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One Degree of Freedom Systems 2In which we explore many facets of the analysis of mechanical systems in thecontext of a simple one degree of freedom system. . .2.1 DevelopmentThe basic building blocks for models of mechanisms are masses, springs, anddampers (sometimes called dashpots). Automobile shock absorbers and the pistonin screen and storm door closers are common examples of dampers. The motion of amass is governed by the forces applied to it. Figure 1.2 shows a fundamental onedegree of freedom system. The mass can only move in the horizontal ( y) direction.The spring and the damper are in parallel. This is the normal configuration.Rotated 90 counterclockwise this could represent an automobile suspension unit.coil spring and shock absorber (damper) in parallel. I will discuss otherconfigurations later. The diagram shows a mass motion forced by an externalforce. The mass can also be made to move if the support moves. I will addressthis possibility below.2.1.1 An Aside About FrictionFriction is a dissipative mechanism. If you rub your hands together briskly, thefriction between the two will make them warmer. This is an example of slidingfriction, also known as dry friction. Sliding friction is common, but difficultto dealwith analytically. The simplest useful model (which apparently dates back toLeonardo da Vinci) supposes that the force of friction is proportional to the normal

force W between the two sliding objects. The proportionality constant Ê is called thecoefficient of friction. (da Vinci apparently believed it to be a universal constantequal to 1/4.) The laws of dry friction:# Springer International Publishing Switzerland 2015R.F. Gans, Mechanical Systems: A Unified Approach to Vibrations and Controls,DOI 10.1007/978-3-319-08371-1_221. The friction force is proportional to the applied load and independent of thecontact area.. The coefficient of sliding friction is independent of the speed of the motion.are named for Amonton and Coulomb. Dry friction is often called Coulomb friction

.Most models suppose that the coefficient of static friction Ês is larger than thecoefficient of sliding friction Êd. The friction force always opposes the motion.Figure 2.1 shows a block and the forces involved.The friction force will be less than ÊW if f is also less than ÊW. The friction forcecannot induce motion; it can only impede motion. Figure 2.2 shows the frictionforce as a function of the applied force for this simple model. One can see thedropupon the commencement of motion and the subsequent constant friction force. Thefriction force adjusts to balance the applied force until it reaches its staticlimit ÊsW,at which point the block starts to slide. The friction force drops because thecoefficient of sliding friction is smaller than the coefficient of static fricti

on, sothe net force is positive and the block will accelerate.It is relatively easy to measure two friction coefficients for this model usinganinclined plane, as shown in Fig. 2.3. A simple static force balance gives the staticfriction coefficientmg sin ¿ . Êmg cos ¿ ! Ês . tan ¿ d2:1TThe angle ¿ is often called the friction angle. The dynamic friction coefficient can

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be found by observing the fall of the block down the slope. I leave it to the exercisesto show thatÊd . Ês 2sgT2 tan ¿ d2:2Twhere s denotes the distance slid in time T.We cannot incorporate this model of friction into linear equations of motion. Dryfriction does not depend on the motion linearly.Fig. 2.1 Force balance forsliding friction22 2 One Degree of Freedom SystemsThis does not mean that dry friction is unimportant. Dry friction is essential forstopping your car. Disc brakes work by forcing a pair of calipers to grip a diskrigidly attached to the wheel. The disk moves with respect to the calipers as long asthe car is still moving, so the force on the disk depends on the coefficient ofslidingfriction. This friction force exerts a torque on the wheel, which causes it to slowdown. The car slows because of the friction between the tire and the ground.Ultimately (as some tire commercials have pointed out) brakes donft stop your

car: tires do. The patch of tire contacting the ground is stationary with respect to theground (rolling without slipping). It can be approximated by a line contact forthepresent purposes. The appropriate coefficient of friction between the tire and theground is the static coefficient. If you skid the appropriate friction coefficientbecomes the coefficient of sliding friction, which is smaller, and your stoppingpower becomes less. Antilock brakes prevent this from happening, so you can stopmore quickly.The following discussion is a bit simplistic, but it will give a flavor of the design

considerations involved in a braking system. The calipers are driven by a hydrauliccylinder. If the pad area is A, the pressure in the cylinder p and the appropriatecoefficient of friction Ê1, then the force on the disk isFig. 2.2 Friction force as afunction of applied forceFig. 2.3 An inclined planefor measuring frictioncoefficients2.1 Development 23f . 2Ê1Apand the torque on the wheel is then rD times this, where rD denotes the effectiv

eradius of the contact points. The force on the ground will be this torque divided bythe wheel radius rW. This force cannot exceed the static friction force betweentheroad and the tire, which can vary greatly depending on the condition of the road(and the tire). We can write this in equation form asf . 2Ê1Ap Ê2Nwhere N denotes the normal force on the ground and Ê2 the coefficient of statefriction between the tire and the road.

