LINEAR MODELS: INITIAL-VALUE PROBLEMS

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1 5 LINEAR MODELS: INITIAL-VALUE PROBLEMS 9 5 LINEAR MODELS: INITIAL-VALUE PROBLEMS REVIEW MATERIAL Secions 4, 4, and 44 Problems 9 6 in Eercises 4 Problems 7 6 in Eercises 44 INTRODUCTION In his secion we are going o consider several linear dynamical sysems in which each mahemaical model is a second-order differenial equaion wih consan coefficien along wih iniial condiions specified a a ime ha we shall ake o be 0: a d y dy b d d cy g(), y(0) y 0, y (0) y Recall ha he funcion g is he inpu, driving funcion, or forcing funcion of he sysem A soluion y() of he differenial equaion on an inerval I conaining 0 ha saisfies he iniial condiions is called he oupu or response of he sysem 5 SPRING/MASS SYSTEMS: FREE UNDAMPED MOTION l unsreched m equilibrium posiion mg ks = 0 (a) (b) (c) FIGURE 5 l s m moion Spring/mass sysem l + s Hooke s Law Suppose ha a fleible spring is suspended verically from a rigid suppor and hen a mass m is aached o is free end The amoun of srech, or elongaion, of he spring will of course depend on he mass; masses wih differen weighs srech he spring by differing amouns By Hooke s law he spring iself eers a resoring force F opposie o he direcion of elongaion and proporional o he amoun of elongaion s Simply saed, F ks, where k is a consan of proporionaliy called he spring consan The spring is essenially characerized by he number k For eample, if a mass weighing 0 pounds sreches a spring foo, hen 0 k implies k 0 lb/f Necessarily hen, a mass weighing, say, 8 pounds sreches he same spring only foo 5 Newon s Second Law Afer a mass m is aached o a spring, i sreches he spring by an amoun s and aains a posiion of equilibrium a which is weigh W is balanced by he resoring force ks Recall ha weigh is defined by W mg, where mass is measured in slugs, kilograms, or grams and g f/s, 98 m/s, or 980 cm/s, respecively As indicaed in Figure 5(b), he condiion of equilibrium is mg ks or mg ks 0 If he mass is displaced by an amoun from is equilibrium posiion, he resoring force of he spring is hen k( s) Assuming ha here are no rearding forces acing on he sysem and assuming ha he mass vibraes free of oher eernal forces free moion we can equae Newon s second law wih he ne, or resulan, force of he resoring force and he weigh: = 0 < 0 > 0 m d k(s ) mg k mg ks k d zero () m FIGURE 5 Direcion below he equilibrium posiion is posiive The negaive sign in () indicaes ha he resoring force of he spring acs opposie o he direcion of moion Furhermore, we adop he convenion ha displacemens measured below he equilibrium posiion 0 are posiive See Figure 5 Copyrigh 0 Cengage Learning All Righs Reserved May no be copied, scanned, or duplicaed, in whole or in par Due o elecronic righs, some hird pary conen may be suppressed from he ebook and/or echaper(s) Ediorial review has deemed ha any suppressed conen does no maerially affec he overall learning eperience Cengage Learning reserves he righ o remove addiional conen a any ime if subsequen righs resricions require i

2 94 CHAPTER 5 MODELING WITH HIGHER-ORDER DIFFERENTIAL EQUATIONS DE of Free Undamped Moion By dividing () by he mass m, we obain he second-order differenial equaion d d (k m) 0, or d, () d 0 where v k m Equaion () is said o describe simple harmonic moion or free undamped moion Two obvious iniial condiions associaed wih () are (0) 0 and (0), he iniial displacemen and iniial velociy of he mass, respecively For eample, if 0 0, 0, he mass sars from a poin below he equilibrium posiion wih an impared upward velociy When (0) 0, he mass is said o be released from res For eample, if 0 0, 0, he mass is released from res from a poin 0 unis above he equilibrium posiion Equaion of Moion To solve equaion (), we noe ha he soluions of is auiliary equaion m v 0 are he comple numbers m vi, m vi Thus from (8) of Secion 4 we find he general soluion of () o b () c cos c sin () The period of moion described by () is T p v The number T represens he ime (measured in seconds) i akes he mass o eecue one cycle of moion A cycle is one complee oscillaion