached masswill start to oscillate. Rank in order, fromhighest to lowest, the maximum speeds ofthe masses.k(a)(b)12k4mk(c)4m2mk(d)20m151050x (cm)14.4 The Dynamics of SimpleHarmonic MotionOur analysis thus far has been based on the experimental observation that the oscillation of a spring looks sinusoidal. Its time to show that Newtons second law predictssinusoidal motion.A motion diagram will help us visualize the objects acceleration. FiguRE 14.12 showsone cycle of the motion, separating motion to the left and motion to the right to makethe diagram clear. As you can see, the objects velocity is large as it passes through theuequilibrium point at x = 0, but v is not changing at that point. Acceleration measuresuuthe change of the velocity; hence a = 0 at x = 0.Motion diagram of simple harmonic motion. The left and right motions areseparated vertically for clarity but really occur along the same line.FiguRE 14.12To the rightrararra0rvrvrrrraavPosition x0EquilibriumTo the left02AraSamepointxATA02TTtAAcceleration axraSamepointPosition and accelerationgraphs for an oscillating spring. Wevechosen f0 = 0.FiguRE 14.13vrrav2A when xamaxAamax02TTtamaxaminv2A when xAIn contrast, the velocity is changing rapidly at the turning points. At the right turnuing point, v changes from a right-pointing vector to a left-pointing vector. Thus theuacceleration a at the right turning point is large and to the left. In one-dimensionalmotion, the acceleration component ax has a large negative value at the right turningupoint. Similarly, the acceleration a at the left turning point is large and to the right.Consequently, ax has a large positive value at the left turning point.Our motion-diagram analysis suggests that the acceleration ax is most positivewhen the displacement is most negative, most negative when the displacement is amaximum, and zero when x = 0. This is confirmed by taking the derivative of thevelocity:ax =dvxd= ( - vA sin vt ) = - v2 A cos vtdtdt(14.27)then graphing it.FiguRE 14.13 shows the position graph that we started with in Figure 14.4 and the corresponding acceleration graph. Comparing the two, you can see that the acceleration14.4 . The Dynamics of Simple Harmonic Motiongraph looks like an upside-down position graph. In fact, because x = A cos vt, Equation 14.27 for the acceleration can be writtenax = - v2 x(Fsp )x = - k x(14.29)The minus sign indicates that the spring force is a restoring force, a force that always points back toward the equilibrium position. If we place the origin of the coordinate system at the equilibrium position, as weve done throughout this chapter, thenx = x and Hookes law is simply ( Fsp )x = - kx.The x-component of Newtons second law for the object attached to the spring is(Fnet )x = (Fsp )x = - kx = max(14.30)Equation 14.30 is easily rearranged to readYou can see that Equation 14.31 is identical to Equation 14.28 if the system oscillateswith angular frequency v = 1k/m . We previously found this expression for v froman energy analysis. Our experimental observation that the acceleration is proportionalto the negative of the displacement is exactly what Hookes law would lead us to expect. Thats the good news.The bad news is that ax is not a constant. As the objects position changes, so doesthe acceleration. Nearly all of our kinematic tools have been based on constant acceleration. We cant use those tools to analyze oscillations, so we must go back to thevery definition of acceleration:ax = -ax =kxm(14.31)dvx d 2 x=2dtdtAcceleration is the second derivative of position with respect to time. If we use thisdefinition in Equation 14.31, it becomeskd 2x= - x (equation of motion for a mass on a spring)mdt 2The prototype of simpleharmonic motion: a mass oscillating ona horizontal spring without friction.FiguRE 14.14(14.28)That is, the acceleration is proportional to the negative of the displacement. Theacceleration is, indeed, most positive when the displacement is most negative and ismost negative when the displacement is most positive.Recall that the acceleration is related to the net force by Newtons second law. Consider again our prototype mass on a spring, shown in FiguRE 14.14. This is the simplestpossible oscillation, with no distractions due to friction or gravitational forces. We willassume the spring itself to be massless.As you learned in Chapter 10, the spring force is given by Hookes law:(14.32)Equation 14.32, which is called the equation of motion, is a second-order differentialequation. Unlike other equations weve dealt with, Equation 14.32 cannot be solvedby direct integration. Well need to take a different approach.Solving the Equation of MotionThe solution to an algebraic equation such as x 2 = 4 is a number. The solution to adifferential equation is a function. The x in Equation 14.32 is really x ( t ), the positionas a function of time. The solution to this equation is a function x ( t ) whose secondderivative is the function itself multiplied by ( - k/m ).One important property of differential equations that you will learn about in mathis that the solutions are unique. That is, there is only one solution to Equation 14.32that satisfies the initial conditions. If we were able to guess a solution, the uniquenessproperty would tell us that we had found the only solution. That might seem a rather387rSpringconstant kFspm2A0xOscillationxA388 c h a p t e r 14 . Oscillationsstrange way to solve equations, but in fact differential equations are frequently solvedby using your knowledge of what the solution needs to look like to guess an appropriate function. Let us give it a try!We know from experimental evidence that the oscillatory motion of a spring appears to be sinusoidal. Let us guess that the solution to Equation 14.32 should havethe functional formx ( t ) = A cos ( vt + f0 )(14.33)where A, v, and f0 are unspecified constants that we can adjust to any values thatmight be necessary to satisfy the differential equation.If you were to guess that a solution to the algebraic equation x 2 = 4 is x = 2, youwould verify your guess by substituting it into the original equation to see if it works.We need to do the same thing here: Substitute our guess for x ( t ) into Equation 14.32to see if, for an appropriate choice of the three constants, it works. To do so, we needthe second derivative of x ( t ). That is straightforward:An optical technique calledinterferometry reveals the belllikevibrations of a wine glass.x ( t ) = A cos ( vt + f0 )dx= - vA sin( vt + f0 )dt(14.34)d 2x= - v2A cos ( vt + f0 )dt 2If we now substitute the first and third of Equations 14.34 into Equation 14.32, we findk- v2A cos ( vt + f0 ) = - A cos ( vt + f0 )(14.35)mEquation 14.35 will be true at all instants of time if and only if v2 = k/m. There do notseem to be any restrictions on the two constants A and f0they are determined by theinitial conditions.So we have foundby guessing!that the solution to the equation of motion for amass oscillating on a spring iskBmx ( t ) = A cos ( vt + f0 )where the angular frequencyv = 2pf =(14.36)(14.37)is determined by the mass and the spring constant.NOTE Once again we see that the oscillation frequency is independent of theamplitude A. Equations 14.36 and 14.37 seem somewhat anticlimactic because weve beenusing these results for the last several pages. But keep in mind that we had beenassuming x = A cos vt simply because the experimental observations looked like acosine function. Weve now justified that assumption by showing that Equation 14.36really is the solution to Newtons second law for a mass on a spring. The theory ofoscillation, based on Hookes law for a spring and Newtons second law, is in goodagreement with the experimental observations. This conclusion gives an affirmativeanswer to the last of the three questions that we asked early in the chapter, which waswhether the sinusoidal oscillation of SHM is a consequence of Newtons laws.ExAMPLE 14.6Analyzing an oscillatorAt t = 0 s, a 500 g block oscillating on a spring is observed moving to the right at x = 15 cm. It reaches a maximum displacementof 25 cm at t = 0.30 s.a. Draw a position-versus-time graph for one cycle of the motion.b. At what times during the first cycle does the mass pass throughx = 20 cm?14.5 . Vertical OscillationsMODELThe motion is simple harmonic motion.12a. The position equation of the block is x ( t ) = A cos ( vt +f0 ). We know that the amplitude is A = 0.25 m and thatx0 = 0.15 m. From these two pieces of information we obtainthe phase constant:SOLvEf0 = cos -1Positionversustime graphfor the oscillator of Example 14.6.FiguRE 14.15x (cm)2520151050510152025x0= cos -1 (0.60) = { 0.927 radAThe object is initially moving to the right, which tells us thatthe phase constant must be between - p and 0 rad. Thus f0 =- 0.927 rad. The block reaches its maximum displacementxmax = A at time t = 0.30 s. At that instant of timexmax = A = A cos ( vt + f0 )This can be true only if cos ( vt + f0 ) = 1, which requiresvt + f0 = 0. Thus- f0- ( - 0.927 rad)== 3.09 rad/sv=t0.30 sNow that we know v, it is straightforward to compute theperiod:2pT== 2.0 svx ( t ) = (25 cm) cos (3.09t - 0.927), wheret is in s, from t = 0 s to t = 2.0 s.b. From x = A cos ( vt + f0 ), the time at which the mass reachesposition x = 20 cm isFiguRE 14.15 graphst==2.0 sT0.3 0.51.01.52.01 12 2112t (s)2x1cos -1- f0vA120 cmcos -1+ 0.927 rad = 0.51 s3.09 rad/s25 cmA calculator returns only one value of cos -1, in the range 0 top rad, but we noted earlier that cos -1 actually has two values.Indeed, you can see in Figure 14.15 that there are two times atwhich the mass passes x = 20 cm. Because they are symmetrical on either side of t = 0.30 s, when x = A, the first point is(0.51 s - 0.30 s) = 0.21 s before the maximum. Thus the masspasses through x = 20 cm at t = 0.09 s and