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What happens if we interpose a layer of liquid between the block and theground? The layer acts as a lubricant, and it provides no resistance to the initiationof motion. The resistance to motion parallel to the interface will be the integral ofthe shear stress in the liquid over the surface. (This leads to gaquaplaningh andaccidents in the automotive application.) If the layer is thin, then the flow in theliquid will be laminar, and it can be approximated by plane Couette flow over mostof the interface, the velocity varying linearly between the two surfaces. The stress isequal to the viscosity times the shear rate, which is constant for plane Couette flow.The shear rate is given by the speed divided by the thickness of the liquid layer. Theresistance to motion parallel to the plane is thus proportional to the speed ofmotionand the contact area and inversely proportional to the thickness of the liquid layer. Itis independent of the weight of the block. This is called viscous friction, frictionproportional to the speed of motion. Any system where the resistance to motion is

controlled by a liquid forced to pass through a narrow gap or opening where laminarflow is a good approximation will provide viscous friction to resist motion.Examples include lubricated bearings, shock absorbers, and screen door closers.The viscous friction approximation is convenient and a good approximation inmany cases. It is amenable to linear analysis, which dry friction is not. I will use itmore or less universally in this text; thus the model shown in Fig. 1.2 is anappropriate place to start our study of one degree of freedom problems.2.1.2 The One Degree of Freedom Equation of MotionWe can write a single differential equation governing the motion of the mass shownin Fig. 2.1 considering it to be a free body acted on by the force shown and the

spring and damper forces. Figure 2.4 shows a free body diagram of the mass.The rate of change of momentum is equal to the sum of the forces, giving anequation of motionm.y . f f k f cwhere y denotes the departure from equilibrium, positive to the right in the figure,and fk and fc denote the spring force and the damper force, respectively. If the block24 2 One Degree of Freedom Systemsmoves to the right, the spring will be stretched, and it will exert a force on the massto the left. I will suppose that the force in a linear spring is proportional to the

displacement from equilibrium. The force it exerts on the mass is to the left, and theforce it exerts on the wall is to the right. The equation for the mass becomesm.y . f kdy y0T f cwhere y0 denotes the position of the mass when the spring is not stretched orcompressed. If y>y0 the force on the block is to the left and if y

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The damper works the same way, except that the force is proportional to thespeed of the mass. Thus we can writem.y . f ky c_y ) .y tcm _y tkmy .fm . aThe dimensions of k/m are 1/time2, and the dimensions of c/m are 1/time. We canintroduce a natural frequency, Ön, and a damping ratio, Ä, the latter dimensionless1cm . 2ÄÖn,km . Ö2n , Ö2n .km, Ä .c2mÖn .

c2ffiffiffiffiffiffipkm d2:3Tand rewrite the one degree of freedom equation in standard form, where a.f/mdenotes the applied acceleration.y t 2ÄÖn _y t Ö2ny . a d2:4TFig. 2.4 Free body diagramcorresponding to Fig. 1.21 This is the parameter I introduced in Chap. 1 (Eq. 1.12).2.1 Development 25Equation (2.4) contains all that one needs to understand the dynamics of linear

one degree of freedom systems. The equation governing the motion of a massdriven by a moving support can be put in this form, as I will show below, as can theequation governing the small angle motion of a pendulum. Equation (2.4) isinhomogeneous, that is, y.0 is not a solution. Specific problems are defined byadding initial conditions. A complete problem consists of a dynamical equation likeEq. (2.4) and a set of initial conditions. Wefll have Eq. (2.4) andyd0T.y0, y_d0T.v0 d2:5Twhere y0 and v0 are constants, the position and speed of the mass at t.0.Let us attack the general problem, Eq. (2.4) subject to the conditions given inEq. (2.5), in successively more complicated situations, starting with the unforced

system (a.0). Equation (2.4) becomes homogeneous. Homogeneous equationswith constant coefficients can always be solved in terms of exponential functions.This is an important fact to remember. It applies to all systems of homogeneousdifferential equations with constant coefficients no matter the order of the individualequations or the number of equations. We learned in Chap. 1 that there is aconnection between exponential and trigonometric functions. The solution to thesimplest case, where there is no damping, can be found in terms of trigonometricfunctions alone.