of he mass, ha is, he mass m moving from, say, he lowes poin below he equilibrium posiion o he poin highes above he equilibrium posiion and hen back o he lowes poin From a graphical viewpoin T p v seconds is he lengh of he ime inerval beween wo successive maima (or minima) of () Keep in mind ha a maimum of () is a posiive displacemen corresponding o he mass aaining is greaes disance below he equilibrium posiion, whereas a minimum of () is negaive displacemen corresponding o he mass aaining is greaes heigh above he equilibrium posiion We refer o eiher case as an ereme displacemen of he mass The frequency of moion is f T v p and is he number of cycles compleed each second For eample, if () cos p 4 sin p, hen he period is T p p s, and he frequency is f cycles/s From a graphical viewpoin he graph of () repeas every second, ha is, ( ) (), and cycles of he graph are compleed each second (or, equivalenly, hree cycles of he graph are compleed every seconds) The number (measured in radians per second) is called he circular frequency of he sysem Depending on which e you read, boh f v p and v are also referred o as he naural frequency of he sysem Finally, when he iniial condiions are used o deermine he consans c and c in (), we say ha he resuling paricular soluion or response is he equaion of moion k>m EXAMPLE Free Undamped Moion A mass weighing pounds sreches a spring 6 inches A 0 he mass is released 4 from a poin 8 inches below he equilibrium posiion wih an upward velociy of f/s Deermine he equaion of moion SOLUTION Because we are using he engineering sysem of unis, he measuremens given in erms of inches mus be convered ino fee: 6 in ; 8 in f f In addiion, we mus conver he unis of weigh given in pounds ino unis of mass From m W g we have m slug Also, from Hooke s law, k 6 implies ha he spring consan is k 4 lb/f Hence () gives d 6 d 4 or d 64 0 d The iniial displacemen and iniial velociy are (0), (0) 4, where he negaive sign in he las condiion is a consequence of he fac ha he mass is given an iniial velociy in he negaive, or upward, direcion Copyrigh 0 Cengage Learning All Righs Reserved May no be copied, scanned, or duplicaed, in whole or in par Due o elecronic righs, some hird pary conen may be suppressed from he ebook and/or echaper(s) Ediorial review has deemed ha any suppressed conen does no maerially affec he overall learning eperience Cengage Learning reserves he righ o remove addiional conen a any ime if subsequen righs resricions require i

3 5 LINEAR MODELS: INITIAL-VALUE PROBLEMS 95 Now v 64 or v 8, so he general soluion of he differenial equaion is () c cos 8 c sin 8 (4) Applying he iniial condiions o () and () gives c and c 6 Thus he equaion of moion is () (5) cos 8 sin 8 6 Alernaive Forms of () When c 0 and c 0, he acual ampliude A of free vibraions is no obvious from inspecion of equaion () For eample, alhough he mass in Eample is iniially displaced foo beyond he equilibrium posiion, he ampliude of vibraions is a number larger han Hence i is ofen convenien o conver a soluion of form () o he simpler form () A sin( ), (6) c + c c where A c c and f is a phase angle defined b sin c A A c (7) cos c an c To verify his, we epand (6) by he addiion formula for he sine funcion: A sin cos cos sin ( sin )cos ( cos )sin (8) I follows from Figure 5 ha if f is defined b φ c FIGURE 5 A relaionship beween c 0, c 0 and phase angle f sin hen (8) becomes c, c c c A, cos c c c c A A c A cos A c A sin c cos c sin () EXAMPLE Alernaive Form of Soluion (5) In view of he foregoing discussion we can wrie soluion (5) in he alernaive form () A sin(8 f) Compuaion of he ampliude is sraighforward, A ( ) ( 6) f, bu some care should be eercised in compuing he phase angle f defined by (7) Wih c and c we find an f 4, and a calculaor hen gives an 6 ( 4) 6 rad This is no he phase angle, since an ( 4) is locaed in he fourh quadran and herefore conradics he fac ha sin f 0 and cos f 0 because c 0 and c 0 Hence we mus ake f o be he second-quadran angle f p ( 6) 86 rad Thus (5) is he same as () 7 6 The period of his funcion is T p 8 p 4 s sin(8 86) (9) You should be aware ha some insrucors in science and engineering prefer ha () be epressed as a shifed cosine funcion () A cos(v f), (6 ) Copyrigh 0 Cengage Learning All Righs Reserved May no be copied, scanned, or duplicaed, in whole or in par Due o elecronic righs, some hird pary conen may be suppressed from he ebook and/or echaper(s) Ediorial review has deemed ha any suppressed conen does no maerially affec he overall learning eperience Cengage Learning reserves he righ o remove addiional conen a any ime if subsequen righs resricions require i