again at t = 0.51 s.Stop to think 14.4This is the position graph of a mass on a spring. What can yousay about the velocity and the force at the instant indicated by the dashed line?a.b.c.d.e.f.g.389Velocity is positive; force is to the right.Velocity is negative; force is to the right.Velocity is zero; force is to the right.Velocity is positive; force is to the left.Velocity is negative; force is to the left.Velocity is zero; force is to the left.Velocity and force are both zero.xAt0Extensive use is made ofmultiple representationsplacing different representationsside by side to help studentsdevelop the key skill oftranslating between words,math, and figures. Essential togood problem-solving, this skillis overlooked in most physicstextbooks.A14.5 vertical OscillationsWe have focused our analysis on a horizontally oscillating spring. But the typicaldemonstration youll see in class is a mass bobbing up and down on a spring hungvertically from a support. Is it safe to assume that a vertical oscillation has the samemathematical description as a horizontal oscillation? Or does the additional force ofgravity change the motion? Let us look at this more carefully.FiguRE 14.16 shows a block of mass m hanging from a spring of spring constant k. Animportant fact to notice is that the equilibrium position of the block is not where thespring is at its unstretched length. At the equilibrium position of the block, where ithangs motionless, the spring has stretched by L.Finding L is a static-equilibrium problem in which the upward spring force balances the downward gravitational force on the block. The y-component of the springforce is given by Hookes law:( Fsp )y = - k y = + k L(14.38)FiguRE 14.16spring.Gravity stretches thekUnstretchedspringLrFspmThe block hangingat rest has stretchedthe spring by L.rFG390 c h a p t e r 14 . OscillationsEquation 14.38 makes a distinction between L, which is simply a distance and isa positive number, and the displacement y. The block is displaced downward, soy = - L. Newtons first law for the block in equilibrium is(Fnet ) y = (Fsp ) y + (FG ) y = k L - mg = 0(14.39)from which we can findmg(14.40)kThis is the distance the spring stretches when the block is attached to it.Let the block oscillate around this equilibrium position, as shown in FiguRE 14.17.Weve now placed the origin of the y-axis at the blocks equilibrium position in order tobe consistent with our analyses of oscillations throughout this chapter. If the block movesupward, as the figure shows, the spring gets shorter compared to its equilibrium length,but the spring is still stretched compared to its unstretched length in Figure 14.16. Whenthe block is at position y, the spring is stretched by an amount L - y and hence exertsan upward spring force Fsp = k ( L - y ). The net force on the block at this point isL=The block oscillates aroundthe equilibrium position.FiguRE 14.17Springstretchedby LSpringstretchedby L yrFspAymBlocksequilibriumpositionAFnetr0FGOscillation around theequilibrium positionis symmetrical.(14.41)But k L - mg is zero, from Equation 14.40, so the net force on the block is simply(Fnet ) y = - ky(14.42)Equation 14.42 for vertical oscillations is exactly the same as Equation 14.30 forhorizontal oscillations, where we found (Fnet ) x = - kx. That is, the restoring force forvertical oscillations is identical to the restoring force for horizontal oscillations. Therole of gravity is to determine where the equilibrium position is, but it doesnt affectthe oscillatory motion around the equilibrium position.Because the net force is the same, Newtons second law has exactly the same oscillatory solution:with, again, v = 2k/m. The vertical oscillations of a mass on a spring are thesame simple harmonic motion as those of a block on a horizontal spring. This isan important finding because it was not obvious that the motion would still be simpleharmonic motion when gravity was included.Hand-drawn sketches areincorporated into selectworked examples to provide aclear model of what studentsshould draw during their ownproblem solving.ExAMPLE 14.7(Fnet ) y = (Fsp ) y + (FG ) y = k( L - y ) - mg = ( k L - mg ) - kyrmy ( t ) = A cos ( vt + f0 )(14.43)Bungee oscillationsAn 83 kg student hangs from a bungee cord with spring constant270 N/m. The student is pulled down to a point where the cord is5.0 m longer than its unstretched length, then released. Where isthe student, and what is his velocity 2.0 s later?A student on a bungee cord oscillatesabout the equilibrium position.FiguRE 14.18The bungee cord ismodeled as a spring.A bungee cord can be modeled as a spring. Vertical oscillations on the bungee cord are SHM.MODELviSuALizE FiguRE 14.18shows the situation.Although the cord is stretched by 5.0 m when the studentis released, this is not the amplitude of the oscillation. Oscillations occur around the equilibrium position, so we have to beginby finding the equilibrium point where the student hangs motionless. The cord stretch at equilibrium is given by Equation 14.40:mgL== 3.0 mkStretching the cord 5.0 m pulls the student 2.0 m below the equilibrium point, so A = 2.0 m. That is, the student oscillates withamplitude A = 2.0 m about a point 3.0 m beneath the bungeeSOLvEcords original end point. The students position as a function oftime, as measured from the equilibrium position, isy ( t ) = (2.0 m) cos ( vt + f0 )where v = 2k/m = 1.80 rad/s. The initial conditiony0 = A cos f0 = - Arequires the phase constant to be f0 = p rad. At t = 2.0 s the students position and velocity are14.6 . The Pendulum391y = (2.0 m) cos 1 (1.80 rad/s) (2.0 s) + p rad 2 = 1.8 mvy = - vA sin ( vt + f0 ) = - 1.6 m/sThe student is 1.8 m above the equilibrium position, or 1.2m below the original end of the cord. Because his velocity is negative,hes passed through the highest point and is heading down.14.6 The PendulumNow lets look at another very common oscillator: a pendulum. FiguRE 14.19a shows amass m attached to a string of length L and free to swing back and forth. The pendulums position can be described by the arc of length s, which is zero when the pendulum hangs straight down. Because angles are measured ccw, s and u are positive whenthe pendulum is to the right of center, negative when it is to the left.uuTwo forces are acting on the mass: the string tension T and gravity FG. It will beconvenient to repeat what we did in our study of circular motion: Divide the forcesinto tangential components, parallel to the motion, and radial components parallel tothe string. These are shown on the free-body diagram of FiguRE 14.19b.Newtons second law for the tangential component, parallel to the motion, is(Fnet ) t = a Ft = (FG ) t = - mg sin u = matpendulum.The motion of a(a)u and s arenegative onthe left.u and s arepositive onthe right.uLm(14.44)Using at = d 2 s/dt 2 for acceleration around the circle, and noting that the mass cancels, we can write Equation 14.44 asd 2s= - g sin udt 2FiguRE 14.19sArc length0(b)Centerof circle(14.45)The tension hasonly a radialcomponent.urThis is the equation of motion for an oscillating pendulum. The sine function makesthis equation more complicated than the equation of motion for an oscillatingspring.TThe Small-Angle ApproximationSuppose we restrict the pendulums oscillations to small angles of less than about10 . This restriction allows us to make use of an interesting and important piece ofgeometry.FiguRE 14.20 shows an angle u and a circular arc of length s = r u. A right trianglehas been constructed by dropping a perpendicular from the top of the arc to the axis.The height of the triangle is h = r sin u. Suppose that the angle u is small. In thatcase there is very little difference between h and s. If h s, then r sin u r u. It follows thatsin uuNOTE uThe gravitational forcehas a tangentialcomponent mg sin u.The small-angle approximation is valid only if angle u is in radians! How small does u have to be to justify using the small-angle approximation? Itseasy to use your calculator to find that the small-angle approximation is good to threerFG(FG)rThe geometrical basis ofthe smallangle approximation.FiguRE 14.20r( u in radians)The result that sin u u for small angles is called the small-angle approximation.We can similarly note that l r for small angles. Because l = r cos u, it follows thatcos u 1. Finally, we can take the ratio of sine and cosine to find tan usin u u.Table 14.3 summarizes the small-angle approximation. We will have other occasionsto use the small-angle approximation throughout the remainder of this text.t(FG)tTangentialaxishr sin usruulr cos uSmallangle approximations.u must be in radians.TABLE 14.3sin uucos u1tan usin uu392 c h a p t e r 14 . Oscillationssignificant figures, an error of 0.1%, up to angles of 0.10 rad ( 5 ) . In practice,we will use the approximation up to about 10 , but for angles any larger it rapidly losesvalidity and produces unacceptable results.If we restrict the pendulum to u 6 10 , we can use sin u u. In that case, Equation 14.44 for the net force on the mass ismg(Fnet ) t = - mg sin u- mgu = sLwhere, in the last step, we used the fact that angle u is related to the arc length byu = s/L. Then the equation of motion becomesgd 2s=- s2LdtThe pendulum clock has been usedfor hundreds of years.