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2.2 Mathematical Analysis of the One Degree of FreedomSystems2.2.1 Undamped Free OscillationsSuppose a.0.Ä. Equation (2.4) reduces to.y t Ö2ny . 0There is no external forcing and no damping. The system is homogeneous. Sincethere is no damping, we expect any nontrivial (y 6.0) solution to persist forever.As noted above a differential equation by itself does not define a problem, butaclass of problems. Its solution, the so-called general solution, has as manyundetermined constants as the order of the equation. Side conditions, as manyas the order of the differential equation, are needed to determine these constants.Here we have one second-order differential equation. It needs two side conditions,and these are usually taken to be the initial conditions.the value of y and itsfirstderivative at the beginning of the motion. (If both are zero, there is no motion.) Isuppose the problem to start from t.0, so that Eq. (2.5) defines the initialconditions26 2 One Degree of Freedom Systems

yd0T.y0, y_d0T.v0The general solution of the differential equation can be written in terms of sinesand cosinesy . A cosdÖntTtB sindÖntT d2:6TNote that if we seek exponential solutions directly, y.A exp(st), the differentialequation reduces tos2AexpdstTtÖ2nAexpdstT.0 . s2 t Ö2n AexpdstT

so that we have a nontrivial solution if s2. Ön2 or s. jÖn. We can use theconnection between the exponential and the trigonometric functions given inChap. 1 to convert the general exponential solution to the form of Eq. (2.6). Thisis only possible when s is purely imaginary. There is a similar transformation forcomplex values of s, which we will need when we have damping.The solution given in Eq. (2.6) can also be expressed asy . C sindÖnt t .T d2:7Twhere . denotes a phase angle. You can verify either formula by direct substitutioninto the differential equation. To convert the form of Eq. (2.7) to that of Eq.

(2.6),expand the sine using the usual multiple angle formulassind¿ t ÀT.sin ¿ cos À t cos ¿ sin Àcosd¿ t ÀT.cos ¿ cos À sin ¿ sin Àto obtainy . Csin . cosdÖntTtC cos . sindÖntTso thatA . C sin ., B . C cos . d2:8TTo convert from the form of Eq. (2.6) to that of Eq. (2.7), we can invert theprocess. We find directly from Eq. (2.8)

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!2vuutd2:11TEquation (2.10a) can also be written in terms of exponential functions using thetrigonometric-exponential correspondences given in Chap. 1.y . x012ejÖnt t e jÖnt tv0Ön12jejÖnt e jÖnt .12x0 jv0Ön ejÖnt t1

2x0 t jv0Ön e jÖnt d2:10bT28 2 One Degree of Freedom SystemsThe two terms on the right are complex conjugates, so their sum is real, as it mustbe. We can solve differential equation in terms of complex exponentials and stillarrive at physically meaningful real solutions. Wefll see this shortly when we add

dissipation to the homogeneous problem, but first letfs look at a couple of examples.Example 2.1 Response to an Impulse This simple one degree of freedom systemseems a most artificial picture, but we can use it to examine the response of asimplesystem to an impulsive load. If we hit the system with a hammer some of itsmomentum will be transferred to the mass. In fact, if the hammer rebounds, thenmore than its momentum will be transferred. The hammer stops in a perfectlyelastic collision, and all of the initial momentum of the hammer will be transferredto the mass. Suppose a 2 kg sledgehammer moving at 5 m/s collides elastically witha 10 kg mass attached to a 98,000 N/m spring. The momentum transferred will be

2 5.10 kg m/s. The velocity of the mass will jump ginstantaneouslyh to 1 m/s,before the mass has a chance to move, so we can write our initial conditions asv0.1 and y0.0. The natural frequency of the system is ã9,800.99 rad/s (.15.8 Hz). The response of the system is theny .199sind99tTmThis is a harmonic response with an amplitude of about one cm. The frequency isa little below the threshold of human hearing.

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Figure 2.5 shows a plot of the response over four periods. The amplitude is in cmand the horizontal scale in seconds. (In a real physical system there will be somedamping, and the motion will decay. I will discuss this below.).1.0.0.50.50.05 0.10 0.15 0.20 0.251.0Fig. 2.5 Response of a one degree of freedom to an impulsive load (see text)2.2 Mathematical Analysis of the One Degree of Freedom Systems 29Example 2.2 The Simple Pendulum The simple pendulum is shown in Fig. 1.1.I denote the angle between the pendulum rod and the vertical by Æ. This angle iszero when the pendulum hangs straight down and reckoned positive in the counterclockwisedirection (Æ.Î/2 when the pendulum is extended horizontally to theright from its pivot point). The pendulum is confined to the plane. Let y denote thehorizontal direction, positive to the right, and z denote the vertical, positive upfollowing the convention I introduced in Chap. 1. Neglect the mass of the rod. Themass m, sometimes called the bob, is acted upon by two external forces: gravity