4 96 CHAPTER 5 MODELING WITH HIGHER-ORDER DIFFERENTIAL EQUATIONS where A c c In his case he radian measured angle f is defined in slighly differen manner han in (7): (7 ) For eample, in Eample wih c and c (7 ) indicaes ha an f 6, 4 Because sin f 0 and cos f 0 he angle f lies in he fourh quadran and so rounded o hree decimal places f an ( 4 ) 045 rad From (6 ) we obain a second alernaive form of soluion (5): () 7 6 sin c A A cos c an cos(8 ( 045)) or () 7 6 cos(8 045) Graphical Inerpreaion Figure 54(a) illusraes he mass in Eample going hrough approimaely wo complee cycles of moion Reading lef o righ, he firs five posiions (marked wih black dos) correspond o he iniial posiion of he mass below he equilibrium posiion ( ), he mass passing hrough he equilibrium posiion for he firs ime heading upward ( 0), he mass a is ereme displacemen above he equilibrium posiion ( 7 6), he mass a he equilibrium posiion for he second ime heading downward ( 0), and he mass a is ereme displacemen below he equilibrium posiion ( 7 6) The black dos on he graph of (9), given in Figure 54(b), also agree wih he five posiions jus given Noe, however, ha in Figure 54(b) he posiive direcion in he -plane is he usual upward c c negaive = 0 posiive = 6 7 = 0 = 0 = = 6 7 (a) posiive = 0 (0, ) ampliude A = 6 7 negaive π 4 period (b) FIGURE 54 Simple harmonic moion Copyrigh 0 Cengage Learning All Righs Reserved May no be copied, scanned, or duplicaed, in whole or in par Due o elecronic righs, some hird pary conen may be suppressed from he ebook and/or echaper(s) Ediorial review has deemed ha any suppressed conen does no maerially affec he overall learning eperience Cengage Learning reserves he righ o remove addiional conen a any ime if subsequen righs resricions require i

5 5 LINEAR MODELS: INITIAL-VALUE PROBLEMS 97 direcion and so is opposie o he posiive direcion indicaed in Figure 54(a) Hence he solid blue graph represening he moion of he mass in Figure 54(b) is he reflecion hrough he -ais of he blue dashed curve in Figure 54(a) Form (6) is very useful because i is easy o find values of ime for which he graph of () crosses he posiive -ais (he line 0) We observe ha sin(v f) 0 when v f np, where n is a nonnegaive ineger Sysems wih Variable Spring Consans In he model discussed above we assumed an ideal world a world in which he physical characerisics of he spring do no change over ime In he nonideal world, however, i seems reasonable o epec ha when a spring/mass sysem is in moion for a long period, he spring will weaken; in oher words, he spring consan will vary or, more specificall, decay wih ime In one model for he aging spring he spring consan k in () is replaced by he decreasing funcion K() ke a, k 0, a 0 The linear differenial equaion m ke a 0 canno be solved by he mehods ha were considered in Chaper 4 Neverheless, we can obain wo linearly independen soluions using he mehods in Chaper 6 See Problem 5 in Eercises 5, Eample 4 in Secion 64, and Problems and 9 in Eercises 64 When a spring/mass sysem is subjeced o an environmen in which he emperaure is rapidly decreasing, i migh make sense o replace he consan k wih K() k, k 0, a funcion ha increases wih ime The resuling model, m k 0, is a form of Airy s differenial equaion Like he equaion for an aging spring, Airy s equaion can be solved by he mehods of Chaper 6 See Problem 6 in Eercises 5, Eample 5 in Secion 6, and Problems 4, 5, and 40 in Eercise 64 5 SPRING/MASS SYSTEMS: FREE DAMPED MOTION m (a) The concep of free harmonic moion is somewha unrealisic, since he moion described by equaion () assumes ha here are no rearding