(14.46)This is exactly the same as Equation 14.32 for a mass oscillating on a spring. Thenames are different, with x replaced by s and k/m by g/L, but that does not make it adifferent equation.Because we know the solution to the spring problem, we can immediately write thesolution to the pendulum problem just by changing variables and constants:s ( t ) = A cos ( vt + f0 )u ( t ) = umax cos ( vt + f0 )orThe angular frequencyv = 2pf =gAL(14.47)(14.48)is determined by the length of the string. The pendulum is interesting in that the frequency, and hence the period, is independent of the mass. It depends only on thelength of the pendulum. The amplitude A and the phase constant f0 are determined bythe initial conditions, just as they were for an oscillating spring.ExAMPLE 14.8The maximum angle of a pendulumA 300 g mass on a 30-cm-long string oscillates as a pendulum. Ithas a speed of 0.25 m/s as it passes through the lowest point. Whatmaximum angle does the pendulum reach?Data-based Examples helpstudents with the skill ofdrawing conclusions fromlaboratory data. Designedto supplement lab-basedinstruction, these examplesalso help students in generalwith mathematical reasoning,graphical interpretation, andassessment of results.Assume that the angle remains small, in which case themotion is simple harmonic motion.MODELSOLvEg9.8 m/s 2== 5.72 rad/sBLB 0.30 mThe speed at the lowest point is vmax = vA, so the amplitude isA = smax =The maximum angle, at the maximum arc length smax, isThe angular frequency of the pendulum isv=ExAMPLE 14.9vmax0.25 m/s== 0.0437 mv5.72 rad/sumax =smax0.0437 m== 0.146 rad = 8.3L0.30 mBecause the maximum angle is less than 10 , our analysisbased on the small-angle approximation is reasonable.A SSESSThe gravimeterDeposits of minerals and ore can alter the local value of the freefall acceleration because they tend to be denser than surroundingrocks. Geologists use a gravimeteran instrument that accuratelymeasures the local free-fall accelerationto search for ore deposits. One of the simplest gravimeters is a pendulum. To achievethe highest accuracy, a stopwatch is used to time 100 oscillationsof a pendulum of different lengths. At one location in the field, ageologist makes the following measurements:Length (m)Time (s)0.5001.0001.5002.000141.7200.6245.8283.5What is the local value of g?14.6 . The PendulumAssume the oscillation angle is small, in which case themotion is simple harmonic motion with a period independent ofthe mass of the pendulum. Because the data are known to four significant figures ( { 1 mm on the length and { 0.1 s on the timing,both of which are easily achievable), we expect to determine g tofour significant figures.M ODEL12L 2 4p2T = 2p=LgAgThat is, the square of a pendulums period is proportional to itslength. Consequently, a graph of T 2 versus L should be a straight linepassing through the origin with slope 4p2 /g. We can use the experimentally measured slope to determine g. FiguRE 14.21 is a graph ofthe data, with the period found by dividing the measured time by 100.As expected, the graph is a straight line passing through theorigin. The slope of the best-fit line is 4.021 s 2/m. Consequently,SOLvEFrom Equation 14.48, using f = 1/T, we find2g=4p24p2== 9.818 m/s 2slope 4.021 s 2/mThe fact that the graph is linear and passes through theorigin confirms our model of the situation. Had this not been theASSESS393Graph of the square of the pendulumsperiod versus its length.FiguRE 14.21T 2 (s2)8y4.021x0.00164Best-fit line20L (m)0.00.51.01.52.0case, we would have had to conclude either that our model of thependulum as a simple, small-angle pendulum was not valid or thatour measurements were bad. This is an important reason for having multiple data points rather than using only one length.The Conditions for Simple Harmonic MotionYou can begin to see how, in a sense, we have solved all simple-harmonic-motionproblems once we have solved the problem of the horizontal spring. The restoringforce of a spring, Fsp = - kx, is directly proportional to the displacement x from equilibrium. The pendulums restoring force, in the small-angle approximation, is directlyproportional to the displacement s. A restoring force that is directly proportional to thedisplacement from equilibrium is called a linear restoring force. For any linear restoring force, the equation of motion is identical to the spring equation (other than perhapsusing different symbols). Consequently, any system with a linear restoring force willundergo simple harmonic motion around the equilibrium position.This is why an oscillating spring is the prototype of SHM. Everything that we learnabout an oscillating spring can be applied to the oscillations of any other linear restoring force, ranging from the vibration of airplane wings to the motion of electrons inelectric circuits. Lets summarize this information with a Tactics Box.TACTiCSBOx 14.1identifying and analyzing simple harmonic motion1 If the net force acting on a particle is a linear restoring force, the motion willbe simple harmonic motion around the equilibrium position.2 The position as a function of time is x ( t ) = A cos ( vt + f0 ). The velocityas a function of time is vx ( t ) = - vA sin( vt + f0 ). The maximum speed isvmax = vA. The equations are given here in terms of x, but they can be writtenin terms of y, u, or some other parameter if the situation calls for it.3 The amplitude A and the phase constant f0 are determined by the initial conditions through x0 = A cos f0 and v0x = - vA sin f0.4 The angular frequency v (and hence the period T = 2p/v ) depends on thephysics of the particular situation. But v does not depend on A or f0.5 Mechanical energy is conserved. Thus 1 mvx2 + 1 kx 2 = 1 kA2 = 1 m(vmax )2.2222Energy conservation provides a relationship between position and velocitythat is independent of time.Exercises 712, 1519Tactics Boxes give step-bystep procedures for developingspecific skills (drawing free-bodydiagrams, using ray tracing, etc.).394 c h a p t e r 14 . OscillationsThe Physical PendulumFiguRE 14.22A physical pendulum.ludMoment arm ofgravitational torqueMgDistance frompivot to center ofmassA mass on a string is often called a simple pendulum. But you can also make a pendulum from any solid object that swings back and forth on a pivot under the influence ofgravity. This is called a physical pendulum.FiguRE 14.22 shows a physical pendulum of mass M for which the distance betweenthe pivot and the center of mass is l. The moment arm of the gravitational force actingat the center of mass is d = l sin u, so the gravitational torque ist = - Mgd = - Mgl sin uThe torque is negative because, for positive u, its causing a clockwise rotation. If werestrict the angle to being small ( u 6 10 ) , as we did for the simple pendulum, we canuse the small-angle approximation to writet = - Mglu(14.49)Gravity causes a linear restoring torque on the pendulumthat is, the torque is directly proportional to the angular displacement uso we expect the physical pendulumto undergo SHM.From Chapter 12, Newtons second law for rotational motion isd 2u ta= 2 =Idtwhere I is the objects moment of inertia about the pivot point. Using Equation 14.49for the torque, we find- Mgld 2u=u2IdtMglBI(14.50)Comparison with Equation 14.32 shows that this is again the SHM equation of motion,this time with angular frequencyv = 2pf =(14.51)It appears that the frequency depends on the mass of the pendulum, but recall thatthe moment of inertia is directly proportional to M. Thus M cancels and the frequencyof a physical pendulum, like that of a simple pendulum, is independent of mass.ExAMPLE 14.10A swinging leg as a pendulumA student in a biomechanics lab measures the length of his leg,from hip to heel, to be 0.90 m. What is the frequency of the pendulum motion of the students leg? What is the period?We can model a human leg reasonably well as a rod ofuniform cross section, pivoted at one end (the hip) to form a physical pendulum. The center of mass of a uniform leg is at the midpoint, so l = L/2.MODELMglMg(L /2)3g111=== 0.64 Hz2p B I2p B ML2 /32p B 2LThe moment of inertia of a rod pivoted about one end isI = 1 ML2, so the pendulum frequency is3SOLvEf=Life-science and bioengineeringworked examples andapplications focus on thephysics of life-science situationsin order to serve the needs oflife-science students taking acalculus-based physics class.The corresponding period is T = 1/f = 1.6 s. Notice that we didntneed to know the mass.As you walk, your legs do swing as physical pendulumsas you bring them forward. The frequency is fixed by the lengthof your legs and their distribution of mass; it doesnt depend onamplitude. Consequently, you dont increase your walking speedby taking more rapid stepschanging the frequency is difficult.You simply take longer strides, changing the amplitude but notthe frequency.ASSESSOne person swings on a swing and finds that the period is 3.0 s. Asecond person of equal mass joins him. With two people swinging, the period isStop to think 14.5a. 6.0 sc. 3.0 se. 1.5 sb. 7 3.0 s but not necessarily 6.0 sd. 6 3.0 s but not necessarily 1.5 sf. Cant tell without knowing the length14.7 . Damped Oscillations39514.7 Damped OscillationsA pendulum left to itself gradually slows down and stops. The sound of a ringing bellgradually dies away. All real oscillators do run downsome very slowly but othersquite quicklyas friction or other dissipative forces transform their mechanical energy into the thermal energy of the oscillator and its environment. An oscillation thatruns down and stops is called a damped oscillation.There are many possible reasons for the dissipation of energy, such as air resistance, friction, and internal forces within a metal spring as it flexes. The forces involved in dissipation are complex, but a simple linear drag model gives a quite accurate description of most damped oscillations. That is, well assume a drag force thatdepends linearly on the velocity asuuD = - bv(model of the drag force)(14.52)where the minus sign is the mathematical statement that the force is always oppositein direction to the velocity in order to slow the object.The damping constant b depends in a complicated way on the shape of the objectand on the viscosity of the air or other medium in which the particle moves. The damping constant plays