inthe z direction and a tension in the rod, directed parallel to the rod and in theupward direction, countering gravity. When the rod is vertical these two forcescancel and the pendulum is in static equilibrium.We can write two equations forthemotion of the mass by drawing a free body diagram, Fig. 2.6.We can resolve the forces in the y and z directions to givem.y . T sin Æm.z . T cos Æ mgThis is not the end of the story, because y and z are related. This is a one degreeof freedom system; all the pendulum can do is swing back and forth along itscircular arc. If we take the origin of the coordinate system to be at the base o

f thependulum, where the rod attaches to the support, then we havey2 t z2 . l2Fig. 2.6 Free body diagramof the bob of a simplependulum30 2 One Degree of Freedom Systemswhere l denotes the length of the rod. This expression can be parameterized in termsof the angle Æy . l sin Æ, z . l cos ÆThe two differential equations becomeml .Æ cos Æ _ Æ2

sin Æ One Degree of Freedom Systems 2In which we explore many facets of the analysis of mechanical systems in thecontext of a simple one degree of freedom system. . .2.1 DevelopmentThe basic building blocks for models of mechanisms are masses, springs, anddampers (sometimes called dashpots). Automobile shock absorbers and the pistonin screen and storm door closers are common examples of dampers. The motion of amass is governed by the forces applied to it. Figure 1.2 shows a fundamental one

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degree of freedom system. The mass can only move in the horizontal ( y) direction.The spring and the damper are in parallel. This is the normal configuration.Rotated 90 counterclockwise this could represent an automobile suspension unit.coil spring and shock absorber (damper) in parallel. I will discuss otherconfigurations later. The diagram shows a mass motion forced by an externalforce. The mass can also be made to move if the support moves. I will addressthis possibility below.2.1.1 An Aside About FrictionFriction is a dissipative mechanism. If you rub your hands together briskly, thefriction between the two will make them warmer. This is an example of slidingfriction, also known as dry friction. Sliding friction is common, but difficultto dealwith analytically. The simplest useful model (which apparently dates back toLeonardo da Vinci) supposes that the force of friction is proportional to the normalforce W between the two sliding objects. The proportionality constant Ê is called thecoefficient of friction. (da Vinci apparently believed it to be a universal constantequal to 1/4.) The laws of dry friction:# Springer International Publishing Switzerland 2015R.F. Gans, Mechanical Systems: A Unified Approach to Vibrations and Controls,DOI 10.1007/978-3-319-08371-1_2

21. The friction force is proportional to the applied load and independent of thecontact area.. The coefficient of sliding friction is independent of the speed of the motion.are named for Amonton and Coulomb. Dry friction is often called Coulomb friction.Most models suppose that the coefficient of static friction Ês is larger than thecoefficient of sliding friction Êd. The friction force always opposes the motion.Figure 2.1 shows a block and the forces involved.The friction force will be less than ÊW if f is also less than ÊW. The friction forcecannot induce motion; it can only impede motion. Figure 2.2 shows the frictionforce as a function of the applied force for this simple model. One can see thedrop

upon the commencement of motion and the subsequent constant friction force. Thefriction force adjusts to balance the applied force until it reaches its staticlimit ÊsW,at which point the block starts to slide. The friction force drops because thecoefficient of sliding friction is smaller than the coefficient of static friction, sothe net force is positive and the block will accelerate.It is relatively easy to measure two friction coefficients for this model usinganinclined plane, as shown in Fig. 2.3. A simple static force balance gives the staticfriction coefficientmg sin ¿ . Êmg cos ¿ ! Ês . tan ¿ d2:1T

The angle ¿ is often called the friction angle. The dynamic friction coefficient canbe found by observing the fall of the block down the slope. I leave it to the exercisesto show thatÊd . Ês 2sgT2 tan ¿ d2:2Twhere s denotes the distance slid in time T.We cannot incorporate this model of friction into linear equations of motion. Dr