forces acing on he moving mass Unless he mass is suspended in a perfec vacuum, here will be a leas a resising force due o he surrounding medium As Figure 55 shows, he mass could be suspended in a viscous medium or conneced o a dashpo damping device DE of Free Damped Moion In he sudy of mechanics, damping forces acing on a body are considered o be proporional o a power of he insananeous velociy In paricular, we shall assume hroughou he subsequen discussion ha his force is given by a consan muliple of d d When no oher eernal forces are impressed on he sysem, i follows from Newon s second law ha m d, (0) d k d d m (b) FIGURE 55 Damping devices where b is a posiive damping consan and he negaive sign is a consequence of he fac ha he damping force acs in a direcion opposie o he moion Dividing (0) by he mass m, we find ha he differenial equaion of free damped moion is d d b d m d k m 0 d or, () d d d 0 where () m, k m Copyrigh 0 Cengage Learning All Righs Reserved May no be copied, scanned, or duplicaed, in whole or in par Due o elecronic righs, some hird pary conen may be suppressed from he ebook and/or echaper(s) Ediorial review has deemed ha any suppressed conen does no maerially affec he overall learning eperience Cengage Learning reserves he righ o remove addiional conen a any ime if subsequen righs resricions require i

6 98 CHAPTER 5 MODELING WITH HIGHER-ORDER DIFFERENTIAL EQUATIONS The symbol l is used only for algebraic convenience because he auiliary equaion is m lm v 0, and he corresponding roos are hen m, m We can now disinguish hree possible cases depending on he algebraic sign of l v Since each soluion conains he damping facor e l, l 0, he displacemens of he mass become negligible as ime increases Case I: L V 0 In his siuaion he sysem is said o be overdamped because he damping coefficien b is large when compared o he spring consan k The corresponding soluion of () is () c e m c e m or () e (c e c e ) () FIGURE 56 Moion of an overdamped sysem This equaion represens a smooh and nonoscillaory moion Figure 56 shows wo possible graphs of () Case II: L V 0 The sysem is said o be criically damped because any sligh decrease in he damping force would resul in oscillaory moion The general soluion of () is () c e m c e m or () e (c c ) (4) FIGURE 57 damped sysem Moion of a criically undamped underdamped Some graphs of ypical moion are given in Figure 57 Noice ha he moion is quie similar o ha of an overdamped sysem I is also apparen from (4) ha he mass can pass hrough he equilibrium posiion a mos one ime Case III: L V 0 In his case he sysem is said o be underdamped, since he damping coefficien is small in comparison o he spring consan The roos m and m are now comple: m i, m i FIGURE 58 Moion of an underdamped sysem Thus he general soluion of equaion () is () e (c cos c sin (5) As indicaed in Figure 58, he moion described by (5) is oscillaory; bu because of he coefficien e l, he ampliudes of vibraion : 0 as : ) EXAMPLE Overdamped Moion I is readily verified ha he soluion of he iniial-value proble d d 5 4 0, (0), (0) d d is () 5 (6) e e 4 The problem can be inerpreed as represening he overdamped moion of a mass on a spring The mass is iniially released from a posiion uni below he equilibrium posiion wih a downward velociy of f/s To graph (), we find he value of for which he funcion has an eremum ha is, he value of ime for which he firs derivaive (velociy) is zero Differeniaing (6) gives () 5 e 8 e 4, so () 0 implies ha Copyrigh 0 Cengage Learning All Righs Reserved May no be copied, scanned, or duplicaed, in whole or in par Due o elecronic righs, some hird pary conen may be suppressed from he ebook and/or echaper(s) Ediorial review has deemed ha any suppressed conen does no maerially affec he overall learning eperience Cengage Learning reserves he righ o remove addiional conen a any ime if subsequen righs resricions require i