the same role in our model of air resistance that the coefficient offriction does in our model of friction.The units of b need to be such that they will give units of force when multiplied byunits of velocity. As you can confirm, these units are kg/s. A value b = 0 kg/s corresponds to the limiting case of no resistance, in which case the mechanical energyis conserved. A typical value of b for a spring or a pendulum in air is 0.10 kg/s.Objects moving in a liquid can have significantly larger values of b.FiguRE 14.23 shows a mass oscillating on a spring in the presence of a drag force.With the drag included, Newtons second law isThe shock absorbers in cars and trucksare heavily damped springs. The vehiclesvertical motion, after hitting a rock or apothole, is a damped oscillation.An oscillating mass in thepresence of a drag force.FiguRE 14.23r(Fnet ) x = (Fsp ) x + Dx = - kx - bvx = maxFsp(14.53)Using vx = dx/dt and ax = d 2 x /dt 2, we can write Equation 14.53 asd 2 x b dx k++ x=0dt 2 m dt mSpringconstant k(14.54)rDmrvEquation 14.54 is the equation of motion of a damped oscillator. If you compare it toEquation 14.32, the equation of motion for a block on a frictionless surface, youll seethat it differs by the inclusion of the term involving dx/dt.Equation 14.54 is another second-order differential equation. We will simply assert(and, as a homework problem, you can confirm) that the solution isx ( t ) = Ae -bt/2m cos ( vt + f0 )( damped oscillator)b2kb2= v02 2B m 4mB4m2(14.55)where the angular frequency is given byHere v0 = 2k/m is the angular frequency of an undamped oscillator (b = 0). Theconstant e is the base of natural logarithms, so e -bt/2m is an exponential function.Because e 0 = 1, Equation 14.55 reduces to our previous solution, x ( t ) = A cos ( vt +f0 ), when b = 0. This makes sense and gives us confidence in Equation 14.55. Alightly damped system, which oscillates many times before stopping, is one for whichb/2m V v0. In that case, v v0 is a good approximation. That is, light dampingdoes not affect the oscillation frequency.FiguRE 14.24 is a graph of the position x ( t ) for a lightly damped oscillator, as givenby Equation 14.55. Notice that the term Ae -bt/2m, which is shown by the dashed line,v=Positionversustime graphfor a damped oscillator.FiguRE 14.24x(14.56)A0AA is the initial amplitude.The envelope of theamplitude decaysexponentially:xmax Ae bt/2mt396 c h a p t e r 14 . OscillationsSeveral oscillationenvelopes, corresponding to differentvalues of the damping constant b.acts as a slowly varying amplitude:FiguRE 14.25xmax ( t ) = Ae -bt/2mEnergy is conservedif there is no damping.For massm 1.0 kgA smaller b causesless damping.Amplitudebb0200.3 kg/s0.03 kg/sb00 kg/sbA0.1 kg/s4060t (s)Envelope from Figure 14.24(14.57)where A is the initial amplitude, at t = 0. The oscillation keeps bumping up againstthis line, slowly dying out with time.A slowly changing line that provides a border to a rapid oscillation is called theenvelope of the oscillations. In this case, the oscillations have an exponentiallydecaying envelope. Make sure you study Figure 14.24 long enough to see how boththe oscillations and the decaying amplitude are related to Equation 14.55.Changing the amount of damping, by changing the value of b, affects how quicklythe oscillations decay. FiguRE 14.25 shows just the envelope xmax ( t ) for several oscillators that are identical except for the value of the damping constant b. (You needto imagine a rapid oscillation within each envelope, as in Figure 14.24.) Increasingb causes the oscillations to damp more quickly, while decreasing b makes them lastlonger.A larger b causes the oscillationsto damp more quickly.mathematical asideExponential decayExponential decay occurs in a vast number of physical systems ofimportance in science and engineering. Mechanical vibrations, electric circuits, and nuclear radioactivity all exhibit exponential decay.The number e = 2.71828 p is the base of natural logarithmsin the same way that 10 is the base of ordinary logarithms. It arisesnaturally in calculus from the integraldu3 u = ln uThis integralwhich shows up in the analysis of many physicalsystemsfrequently leads to solutions of the formu = Ae -v/v0 = A exp ( - v/v0 )where exp is the exponential function.uu starts at A.Au decays to 37% ofits initial value at vv0.u decays to 13% of itsinitial value at v 2v0.e 1AA graph of u illustrates what we mean by exponential decay.It starts with u = A at v = 0 (because e 0 = 1 ) and then steadilydecays, asymptotically approaching zero. The quantity v0 is calledthe decay constant. When v = v0, u = e -1A = 0.37A. Whenv = 2v0, u = e -2A = 0.13A.Arguments of functions must be pure numbers, without units.That is, we can evaluate e -2, but e -2 kg makes no sense. If v/v0is a pure number, which it must be, then the decay constant v0must have the same units as v. If v represents position, then v0 is alength; if v represents time, then v0 is a time interval. In a specificsituation, v0 is often called the decay length or the decay time. Itis the length or time in which the quantity decays to 37% of itsinitial value.No matter what the process is or what u represents, a quantity that decays exponentially decays to 37% of its initial valuewhen one decay constant has passed. Thus exponential decayis a universal behavior. Every time you meet a new system thatexhibits exponential decay, its behavior will be exactly the sameas every other exponential decay. The decay curve always looksexactly like the figure shown here. Once youve learned the properties of exponential decay, youll immediately know how to apply this knowledge to a new situation.e 2A00v02v0vEnergy in Damped SystemsWhen considering the oscillators mechanical energy, it is useful to define the timeconstant t (also called the decay time) to bemt=(14.58)bBecause b has units of kg/s, t has units of seconds. With this definition, we can writethe oscillation amplitude as xmax ( t ) = Ae -t/2t.14.7 . Damped Oscillations12397Because of the drag force, the mechanical energy is no longer conserved. At anyparticular time we can compute the mechanical energy from111 2 -t/tk(xmax )2 = k(Ae -t/2t )2 =kA e= E0 e -t/t222E (t) =122 kAm2(14.59)where E0 =is the initial energy at t = 0 and where we used (z ) = z . In otherwords, the oscillators mechanical energy decays exponentially with time constant T.As FiguRE 14.26 shows, the time constant is the amount of time needed for the energyto decay to e -1, or 37%, of its initial value. We say that the time constant t measures the characteristic time during which the energy of the oscillation is dissipated.Roughly two-thirds of the initial energy is gone after one time constant has elapsed,and nearly 90% has dissipated after two time constants have gone by.For practical purposes, we can speak of the time constant as the lifetime of theoscillationabout how long it lasts. Mathematically, there is never a time when theoscillation is over. The decay approaches zero asymptotically, but it never gets therein any finite time. The best we can do is define a characteristic time when the motionis almost over, and that is what the time constant t does.ExAMPLE 14.11The energy has decreased to37% of its initial value at t t.The energy hasdecreased to 13%of its initial valueat t 2t.0.37E00.13E000ttt2tE ( t ) = E0 e -t/tThe time at which an exponential decay is reduced to 1 E0, half2its initial value, has a special name. It is called the half-lifeand given the symbol t1/2. The concept of the half-life is widelyused in applications such as radioactive decay. To relate t1/2 tot, we first write1E (at t = t1/2 ) = E0 = E0 e -t1/2 /t2The E0 cancels, giving1= e -t1/2 /t2Again, we take the natural logarithm of both sides:The motion is a damped oscillation.a. The initial amplitude at t = 0 is xmax = A. At t = 35.0 sthe amplitude is xmax = 1 A. The amplitude of oscillation at2time t is given by Equation 14.57:SOLvExmax ( t ) = Ae -bt/2m = Ae -t/2tIn this case,1A = Ae - (35.0 s)/2t2Notice that we do not need to know A itself because it cancelsout. To solve for t, we take the natural logarithm of both sidesof the equation:12ln121= - ln 2 = ln e -t1/2 /t = - t1/2 /t2Finally, we solve for t1/2:135.0 s= - ln 2 = ln e - (35.0 s)/2t = 22tThis is easily rearranged to give35.0 st== 25.2 s2 ln 2If desired, we could now determine the damping constant to beb = m/t = 0.020 kg/s.t1/2 = t ln 2 = 0.693tThis result that t1/2 is 69% of t is valid for any exponentialdecay. In this particular problem, half the energy is gone att1/2 = (0.693) (25.2 s) = 17.5 sThe oscillator loses energy faster than it loses amplitude.This is what we should expect because the energy depends on thesquare of the amplitude.ASSESSRank in order, from largest to smallest, the time constants ta to tdof the decays shown in the figure. All the graphs have the same scale.Stop to think 14.6EEEt(a)E0b. The energy at time t is given bya. What is the time constant for this oscillator?b. At what time will the energy have decayed to half its initialvalue?lnThe oscillator startswith energy E0.EnergyA damped pendulumA 500 g mass swings on a 60-cm-string as a pendulum. The amplitude is observed to decay to half its initial value after 35.0 s.MODELExponential decay of themechanical energy of an oscillator.FiguRE 14.262mEt(b)t(c)t(d)t398 c h a p t e r 14 . Oscillations14.8 Driven Oscillations and ResonanceThe response curve showsthe amplitude of a driven oscillator atfrequencies near its natural frequencyof 2.0 Hz.FiguRE 14.27AmplitudeThe oscillation hasmaximum amplitudewhen fext f0. Thisis resonance.The oscillation hasonly a small amplitudewhen fext differssubstantially from f0.12fext (Hz)3This is the naturalfrequency.The resonance amplitudebecomes higher and narrower as thedamping constant