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yfriction does not depend on the motion linearly.Fig. 2.1 Force balance forsliding friction22 2 One Degree of Freedom SystemsThis does not mean that dry friction is unimportant. Dry friction is essential forstopping your car. Disc brakes work by forcing a pair of calipers to grip a diskrigidly attached to the wheel. The disk moves with respect to the calipers as long asthe car is still moving, so the force on the disk depends on the coefficient ofslidingfriction. This friction force exerts a torque on the wheel, which causes it to slowdown. The car slows because of the friction between the tire and the ground.Ultimately (as some tire commercials have pointed out) brakes donft stop yourcar: tires do. The patch of tire contacting the ground is stationary with respect to theground (rolling without slipping). It can be approximated by a line contact forthepresent purposes. The appropriate coefficient of friction between the tire and theground is the static coefficient. If you skid the appropriate friction coefficient

becomes the coefficient of sliding friction, which is smaller, and your stoppingpower becomes less. Antilock brakes prevent this from happening, so you can stopmore quickly.The following discussion is a bit simplistic, but it will give a flavor of the designconsiderations involved in a braking system. The calipers are driven by a hydrauliccylinder. If the pad area is A, the pressure in the cylinder p and the appropriatecoefficient of friction Ê1, then the force on the disk isFig. 2.2 Friction force as afunction of applied forceFig. 2.3 An inclined plane

for measuring frictioncoefficients2.1 Development 23f . 2Ê1Apand the torque on the wheel is then rD times this, where rD denotes the effectiveradius of the contact points. The force on the ground will be this torque divided bythe wheel radius rW. This force cannot exceed the static friction force betweentheroad and the tire, which can vary greatly depending on the condition of the road(and the tire). We can write this in equation form asf . 2Ê1Ap Ê2N

where N denotes the normal force on the ground and Ê2 the coefficient of statefriction between the tire and the road.What happens if we interpose a layer of liquid between the block and theground? The layer acts as a lubricant, and it provides no resistance to the initiationof motion. The resistance to motion parallel to the interface will be the integral ofthe shear stress in the liquid over the surface. (This leads to gaquaplaningh andaccidents in the automotive application.) If the layer is thin, then the flow in the

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liquid will be laminar, and it can be approximated by plane Couette flow over mostof the interface, the velocity varying linearly between the two surfaces. The stress isequal to the viscosity times the shear rate, which is constant for plane Couette flow.The shear rate is given by the speed divided by the thickness of the liquid layer. Theresistance to motion parallel to the plane is thus proportional to the speed ofmotionand the contact area and inversely proportional to the thickness of the liquid layer. Itis independent of the weight of the block. This is called viscous friction, frictionproportional to the speed of motion. Any system where the resistance to motion iscontrolled by a liquid forced to pass through a narrow gap or opening where laminarflow is a good approximation will provide viscous friction to resist motion.Examples include lubricated bearings, shock absorbers, and screen door closers.The viscous friction approximation is convenient and a good approximation inmany cases. It is amenable to linear analysis, which dry friction is not. I will use itmore or less universally in this text; thus the model shown in Fig. 1.2 is an

appropriate place to start our study of one degree of freedom problems.2.1.2 The One Degree of Freedom Equation of MotionWe can write a single differential equation governing the motion of the mass shownin Fig. 2.1 considering it to be a free body acted on by the force shown and thespring and damper forces. Figure 2.4 shows a free body diagram of the mass.The rate of change of momentum is equal to the sum of the forces, giving anequation of motionm.y . f f k f cwhere y denotes the departure from equilibrium, positive to the right in the figure,and fk and fc denote the spring force and the damper force, respectively. If the block

24 2 One Degree of Freedom Systemsmoves to the right, the spring will be stretched, and it will exert a force on the massto the left. I will suppose that the force in a linear spring is proportional to thedisplacement from equilibrium. The force it exerts on the mass is to the left, and theforce it exerts on the wall is to the right. The equation for the mass becomesm.y . f kdy y0T f cwhere y0 denotes the position of the mass when the spring is not stretched orcompressed. If y>y0 the force on the block is to the left and if y

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y .fm . aThe dimensions of k/m are 1/time2, and the dimensions of c/m are 1/time. We canintroduce a natural frequency, Ön, and a damping ratio, Ä, the latter dimensionless1cm . 2ÄÖn,km . Ö2n , Ö2n .km, Ä .c2mÖn .c2ffiffiffiffiffiffipkm d2:3Tand rewrite the one degree of freedom equation in standard form, where a.f/mdenotes the applied acceleration.y t 2ÄÖn _y t Ö2ny . a d2:4T