7 5 LINEAR MODELS: INITIAL-VALUE PROBLEMS 99 5 = e e 4 (a) () e 8 or ln I follows from he firs derivaive es, as well as our physical inuiion, ha (057) 069 f is acually a maimum In oher words, he mass aains an ereme displacemen of 069 fee below he equilibrium posiion We should also check o see wheher he graph crosses he -ais ha is, wheher he mass passes hrough he equilibrium posiion This canno happen in his insance because he equaion () 0, or e, has he physically irrelevan soluion ln The graph of (), along wih some oher perinen daa, is given in Figure 59 FIGURE 59 Eample = (b) Overdamped sysem in 4 maimum heigh above equilibrium posiion FIGURE 50 Criically damped sysem in Eample 4 EXAMPLE 4 Criically Damped Moion A mass weighing 8 pounds sreches a spring fee Assuming ha a damping force numerically equal o imes he insananeous velociy acs on he sysem, deermine he equaion of moion if he mass is iniially released from he equilibrium posiion wih an upward velociy of f/s SOLUTION From Hooke s law we see ha 8 k() gives k 4 lb/f and ha W mg gives m 8 4 slug The differenial equaion of moion is hen 4 d d 4 d (7) d or d d d d The auiliary equaion for (7) is m 8m 6 (m 4) 0, so m m 4 Hence he sysem is criically damped, and () c e 4 c e 4 (8) Applying he iniial condiions (0) 0 and (0), we find, in urn, ha c 0 and c Thus he equaion of moion is () e 4 (9) To graph (), we proceed as in Eample From () e 4 ( 4) we see ha () 0 when The corresponding ereme displacemen is ( 4) ( 4)e f As shown in Figure 50, we inerpre his value o mean ha he mass reaches a maimum heigh of 076 foo above he equilibrium posiion EXAMPLE 5 Underdamped Moion A mass weighing 6 pounds is aached o a 5-foo-long spring A equilibrium he spring measures 8 fee If he mass is iniially released from res a a poin fee above he equilibrium posiion, find he displacemens () if i is furher known ha he surrounding medium offers a resisance numerically equal o he insananeous velociy SOLUTION The elongaion of he spring afer he mass is aached is 8 5 f, so i follows from Hooke s law ha 6 k() or k 5 lb/f In addiion, m 6 slug, so he differenial equaion is given by d d 5 d d or Proceeding, we find ha he roos of m m 0 0 are m i and m i, which hen implies ha he sysem is underdamped, and () e (c cos c sin ) d d 0 0 (0) d d () Copyrigh 0 Cengage Learning All Righs Reserved May no be copied, scanned, or duplicaed, in whole or in par Due o elecronic righs, some hird pary conen may be suppressed from he ebook and/or echaper(s) Ediorial review has deemed ha any suppressed conen does no maerially affec he overall learning eperience Cengage Learning reserves he righ o remove addiional conen a any ime if subsequen righs resricions require i

8 00 CHAPTER 5 MODELING WITH HIGHER-ORDER DIFFERENTIAL EQUATIONS Finally, he iniial condiions (0) and (0) 0 yield c and c, so he equaion of moion is () e cos sin () Alernaive Form of () In a manner idenical o he procedure used on page 95, we can wrie any soluion in he alernaive form where A c c () e (c cos () Ae sin(, () and he phase angle f is deermined from he equaions sin c A, cos c A, an c The coefficien Ae l is someimes called he damped ampliude of vibraions Because () is no a periodic funcion, he number is called he quasi period and is he quasi frequency The quasi period is he ime inerval beween wo successive maima of () You should verify, for he equaion of moion in Eample 5, ha A 0 and f 49 Therefore an equivalen form of () is () 0 c sin e sin( 49) 5 SPRING/MASS SYSTEMS: DRIVEN MOTION DE of Driven Moion wih Damping Suppose we now ake ino consideraion an eernal force f() acing on a vibraing mass on a spring For eample, f() could represen a driving force causing an oscillaory verical moion of he suppor of he spring See Figure 5 The inclusion of f() in he formulaion of Newon s second law gives he differenial equaion of driven or forced moion: m d (4) d k d f() d ) c ) m FIGURE 5 Oscillaory verical moion of he suppor Dividing (4) by m gives d d d, (5) d F() where F() f() m and, as in he preceding secion, l b m, v k m To solve he laer nonhomogeneous equaion, we can use eiher he mehod of undeermined coefficiens or variaion of parameers EXAMPLE 6 Inerpreaion of an Iniial-Value Problem Inerpre and solve he iniial-value problem 5 d d d d 5 cos 4, (0), (0) 0 (6) SOLUTION We can inerpre he problem o represen a vibraional sysem consising of a mass ( m 5 slug or kilogram) aached o a spring (k lb/f or N/m) Copyrigh 0 Cengage Learning All Righs Reserved May no be copied, scanned, or duplicaed, in whole or in par Due o elecronic righs, some hird pary conen may be suppressed from he ebook and/or echaper(s) Ediorial review has deemed ha any suppressed conen does no maerially affec he overall learning eperience Cengage Learning reserves he righ o remove addiional conen a any ime if subsequen righs resricions require i