decreases.FiguRE 14.28Amplitudef0b0.08 kg/sb0.20 kg/sb12.0 HzA lightly damped systemhas a very tall and verynarrow response curve.0.80 kg/sA heavily dampedsystem has littleresponse.23fext (Hz)Thus far we have focused on the free oscillations of an isolated system. Some initialdisturbance displaces the system from equilibrium, and it then oscillates freely until itsenergy is dissipated. These are very important situations, but they do not exhaust thepossibilities. Another important situation is an oscillator that is subjected to a periodicexternal force. Its motion is called a driven oscillation.A simple example of a driven oscillation is pushing a child on a swing, where yourpush is a periodic external force applied to the swing. A more complex example is acar driving over a series of equally spaced bumps. Each bump causes a periodic upward force on the cars shock absorbers, which are big, heavily damped springs. Theelectromagnetic coil on the back of a loudspeaker cone provides a periodic magneticforce to drive the cone back and forth, causing it to send out sound waves. Air turbulence moving across the wings of an aircraft can exert periodic forces on the wings andother aerodynamic surfaces, causing them to vibrate if they are not properly designed.As these examples suggest, driven oscillations have many important applications.However, driven oscillations are a mathematically complex subject. We will simplyhint at some of the results, saving the details for more advanced classes.Consider an oscillating system that, when left to itself, oscillates at a frequency f0.We will call this the natural frequency of the oscillator. The natural frequency fora mass on a spring is 1k/m /2p, but it might be given by some other expression foranother type of oscillator. Regardless of the expression, f0 is simply the frequency ofthe system if it is displaced from equilibrium and released.Suppose that this system is subjected to a periodic external force of frequency fext.This frequency, which is called the driving frequency, is completely independentof the oscillators natural frequency f0. Somebody or something in the environmentselects the frequency fext of the external force, causing the force to push on the systemfext times every second.Although it is possible to solve Newtons second law with an external drivingforce, we will be content to look at a graphical representation of the solution. Themost important result is that the oscillation amplitude depends very sensitively on thefrequency fext of the driving force. The response to the driving frequency is shownin FiguRE 14.27 for a system with m = 1.0 kg, a natural frequency f0 = 2.0 Hz, and adamping constant b = 0.20 kg/s. This graph of amplitude versus driving frequency,called the response curve, occurs in many different applications.When the driving frequency is substantially different from the oscillators naturalfrequency, at the right and left edges of Figure 14.27, the system oscillates but the amplitude is very small. The system simply does not respond well to a driving frequencythat differs much from f0. As the driving frequency gets closer and closer to the naturalfrequency, the amplitude of the oscillation rises dramatically. After all, f0 is the frequency at which the system wants to oscillate, so it is quite happy to respond to adriving frequency near f0. Hence the amplitude reaches a maximum when the drivingfrequency exactly matches the systems natural frequency: fext = f0.The amplitude can become exceedingly large when the frequencies match, especially if the damping constant is very small. FiguRE 14.28 shows the same oscillatorwith three different values of the damping constant. Theres very little response if thedamping constant is increased to 0.80 kg/s, but the amplitude for fext = f0 becomesvery large when the damping constant is reduced to 0.08 kg/s. This large-amplituderesponse to a driving force whose frequency matches the natural frequency of the system is a phenomenon called resonance. The condition for resonance isfext = f0 (resonance condition)(14.60)Within the context of driven oscillations, the natural frequency f0 is often called theresonance frequency.An important feature of Figure 14.28 is how the amplitude and width of the resonance depend on the damping constant. A heavily damped system responds fairlyChallenge Examplelittle, even at resonance, but it responds to a wide range of driving frequencies. Verylightly damped systems can reach exceptionally high amplitudes, but notice that therange of frequencies to which the system responds becomes narrower and narroweras b decreases.This allows us to understand why a few singers can break crystal goblets but notinexpensive, everyday glasses. An inexpensive glass gives a thud when tapped, but afine crystal goblet rings for several seconds. In physics terms, the goblet has a muchlonger time constant than the glass. That, in turn, implies that the goblet is very lightlydamped while the ordinary glass is heavily damped (because the internal forces withinthe glass are not those of a high-quality crystal structure).The singer causes a sound wave to impinge on the goblet, exerting a small drivingforce at the frequency of the note she is singing. If the singers frequency matches thenatural frequency of the gobletresonance! Only the lightly damped goblet, like thetop curve in Figure 14.28, can reach amplitudes large enough to shatter. The restriction, though, is that its natural frequency has to be matched very precisely. The soundalso has to be very loud.CHALLENgE E xAMPLE 14.12A swinging pendulumA pendulum consists of a massless, rigid rod with a mass at oneend. The other end is pivoted on a frictionless pivot so that the rodcan rotate in a complete circle. The pendulum is inverted, withthe mass directly above the pivot point, then released. The speedof the mass as it passes through the lowest point is 5.0 m/s. If thependulum later undergoes small-amplitude oscillations at the bottom of the arc, what will its frequency be?This is a simple pendulum because the rod is massless.However, our analysis of a pendulum used the small-angle approximation. It applies only to the small-amplitude oscillations atthe end, not to the pendulum swinging down from the invertedposition. Fortunately, energy is conserved throughout, so we cananalyze the big swing using conservation of mechanical energy.A singer or musical instrument can shattera crystal goblet by matching the gobletsnatural oscillation frequency.The frequency of a simple pendulum is f = 1g/L /2p.Were not given L, but we can find it by analyzing the pendulumsswing down from an inverted position. Mechanical energy is conserved, and the only potential energy is gravitational potentialenergy. Conservation of mechanical energy Kf + Ugf = Ki + Ugi,with Ug = mgy, isSOLvE121mv + mgyf = mvi2 + mgyi2f2MODELis a pictorial representation of the pendulum swinging down from the inverted position. The pendulumlength is L, so the initial height is 2L.399The mass cancels, which is good since we dont know it, and twoterms are zero. Thus12v = g(2L) = 2gL2fviSuALizE FiguRE 14.29Solving for L, we findL=Beforeandafter pictorial representation of thependulum swinging down from an inverted position.FiguRE 14.29vf2(5.0 m/s)2== 0.638 m4g 4(9.80 m/s 2)g19.80 m/s 21== 0.62 Hz2p B L 2p B 0.638 mNow we can calculate the frequency:f=The frequency corresponds to a period of about 1.5 s,which seems reasonable.ASSESSA consistent 4-step approachprovides a problem-solvingframework throughout thebook (and all supplements):students learn the importanceof making assumptions (inthe MODEL step), gatheringinformation, and makingsketches (in the VISUALIZEstep) before treating theproblem mathematically(SOLVE) and then analyzingtheir result (ASSESS).Challenge Examples illustratehow to integrate multipleconcepts and use moresophisticated reasoning inproblem-solving, ensuringan optimal range of workedexamples for students to studyin preparation for homeworkproblems.400 c h a p t e r 14 . OscillationsSuMMARyThe goal of Chapter 14 has been to understand systems that oscillate with simple harmonic motion.general PrinciplesDynamicsEnergySHM occurs when a linear restoring force acts to return asystem to an equilibrium position.If there is no frictionor dissipation, kineticand potential energy arealternately transformedinto each other, but thetotal mechanical energyE = K + U is conserved.11E = mv 2 + k x 2221= m(vmax) 221= k A22kHorizontal springm(Fnet )x = - kx0xVertical springThe origin is at the equilibriumposition L = mg/k.Unique and critically acclaimedvisual chapter summariesconsolidate understandingby providing each concept inwords, math, and figures andorganizing these into a verticalhierarchyfrom GeneralPrinciples (top) to Applications(bottom).kBoth: v =BmmT = 2pBk(Fnet )y = - kyPendulumgv=BL(Fnet )t = -12k0myLT = 2pBgmgsL0AAxAll potentialEE0In a damped system, theenergy decays exponentiallyL0All kinetic0.37E0E = E0 e -t/t0t0twhere t is the time constant.simportant ConceptsSimple harmonic motion (SHM) is a sinusoidal oscillation withperiod T and amplitude A.Frequency f =1TAngular frequency2pv = 2pf =Tx12pt+ f0T2yf = vt + f0 is the phaseAPosition x ( t ) = A cos ( vt + f0 )= A cosTSHM is the projectiononto the x-axis ofuniform circular motion.t0AAfThe position at time t is0xx ( t ) = A cos f= A cos ( vt + f0 )f0xA cos fx0 A cos f0The phase constant f0determines the initial conditions:x0 = A cos f0Velocity vx ( t ) = - vmax sin ( vt + f0 ) with maximum speedvmax = vAv0x = - vA sin f0Acceleration ax ( t ) = - v2 x ( t ) = - v2Acos ( vt + f0 )ApplicationsResonanceWhen a system is driven bya periodic external force, itresponds with a large-amplitudeoscillation if fext f0, wheref0 is the systems naturaloscillation frequency, orresonant