Fig. 2.4 Free body diagramcorresponding to Fig. 1.21 This is the parameter I introduced in Chap. 1 (Eq. 1.12).2.1 Development 25Equation (2.4) contains all that one needs to understand the dynamics of linearone degree of freedom systems. The equation governing the motion of a massdriven by a moving support can be put in this form, as I will show below, as can theequation governing the small angle motion of a pendulum. Equation (2.4) isinhomogeneous, that is, y.0 is not a solution. Specific problems are defined byadding initial conditions. A complete problem consists of a dynamical equation likeEq. (2.4) and a set of initial conditions. Wefll have Eq. (2.4) and

yd0T.y0, y_d0T.v0 d2:5Twhere y0 and v0 are constants, the position and speed of the mass at t.0.Let us attack the general problem, Eq. (2.4) subject to the conditions given inEq. (2.5), in successively more complicated situations, starting with the unforcedsystem (a.0). Equation (2.4) becomes homogeneous. Homogeneous equationswith constant coefficients can always be solved in terms of exponential functions.This is an important fact to remember. It applies to all systems of homogeneousdifferential equations with constant coefficients no matter the order of the individualequations or the number of equations. We learned in Chap. 1 that there is aconnection between exponential and trigonometric functions. The solution to the

simplest case, where there is no damping, can be found in terms of trigonometricfunctions alone.2.2 Mathematical Analysis of the One Degree of FreedomSystems2.2.1 Undamped Free OscillationsSuppose a.0.Ä. Equation (2.4) reduces to.y t Ö2ny . 0There is no external forcing and no damping. The system is homogeneous. Sincethere is no damping, we expect any nontrivial (y 6.0) solution to persist foreve

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r.As noted above a differential equation by itself does not define a problem, butaclass of problems. Its solution, the so-called general solution, has as manyundetermined constants as the order of the equation. Side conditions, as manyas the order of the differential equation, are needed to determine these constants.Here we have one second-order differential equation. It needs two side conditions,and these are usually taken to be the initial conditions.the value of y and itsfirstderivative at the beginning of the motion. (If both are zero, there is no motion.) Isuppose the problem to start from t.0, so that Eq. (2.5) defines the initialconditions26 2 One Degree of Freedom Systemsyd0T.y0, y_d0T.v0The general solution of the differential equation can be written in terms of sinesand cosinesy . A cosdÖntTtB sindÖntT d2:6TNote that if we seek exponential solutions directly, y.A exp(st), the differentialequation reduces to

s2AexpdstTtÖ2nAexpdstT.0 . s2 t Ö2n AexpdstTso that we have a nontrivial solution if s2. Ön2 or s. jÖn. We can use theconnection between the exponential and the trigonometric functions given inChap. 1 to convert the general exponential solution to the form of Eq. (2.6). Thisis only possible when s is purely imaginary. There is a similar transformation forcomplex values of s, which we will need when we have damping.

The solution given in Eq. (2.6) can also be expressed asy . C sindÖnt t .T d2:7Twhere . denotes a phase angle. You can verify either formula by direct substitutioninto the differential equation. To convert the form of Eq. (2.7) to that of Eq.(2.6),expand the sine using the usual multiple angle formulassind¿ t ÀT.sin ¿ cos À t cos ¿ sin Àcosd¿ t ÀT.cos ¿ cos À sin ¿ sin Àto obtainy . Csin . cosdÖntTtC cos . sindÖntTso thatA . C sin ., B . C cos . d2:8T

To convert from the form of Eq. (2.6) to that of Eq. (2.7), we can invert theprocess. We find directly from Eq. (2.8)C2 . A2 t B2, tan . .AB d2:9TOne has to be careful in calculating the phase. The inverse tangent is ambiguous.One can choose the appropriate quadrant by noting that2.2 Mathematical Analysis of the One Degree of Freedom Systems 27sin . .

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AC, cos . .BCThe sine and cosine are both positive in the first quadrant (0

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ejÖnt t e jÖnt tv0Ön12jejÖnt e jÖnt .12x0 jv0Ön ejÖnt t12x0 t jv0Ön e jÖnt d2:10bT28 2 One Degree of Freedom SystemsThe two terms on the right are complex conjugates, so their sum is real, as it m

ustbe. We can solve differential equation in terms of complex exponentials and stillarrive at physically meaningful real solutions. Wefll see this shortly when we adddissipation to the homogeneous problem, but first letfs look at a couple of examples.Example 2.1 Response to an Impulse This simple one degree of freedom systemseems a most artificial picture, but we can use it to examine the response of asimplesystem to an impulsive load. If we hit the system with a hammer some of itsmomentum will be transferred to the mass. In fact, if the hammer rebounds, thenmore than its momentum will be transferred. The hammer stops in a perfectly

elastic collision, and all of the initial momentum of the hammer will be transferredto the mass. Suppose a 2 kg sledgehammer moving at 5 m/s collides elastically witha 10 kg mass attached to a 98,000 N/m spring. The momentum transferred will be2 5.10 kg m/s. The velocity of the mass will jump ginstantaneouslyh to 1 m/s,before the mass has a chance to move, so we can write our initial conditions asv0.1 and y0.0. The natural frequency of the system is ã9,800.99 rad/s (.15.8 Hz). The response of the system is theny .199sind99tTm