frequency.DampingAmplitudeuuIf there is a drag force D = - bv ,where b is the damping constant,then (for lightly damped systems)x ( t ) = Ae -bt/2m cos ( vt + f0 )f0fextA0The time constant for energy lossis t = m/b.AxtConceptual Questions401Terms and Notationoscillatory motionoscillatorperiod, Tfrequency, fhertz, Hzsimple harmonic motion,SHMlinear restoring forcedamped oscillationdamping constant, benvelopetime constant, thalf-life, t1/2driven oscillationamplitude, Aangular frequency, vphase, fphase constant, f0restoring forceequation of motionsmall-angle approximationnatural frequency, f0driving frequency, fextresponse curveresonanceresonance frequency, f0CONCEPTuAL QuESTiONS1. A block oscillating on a spring has period T = 2 s. What is theperiod if:a. The blocks mass is doubled? Explain. Note that you do notknow the value of either m or k, so do not assume any particular values for them. The required analysis involves thinkingabout ratios.b. The value of the spring constant is quadrupled?c. The oscillation amplitude is doubled while m and k areunchanged?2. A pendulum on Planet X, where the value of g is unknown, oscillates with a period T = 2 s. What is the period of this pendulum if:a. Its mass is doubled? Explain. Note that you do not know thevalue of m, L, or g, so do not assume any specific values. Therequired analysis involves thinking about ratios.b. Its length is doubled?c. Its oscillation amplitude is doubled?x (cm)3. FiguRE Q14.3 shows a positionversus-time graph for a particle in10SHM. What are (a) the amplitudeA, (b) the angular frequency v, andt (s)0(c) the phase constant f0? Explain.2410FiguRE Q14.32t0AFiguRE Q14.7vvmax2t01vmax3FiguRE Q14.89.shows the potential-energy diagram and the totalenergy line of a particle oscillating on a spring.a. What is the springs equilibrium length?b. Where are the turning points of the motion? Explain.c. What is the particles maximum kinetic energy?d. What will be the turning points if the particles total energy isdoubled?FiguRE Q14.9Energy (J)x1FiguRE Q14.8 shows a velocity-versus-time graph for a particle inSHM.a. What is the phase constant f0? Explain.b. What is the phase of the particle at each of the three numbered points on the graph?204. Equation 14.25 states that 1 kA2 = 1 m ( vmax)2. What does this22mean? Write a couple of sentences explaining how to interpretthis equation.5. A block oscillating on a spring has an amplitude of 20 cm. Whatwill the amplitude be if the total energy is doubled? Explain.6. A block oscillating on a spring has a maximum speed of 20 cm/s.What will the blocks maximum speed be if the total energy isdoubled? Explain.7. FiguRE Q14.7 shows a position-versus-time graph for a particle inSHM.a. What is the phase constant f0? Explain.b. What is the phase of the particle at each of the three numbered points on the graph?A8.3PE1510TE501216202428x (cm)FiguRE Q14.910. Suppose the damping constant b of an oscillator increases.a. Is the medium more resistive or less resistive?b. Do the oscillations damp out more quickly or less quickly?c. Is the time constant t increased or decreased?11. a. Describe the difference between t and T. Dont just namethem; say what is different about the physical concepts theyrepresent.b. Describe the difference between t and t1/2.12. What is the difference between the driving frequency and thenatural frequency of an oscillator?Conceptual Questions requirecareful reasoning and can beused for group discussions orindividual work.402 c h a p t e r 14 . OscillationsThe end-of-chapter problemsare rated by students toshow difficulty level with thevariety expanded to includemore real-world, challenging,and explicitly calculus-basedproblems.Exercises (for each section)allow students to buildtheir skills and confidencewith straightforward, one-stepquestions.ExErcisEs and ProblEmsProblems labeledintegrate material from earlier chapters.Section 14.4 The Dynamics of Simple Harmonic MotionExercises11.Section 14.1 Simple Harmonic MotionWhen a guitar string plays the note A, the string vibrates at440 Hz. What is the period of the vibration?2. | An air-track glider attached to a spring oscillates between the10 cm mark and the 60 cm mark on the track. The glider completes 10 oscillations in 33 s. What are the (a) period, (b) frequency, (c) angular frequency, (d) amplitude, and (e) maximumspeed of the glider?3. || An air-track glider is attached to a spring. The glider is pulledto the right and released from rest at t = 0 s. It then oscillateswith a period of 2.0 s and a maximum speed of 40 cm/s.a. What is the amplitude of the oscillation?b. What is the gliders position at t = 0.25 s?1.Section 14.3 Energy in Simple Harmonic Motion|12.13.Section 14.2 Simple Harmonic Motion and Circular Motion4.What are the (a) amplitude, (b) frequency, and (c) phase constant of the oscillation shown in FigurE Ex14.4?|14.x (cm)20100210FigurE Ex14.45.46t (s)815.20What are the (a) amplitude, (b) frequency, and (c) phase constant of the oscillation shown in FigurE Ex14.5?||x (cm)A block attached to a spring with unknown spring constantoscillates with a period of 2.0 s. What is the period ifa. The mass is doubled?b. The mass is halved?c. The amplitude is doubled?d. The spring constant is doubled?Parts a to d are independent questions, each referring to the initial situation.|| A 200 g air-track glider is attached to a spring. The glider ispushed in 10 cm and released. A student with a stopwatch findsthat 10 oscillations take 12.0 s. What is the spring constant?|| A 200 g mass attached to a horizontal spring oscillates at afrequency of 2.0 Hz. At t = 0 s, the mass is at x = 5.0 cm andhas vx = - 30 cm/s. Determine:a. The period.b. The angular frequency.c. The amplitude.d. The phase constant.e. The maximum speed.f. The maximum acceleration.g. The total energy.h. The position at t = 0.40 s.| The position of a 50 g oscillating mass is given by x ( t ) =( 2.0 cm) cos (10t - p/4), where t is in s. Determine:a. The amplitude.b. The period.c. The spring constant.d. The phase constant.e. The initial conditions.f. The maximum speed.g. The total energy.h. The velocity at t = 0.40 s.|| A 1.0 kg block is attached to a spring with spring constant16 N/m. While the block is sitting at rest, a student hits it with ahammer and almost instantaneously gives it a speed of 40 cm/s.What area. The amplitude of the subsequent oscillations?b. The blocks speed at the point where x = 1 A?2|10505FigurE Ex14.56.7.8.9.10.1234t (s)10|| An object in simple harmonic motion has an amplitude of4.0 cm, a frequency of 2.0 Hz, and a phase constant of 2p/3 rad.Draw a position graph showing two cycles of the motion.|| An object in simple harmonic motion has an amplitude of8.0 cm, a frequency of 0.25 Hz, and a phase constant of - p/2 rad.Draw a position graph showing two cycles of the motion.| An object in simple harmonic motion has amplitude 4.0 cmand frequency 4.0 Hz, and at t = 0 s it passes through the equilibrium point moving to the right. Write the function x ( t ) thatdescribes the objects position.| An object in simple harmonic motion has amplitude 8.0 cmand frequency 0.50 Hz. At t = 0 s it has its most negative position. Write the function x ( t ) that describes the objects position.|| An air-track glider attached to a spring oscillates with a periodof 1.5 s. At t = 0 s the glider is 5.00 cm left of the equilibriumposition and moving to the right at 36.3 cm/s.a. What is the phase constant?b. What is the phase at t = 0 s, 0.5 s, 1.0 s, and 1.5 s?Section 14.5 Vertical Oscillations16.| A spring is hanging from the ceiling. Attaching a 500 gphysics book to the spring causes it to stretch 20 cm in order tocome to equilibrium.a. What is the spring constant?b. From equilibrium, the book is pulled down 10 cm and released. What is the period of oscillation?c. What is the books maximum speed?17. || A spring with spring constant 15 N/m hangs from the ceiling.A ball is attached to the spring and allowed to come to rest. It isthen pulled down 6.0 cm and released. If the ball makes 30 oscillations in 20 s, what are its (a) mass and (b) maximum speed?18. || A spring is hung from the ceiling. When a block is attachedto its end, it stretches 2.0 cm before reaching its new equilibrium length. The block is then pulled down slightly and released.What is the frequency of oscillation?Section 14.6 The Pendulum19.| A mass on a string of unknown length oscillates as a pendulum with a period of 4.0 s. What is the period ifa. The mass is doubled?Exercises and Problems20.21.22.23.24.b. The string length is doubled?c. The string length is halved?d. The amplitude is doubled?Parts a to d are independent questions, each referring to the initial situation.|| A 200 g ball is tied to a string. It is pulled to an angle of 8.0 andreleased to swing as a pendulum. A student with a stopwatchfinds that 10 oscillations take 12 s. How long is the string?| What is the period of a 1.0-m-long pendulum on (a) the earthand (b) Venus?| What is the length of a pendulum whose period on the moonmatches the period of a 2.0-m-long pendulum on the earth?| Astronauts on the first trip to Mars take along a pendulum thathas a period on earth of 1.50 s. The period on Mars turns out tobe 2.45 s. What is the free-fall acceleration on Mars?|| A uniform steel bar swings from a pivot at one end with aperiod of 1.2 s. How long is the bar?Section 14.7 Damped OscillationsSection 14.8 Driven Oscillations and Resonance25.26.27.28.29.BIO| A 2.0 g spider is dangling at the end of a silk thread. You canmake the spider bounce up and down on the thread by tappinglightly on his feet with a pencil. You soon discover that you cangive the spider the largest amplitude on his little bungee cord ifyou tap exactly once every second. What is the spring constantof the silk thread?|| The amplitude of an oscillator decreases to 36.8% of its initialvalue in 10.0 s. What is the value of the time constant?|| Sketch a position graph from t = 0 s to t = 10 s of a dampedoscillator having a frequency of 1.0 Hz and a time constantof 4.0 s.| In a science museum, a 110 kg brass pendulum bob swingsat the end of a 15.0-m-long wire. The pendulum is started at exactly 8:00 a.m. every morning by pulling it 1.5 m to the side andreleasing it. Because of its compact shape and smooth surface,the pendulums damping constant is only 0.010 kg/s. At exactly12:00 noon, how many oscillations will the pendulum have completed and what is its amplitude?|| Vision is blurred if the head is vibrated at 29 Hz because thevibrations are resonant with the natural frequency of the eyeballin its socket. If the mass of the eyeball is 7.5 g, a typical value,what is the effective spring constant of the musculature thatholds the eyeball in the socket?Problems30.FiguRE P14.30 is the velocity-versus-time graph of a particle insimple harmonic motion.a. What is the amplitude of the oscillation?b. What is the phase constant?c. What is the position at t = 0 s?||vx (cm/s)6030030FiguRE P14.306036912t (s)31.403| FiguRE P14.31 is the position-versus-time graph of a particle insimple harmonic motion.a. What is the phase constant?b. What is the velocity at t = 0 s?c. What is vmax?yx (cm)A10B5051234t (s)0612t (s)10FiguRE P14.3132.33.34.35.BIOFiguRE P14.32The two graphs in FiguRE P14.32 are for two different verticalmass-spring systems. If both systems have the same mass, whatis the ratio kA /kB of their spring constants?||| An object in SHM oscillates with a period of 4.0 s and an amplitude of 10 cm. How long does the object take to move fromx = 0.0 cm to x = 6.0 cm?|| A 1.0 kg block oscillates on a spring with spring constant20 N/m. At t = 0 s the block is 20 cm to the right of the equilibrium position and moving to the left at a speed of 100 cm/s.Determine (a) the period and (b) the amplitude.|| Astronauts in space cannot weigh themselves by standing on abathroom scale. Instead, they determine their mass by oscillatingon a large spring. Suppose an astronaut attaches one end of alarge spring to her belt and the other end to a hook on the wall ofthe space capsule. A fellow astronaut then pulls her away fromthe wall and releases her. The springs length as a function oftime is shown in FiguRE P14.35.a. What is her mass if the spring constant is 240 N/m?b. What is her speed when the springs length is 1.2 m?||L (m)1.41.21.00.80.60.40.20.0FiguRE P14.35036t (s)|| The motion of a particle is given by x ( t ) = (25 cm)cos (10t ),where t is in s. At what time is the kinetic energy twice the potential energy?37. || a. When the displacement of a mass on a spring is 1 A, what2fraction of the energy is kinetic energy and what fraction ispotential energy?b. At what displacement, as a fraction of A, is the energy halfkinetic and half potential?38. || For a particle in simple harmonic motion, show that vmax =( p/2) vavg where vavg is the average speed during one cycle of themotion.39. || A 100 g block attached to a spring with spring constant2.5 N/m oscillates horizontally on a frictionless table. Its velocity is 20 cm/s when x = - 5.0 cm.a. What is the amplitude of oscillation?b. What is the blocks maximum acceleration?c. What is the blocks position when the acceleration is maximum?d. What is the speed of the block when x = 3.0 cm?36.Problems (spanning conceptsfrom the whole chapter),require in-depth reasoningand planning, and allowstudents to practice theirproblem-solving strategies.Context-rich problems requirestudents to simplify and modelmore complex real-worldsituations. Specifically labeledproblems integr ate conceptsfrom multiple previouschapters.404 c h a p t e r 14 . Oscillations40.41.42.BIOBio problems are set in lifescience, bioengineering, orbiomedical contexts.43.44.A block on a spring is pulled to the right and released att = 0 s. It passes x = 3.00 cm at t = 0.685 s, and it passesx = - 3.00 cm at t = 0.886 s.a. What is the angular frequency?b. What is the amplitude?Hint: cos ( p - u ) = - cos u.||| A 300 g oscillator has a speed of 95.4 cm/s when its displacement is 3.0 cm and 71.4 cm/s when its displacement is 6.0 cm.What is the oscillators maximum speed?|| An ultrasonic transducer, of the type used in medical ultrasound imaging, is a very thin disk (m = 0.10 g) driven back andforth in SHM at 1.0 MHz by an electromagnetic coil.a. The maximum restoring force that can be applied to the diskwithout breaking it is 40,000 N. What is the maximum oscillation amplitude that wont rupture the disk?b. What is the disks maximum speed at this amplitude?|| A 5.0 kg block hangs from a spring with spring constant2000 N/m. The block is pulled down 5.0 cm from the equilibrium position and given an initial velocity of 1.0 m/s back towardequilibrium. What are the (a) frequency, (b) amplitude, and(c) total mechanical energy of the motion?|| Your lab instructor has asked you to measure a spring constantusing a dynamic methodletting it oscillaterather than a static method of stretching it. You and your lab partner suspend thespring from a hook, hang different masses on the lower end, andstart them oscillating. One of you uses a meter stick to measurethe amplitude, the other uses a stopwatch to time 10 oscillations.Your data are as follows:||a. What is the spring constant of each spring if the empty carbounces up and down 2.0 times each second?b. What will be the cars oscillation frequency while carryingfour 70 kg passengers?49. || The two blocks in FiguRE P14.49 oscillate on a frictionless surface with a period of 1.5 s. The upper block just begins to slipwhen the amplitude is increased to 40 cm. What is the coefficientof static friction between the two blocks?FiguRE P14.4950.BIO||| It has recently become possible to weigh DNA moleculesby measuring the influence of their mass on a nano-oscillator.FiguRE P14.50 shows a thin rectangular cantilever etched out ofsilicon (density 2300 kg/m3 ) with a small gold dot at the end. Ifpulled down and released, the end of the cantilever vibrates withsimple harmonic motion, moving up and down like a divingboard after a jump. When bathed with DNA molecules whoseends have been modified to bind with gold, one or more molecules may attach to the gold dot. The addition of their masscauses a very slightbut measurabledecrease in the oscillation frequency.4000 nmMass (g)Time (s)1006.57.8150Data-based problems allowstudents to practice drawingconclusions from data (asdemonstrated in the new databased examples in the text).Amplitude (cm)5.59.8400 nmThickness = 100 nm20046.47.48.10.925045.6.0FiguRE P14.503.512.4A vibrating cantilever of mass M can be modeled as a block ofmass 1 M attached to a spring. (The factor of 1 arises from the mo33ment of inertia of a bar pivoted at one end.) Neither the mass northe spring constant can be determined very accurately perhapsto only two significant figuresbut the oscillation frequency canbe measured with very high precision simply by counting the oscillations. In one experiment, the cantilever was initially vibratingat exactly 12 MHz. Attachment of a DNA molecule caused thefrequency to decrease by 50 Hz. What was the mass of the DNA?|| It is said that Galileo discovered a basic principle of thependulumthat the period is independent of the amplitudebyusing his pulse to time the period of swinging lamps in the cathedral as they swayed in the breeze. Suppose that one oscillation ofa swinging lamp takes 5.5 s.a. How long is the lamp chain?b. What maximum speed does the lamp have if its maximumangle from vertical is 3.0 ?|| A 100 g mass on a 1.0-m-long string is pulled 8.0 to one sideand released. How long does it take for the pendulum to reach4.0 on the opposite side?|| Orangutans can move by brachiation, swinging like a pendulum beneath successive handholds. If an orangutan has arms thatare 0.90 m long and repeatedly swings to a 20 angle, taking oneswing after another, estimate its speed of forward motion in m/s.While this is somewhat beyond the range of validity of the smallangle approximation, the standard results for a pendulum areadequate for making an estimate.Use the best-fit line of an appropriate graph to determine thespring constant.||| A 200 g block hangs from a spring with spring constant10 N/m. At t = 0 s the block is 20 cm below the equilibriumpoint and moving upward with a speed of 100 cm/s. What arethe blocksa. Oscillation frequency?b. Distance from equilibrium when the speed is 50 cm/s?c. Distance from equilibrium at t = 1.0 s?|| A spring with spring constant k is suspended vertically froma support and a mass m is attached. The mass is held at the pointwhere the spring is not stretched. Then the mass is released andbegins to oscillate. The lowest point in the oscillation is 20 cmbelow the point where the mass was released. What is the oscillation frequency?|| While grocery shopping, you put several apples in the springscale in the produce department. The scale reads 20 N, and youuse your ruler (which you always carry with you) to discoverthat the pan goes down 9.0 cm when the apples are added. If youtap the bottom of the apple-filled pan to make it bounce up anddown a little, what is its oscillation frequency? Ignore the massof the pan.|| A compact car has a mass of 1200 kg. Assume that the car hasone spring on each wheel, that the springs are identical, and thatthe mass is equally distributed over the four springs.51.52.53.BIOExercises and Problems54.