This is a harmonic response with an amplitude of about one cm. The frequency isa little below the threshold of human hearing.Figure 2.5 shows a plot of the response over four periods. The amplitude is in cmand the horizontal scale in seconds. (In a real physical system there will be somedamping, and the motion will decay. I will discuss this below.).1.0.0.50.5

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0.05 0.10 0.15 0.20 0.251.0Fig. 2.5 Response of a one degree of freedom to an impulsive load (see text)2.2 Mathematical Analysis of the One Degree of Freedom Systems 29Example 2.2 The Simple Pendulum The simple pendulum is shown in Fig. 1.1.I denote the angle between the pendulum rod and the vertical by Æ. This angle iszero when the pendulum hangs straight down and reckoned positive in the counterclockwisedirection (Æ.Î/2 when the pendulum is extended horizontally to theright from its pivot point). The pendulum is confined to the plane. Let y denote thehorizontal direction, positive to the right, and z denote the vertical, positive upfollowing the convention I introduced in Chap. 1. Neglect the mass of the rod. Themass m, sometimes called the bob, is acted upon by two external forces: gravityinthe z direction and a tension in the rod, directed parallel to the rod and in theupward direction, countering gravity. When the rod is vertical these two forcescancel and the pendulum is in static equilibrium.We can write two equations forthemotion of the mass by drawing a free body diagram, Fig. 2.6.We can resolve the forces in the y and z directions to givem.y . T sin Æ

m.z . T cos Æ mgThis is not the end of the story, because y and z are related. This is a one degreeof freedom system; all the pendulum can do is swing back and forth along itscircular arc. If we take the origin of the coordinate system to be at the base of thependulum, where the rod attaches to the support, then we havey2 t z2 . l2Fig. 2.6 Free body diagramof the bob of a simplependulum30 2 One Degree of Freedom Systemswhere l denotes the length of the rod. This expression can be parameterized in t

ermsof the angle Æy . l sin Æ, z . l cos ÆThe two differential equations becomeml .Æ cos Æ _ Æ2sin Æ

Creo Parametric 3.0 Basics - Part 5 (Sheet Metal Design and Surface Design)CreoParametric 3.0 Basics - Part 5 (Sheet Metal Design and Surface Design)Creo Parametric 3.0 Basics - Part 5 (Sheet Metal Design and Surface Design)Creo Parametric 3.0 Basics - Part 5 (Sheet Metal Design and Surface Design)Creo Parametric 3.0Basics - Part 5 (Sheet Metal Design and Surface Design)Creo Parametric 3.0 Basic

s - Part 5 (Sheet Metal Design and Surface Design)Creo Parametric 3.0 Basics - Part 5 (Sheet Metal Design and Surface Design)Creo Parametric 3.0 Basics - Part 5 (Sheet Metal Design and Surface Design)Creo Parametric 3.0 Basics - Part 5 (Sheet Metal Design and Surface Design)Creo Parametric 3.0 Basics - Part 5 (Sheet Metal Design and Surface Design)Creo Parametric 3.0 Basics - Part 5 (Sheet Metal Design and Surface Design)Creo Parametric 3.0 Basics - Part 5 (Sheet Metal Design and Surface Design)Creo Parametric 3.0 Basics - Part 5 (Sheet Metal Design andSurface Design)Creo Parametric 3.0 Basics - Part 5 (Sheet Metal Design and Surface Design)Creo Parametric 3.0 Basics - Part 5 (Sheet Metal Design and Surface Design)Creo Parametric 3.0 Basics - Part 5 (Sheet Metal Design and Surface Design)