| Show that Equation 14.51 for the angular frequency of a physical pendulum gives Equation 14.48 when applied to a simplependulum of a mass on a string.55. ||| A 15@cm@long, 200 g rod is pivoted at one end. A 20 g ball ofclay is stuck on the other end. What is the period if the rod andclay swing as a pendulum?56. ||| A uniform rod of mass M and length L swings as a pendulumon a pivot at distance L/4 from one end of the rod. Find an expression for the frequency f of small-angle oscillations.57. ||| A solid sphere of mass M and radius R is suspended from athin rod, as shown in FiguRE P14.57. The sphere can swing backand forth at the bottom of the rod. Find an expression for thefrequency f of small-angle oscillations.Potential energy (J)4Rubber bandsFiguRE P14.6263.RFiguRE P14.57|| A geologist needs to determine the local value of g . Unfortunately, his only tools are a meter stick, a saw, and a stopwatch.He starts by hanging the meter stick from one end and measuringits frequency as it swings. He then saws off 20 cmusing thecentimeter markingsand measures the frequency again. Aftertwo more cuts, these are his data:Length (cm)10060.61.62.0.794059.0.676064.0.6180BIOFrequency (Hz)0.96Use the best-fit line of an appropriate graph to determine thelocal value of g.|| Interestingly, there have been several studies using cadaversto determine the moments of inertia of human body parts, information that is important in biomechanics. In one study, the center of mass of a 5.0 kg lower leg was found to be 18 cm from theknee. When the leg was allowed to pivot at the knee and swingfreely as a pendulum, the oscillation frequency was 1.6 Hz. Whatwas the moment of inertia of the lower leg about the knee joint?|| A 500 g air-track glider attached to a spring with spring constant 10 N/m is sitting at rest on a frictionless air track. A 250 gglider is pushed toward it from the far end of the track at a speedof 120 cm/s. It collides with and sticks to the 500 g glider. Whatare the amplitude and period of the subsequent oscillations?|| A 200 g block attached to a horizontal spring is oscillatingwith an amplitude of 2.0 cm and a frequency of 2.0 Hz. Just as itpasses through the equilibrium point, moving to the right, a sharpblow directed to the left exerts a 20 N force for 1.0 ms. What arethe new (a) frequency and (b) amplitude?|| FiguRE P14.62 is a top view of an object of mass m connectedbetween two stretched rubber bands of length L. The object restson a frictionless surface. At equilibrium, the tension in each rubber band is T. Find an expression for the frequency of oscillations perpendicular to the rubber bands. Assume the amplitudeis sufficiently small that the magnitude of the tension in the rubber bands is essentially unchanged as the mass oscillates.L65.66.67.68.69.191019210191L103Pivot58.40510190.08 0.10 0.12 0.14 0.16Bond length (nm)FiguRE P14.63|| A molecular bond can be modeled as a spring between twoatoms that vibrate with simple harmonic motion. FiguRE P14.63shows an SHM approximation for the potential energy of anHCl molecule. For E 6 4 * 10-19 J it is a good approximation tothe more accurate HCl potential-energy curve that was shown inFigure 10.31. Because the chlorine atom is so much more massive than the hydrogen atom, it is reasonable to assume that thehydrogen atom (m = 1.67 * 10-27 kg) vibrates back and forthwhile the chlorine atom remains at rest. Use the graph to estimate the vibrational frequency of the HCl molecule.|| An ice cube can slide around the inside of a vertical circular hoop of radius R. It undergoes small-amplitude oscillationsif displaced slightly from the equilibrium position at the lowestpoint. Find an expression for the period of these small-amplitudeoscillations.|| A penny rides on top of a piston as it undergoes vertical simpleharmonic motion with an amplitude of 4.0 cm. If the frequencyis low, the penny rides up and down without difficulty. If thefrequency is steadily increased, there comes a point at which thepenny leaves the surface.a. At what point in the cycle does the penny first lose contactwith the piston?b. What is the maximum frequency for which the penny justbarely remains in place for the full cycle?|| On your first trip to Planet X you happen to take along a200 g mass, a 40-cm-long spring, a meter stick, and a stopwatch.Youre curious about the free-fall acceleration on Planet X,where ordinary tasks seem easier than on earth, but you cantfind this information in your Visitors Guide. One night you suspend the spring from the ceiling in your room and hang the massfrom it. You find that the mass stretches the spring by 31.2 cm.You then pull the mass down 10.0 cm and release it. With thestopwatch you find that 10 oscillations take 14.5 s. Based on thisinformation, what is g?|| The 15 g head of a bobble-head doll oscillates in SHM at afrequency of 4.0 Hz.a. What is the spring constant of the spring on which the head ismounted?b. The amplitude of the heads oscillations decreases to 0.5 cmin 4.0 s. What is the heads damping constant?|| An oscillator with a mass of 500 g and a period of 0.50 s hasan amplitude that decreases by 2.0% during each complete oscillation. If the initial amplitude is 10 cm, what will be the amplitude after 25 oscillations?|| A spring with spring constant 15.0 N/m hangs from the ceiling.A 500 g ball is attached to the spring and allowed to come to rest. Itis then pulled down 6.0 cm and released. What is the time constantif the balls amplitude has decreased to 3.0 cm after 30 oscillations?An Increased emphasis onsymbolic answers encouragesstudents to work algebraically.The Student Workbook alsocontains new exercises tohelp students work throughsymbolic solutions.406 c h a p t e r 14 . Oscillations||| A 250 g air-track glider is attached to a spring with springconstant 4.0 N/m. The damping constant due to air resistance is0.015 kg/s. The glider is pulled out 20 cm from equilibrium andreleased. How many oscillations will it make during the time inwhich the amplitude decays to e -1 of its initial value?71. || A 200 g oscillator in a vacuum chamber has a frequency of2.0 Hz. When air is admitted, the oscillation decreases to 60%of its initial amplitude in 50 s. How many oscillations will havebeen completed when the amplitude is 30% of its initial value?72. || Prove that the expression for x ( t ) in Equation 14.55 is a solution to the equation of motion for a damped oscillator, Equation 14.54, if and only if the angular frequency v is given by theexpression in Equation 14.56.73. || A block on a frictionless table is connected as shown in FiguRE P14.73 to two springs having spring constants k1 and k2 .Show that the blocks oscillation frequency is given by70.f = 2f12 + f22where f1 and f2 are the frequencies at which it would oscillate ifattached to spring 1 or spring 2 alone.k1k2k1mFiguRE P14.7374.k2mFiguRE P14.74A block on a frictionless table is connected as shown into two springs having spring constants k1 and k2.Find an expression for the blocks oscillation frequency f interms of the frequencies f1 and f2 at which it would oscillate ifattached to spring 1 or spring 2 alone.||FiguRE P14.7476. A 1.00 kg block is attached to a horizontal spring with springconstant 2500 N/m. The block is at rest on a frictionless surface.A 10 g bullet is fired into the block, in the face opposite thespring, and sticks. What was the bullets speed if the subsequentoscillations have an amplitude of 10.0 cm?77. A spring is standing upright on a table with its bottom end fastened to the table. A block is dropped from a height 3.0 cm abovethe top of the spring. The block sticks to the top end of the springand then oscillates with an amplitude of 10 cm. What is the oscillation frequency?78. The analysis of a simple pendulum assumed that the mass wasa particle, with no size. A realistic pendulum is a small, uniformsphere of mass M and radius R at the end of a massless string, withL being the distance from the pivot to the center of the sphere.a. Find an expression for the period of this pendulum.b. Suppose M = 25 g, R = 1.0 cm, and L = 1.0 m, typical values for a real pendulum. What is the ratio Treal /Tsimple, whereTreal is your expression from part a and Tsimple is the expression derived in this chapter?79. a. A mass m oscillating on a spring has period T. Supposethe mass changes very slightly from m to m + m, wherem V m. Find an expression for T, the small change in theperiod. Your expression should involve T, m, and m but notthe spring constant.b. Suppose the period is 2.000 s and the mass increases by 0.1%.What is the new period?80. FiguRE CP14.80 shows a 200 g uniform rod pivoted at one end.The other end is attached to a horizontal spring. The spring isneither stretched nor compressed when the rod hangs straightdown. What is the rods oscillation period? You can assume thatthe rods angle from vertical is always small.Axle20 cmChallenge Problems75. A block hangs in equilibrium from a vertical spring. When a second identical block is added, the original block sags by 5.0 cm.What is the oscillation frequency of the two-block system?k3.0 N/mFiguRE CP14.80STOP TO THiNK ANSwERSStop to Think 14.1: c. vmax = 2pA/T. Doubling A and T leaves vmaxunchanged.Stop to Think 14.3: c + b + ad. Energy conservation=122 m(vmax ) gives vmax = 1k/m A. k or m has to be increased or decreased by a factor of 4 to have the same effect as increasing or decreasing A by a factor of 2.Stop to Think 14.2: d. Think of circular motion. At 45 , the particleis in the first quadrant (positive x) and moving to the left (negative vx ) .122 kAStop to Think 14.4: c. vx = 0 because the slope of the position graphis zero. The negative value of x shows that the particle is left of theequilibrium position, so the restoring force is to the right.Stop to Think 14.5: c. The period of a pendulum does not dependon its mass.Stop to Think 14.6: Td + TbTc + Ta. The time constant is thetime to decay to 37% of the initial height. The time constant is independent of the initial height....

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