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Creo Parametric 3.0 Basics - Part 5 (Sheet Metal Design and Surface Design)CreoParametric 3.0 Basics - Part 5 (Sheet Metal Design and Surface Design)Creo Parametric 3.0 Basics - Part 5 (Sheet Metal Design and Surface Design)Creo Parametric 3.0 Basics - Part 5 (Sheet Metal Design and Surface Design)Creo Parametric 3.0Basics - Part 5 (Sheet Metal Design and Surface Design)Creo Parametric 3.0 Basics - Part 5 (Sheet Metal Design and Surface Design)Creo Parametric 3.0 Basics - Part 5 (Sheet Metal Design and Surface Design)Creo Parametric 3.0 Basics - Part 5 (Sheet Metal Design and Surface Design)Creo Parametric 3.0 Basics - Part 5 (Sheet Metal Design and Surface Design)Creo Parametric 3.0 Basics - Part 5 (Sheet Metal Design and Surface Design)Creo Parametric 3.0 Basics - Part 5 (Sheet Metal Design and Surface Design)Creo Parametric 3.0 Basics - Part 5 (Sheet Metal Design and Surface Design)Creo Parametric 3.0 Basics - Part 5 (Sheet Metal Design andSurface Design)Creo Parametric 3.0 Basics - Part 5 (Sheet Metal Design and Surface Design)Creo Parametric 3.0 Basics - Part 5 (Sheet Metal Design and Surface Design)Creo Parametric 3.0 Basics - Part 5 (Sheet Metal Design and Surface Design)Creo Parametric 3.0 Basics - Part 5 (Sheet Metal Design and Surface Design)CreoParametric 3.0 Basics - Part 5 (Sheet Metal Design and Surface Design)Creo Parametric 3.0 Basics - Part 5 (Sheet Metal Design and Surface Design)Creo Parametric 3.0 Basics - Part 5 (Sheet Metal Design and Surface Design)Creo Parametric 3.0Basics - Part 5 (Sheet Metal Design and Surface Design)Creo Parametric 3.0 Basics - Part 5 (Sheet Metal Design and Surface Design)Creo Parametric 3.0 Basics - Part 5 (Sheet Metal Design and Surface Design)Creo Parametric 3.0 Basics - Part 5 (Sheet Metal Design and Surface Design)Creo Parametric 3.0 Basics - Part 5 (Sheet Metal Design and Surface Design)Creo Parametric 3.0 Basics - Part 5 (Sheet Me

tal Design and Surface Design)Creo Parametric 3.0 Basics - Part 5 (Sheet Metal Design and Surface Design)Creo Parametric 3.0 Basics - Part 5 (Sheet Metal Design and Surface Design)Creo Parametric 3.0 Basics - Part 5 (Sheet Metal Design andSurface Design)Creo Parametric 3.0 Basics - Part 5 (Sheet Metal Design and Surface Design)Creo Parametric 3.0 Basics - Part 5 (Sheet Metal Design and Surface Design)Creo Parametric 3.0 Basics - Part 5 (Sheet Metal Design and Surface Design)Creo Parametric 3.0 Basics - Part 5 (Sheet Metal Design and Surface Design)CreoParametric 3.0 Basics - Part 5 (Sheet Metal Design and Surface Design)Creo Parametric 3.0 Basics - Part 5 (Sheet Metal Design and Surface Design)Creo Parametric 3.0 Basics - Part 5 (Sheet Metal Design and Surface Design)Creo Parametric 3.0Basics - Part 5 (Sheet Metal Design and Surface Design)Creo Parametric 3.0 Basics - Part 5 (Sheet Metal Design and Surface Design)Creo Parametric 3.0 Basics - Part 5 (Sheet Metal Design and Surface Design)Creo Parametric 3.0 Basics - Part 5

(Sheet Metal Design and Surface Design)Creo Parametric 3.0 Basics - Part 5 (Sheet Metal Design and Surface Design)Creo Parametric 3.0 Basics - Part 5 (Sheet Metal Design and Surface Design)Creo Parametric 3.0 Basics - Part 5 (Sheet Metal Design and Surface Design)Creo Parametric 3.0 Basics - Part 5 (Sheet Metal Design and Surface Design)Creo Parametric 3.0 Basics - Part 5 (Sheet Metal Design andSurface Design)Creo Parametric 3.0 Basics - Part 5 (Sheet Metal Design and Surface Design)Creo Parametric 3.0 Basics - Part 5 (Sheet Metal Design and Surface Design)Creo Parametric 3.0 Basics - Part 5 (Sheet Metal Design and Surface Design)Creo Parametric 3.0 Basics - Part 5 (Sheet Metal Design and Surface Design)CreoParametric 3.0 Basics - Part 5 (Sheet Metal Design and Surface Design)Creo Parametric 3.0 Basics - Part 5 (Sheet Metal Design and Surface Design)Creo Parametric 3.0 Basics - Part 5 (Sheet Metal Design and Surface Design)Creo Parametric 3.0Basics - Part 5 (Sheet Metal Design and Surface Design)Creo Parametric 3.0 Basic


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s - Part 5 (Sheet Metal Design and Surface Design)Creo Parametric 3.0 Basics - Part 5 (Sheet Metal Design and Surface Design)Creo Parametric 3.0 Basics - Part 5 (Sheet Metal Design and Surface Design)Creo Parametric 3.0 Basics - Part 5 (Sheet Metal Design and Surface Design)Creo Parametric 3.0 Basics - Part 5 (Sheet Metal Design and Surface Design)

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