9 5 LINEAR MODELS: INITIAL-VALUE PROBLEMS 0 seady sae p () ransien FIGURE 5 (8) of Eample 6 (a) (b) π / ()=ransien + seady sae π / Graph of soluion in The mass is iniially released from res uni (foo or meer) below he equilibrium posiion The moion is damped (b ) and is being driven by an eernal periodic (T p s) force beginning a 0 Inuiively, we would epec ha even wih damping, he sysem would remain in moion unil such ime as he forcing funcion was urned off, in which case he ampliudes would diminish However, as he problem is given, f() 5 cos 4 will remain on forever We firs muliply he di ferenial equaion in (6) by 5 and solve by he usual mehods Because m i, m i, i follows ha c () e (c cos c sin ) Using he mehod of undeermined coefficiens, we assume a paricular soluion of he form p () A cos 4 B sin 4 Differeniaing p () and subsiuing ino he DE gives p 6 p 0 p ( 6A 4B) cos 4 ( 4A 6B) sin 4 5 cos 4 The resuling sysem of equaions yields A 5 and B 50 I follows ha 0 6A 4B 5, 4A 6B 0 5 d d d d () e (c cos c sin ) 5 0 sin 4 (7) When we se 0 in he above equaion, we obain c 8 5 By differeniaing he epression and hen seing 0, we also find ha c 86 5 Therefore he equaion of moion is () e 8 86 cos 5 5 sin 5 0 cos cos sin 4 (8) Transien and Seady-Sae Terms When F is a periodic funcion, such as F() F 0 sin g or F() F 0 cos g, he general soluion of (5) for l 0 is he sum of a nonperiodic funcion c () and a periodic funcion p () Moreover, c () dies off as ime increases ha is, lim : c () 0 Thus for large values of ime, he displacemens of he mass are closely approimaed by he paricular soluion p () The complemenary funcion c () is said o be a ransien erm or ransien soluion, and he funcion p (), he par of he soluion ha remains afer an inerval of ime, is called a seady-sae erm or seady-sae soluion Noe herefore ha he effec of he iniial condiions on a spring/mass sysem driven by F is ransien In he paricular soluion (8), e ( 8 86 is a ransien erm, and cos 4 5 sin 4 5 cos 5 sin ) p () is a seady-sae erm The graphs of hese wo erms and he soluion (8) are given in Figures 5(a) and 5(b), respecively =7 = =0 =_ EXAMPLE 7 The soluion of he iniial-value problem Transien/Seady-Sae Soluions d d, d d 4 cos sin, (0) 0, (0) where is consan, is given by () ( ) e sin sin π π FIGURE 5 Graph of soluion in Eample 7 for various iniial velociies ransien seady-sae Soluion curves for seleced values of he iniial velociy are shown in Figure 5 The graphs show ha he influence of he ransien erm is negligible for abou p Copyrigh 0 Cengage Learning All Righs Reserved May no be copied, scanned, or duplicaed, in whole or in par Due o elecronic righs, some hird pary conen may be suppressed from he ebook and/or echaper(s) Ediorial review has deemed ha any suppressed conen does no maerially affec he overall learning eperience Cengage Learning reserves he righ o remove addiional conen a any ime if subsequen righs resricions require i

10 0 CHAPTER 5 MODELING WITH HIGHER-ORDER DIFFERENTIAL EQUATIONS DE of Driven Moion wihou Damping Wih a periodic impressed force and no damping force, here is no ransien erm in he soluion of a problem Also, we shall see ha a periodic impressed force wih a frequency near or he same as he frequency of free undamped vibraions can cause a severe problem in any oscillaory mechanical sysem EXAMPLE 8 Undamped Forced Moion Solve he iniial-value problem where F 0 is a consan and g v d, (9) d F 0 sin, (0) 0, (0) 0 SOLUTION The complemenary funcion is c () c cos v c sin v To obain a paricular soluion, we assume p () A cos g B sin g so ha p p A( ) cos B( ) sin F 0 sin Equaing coefficiens immediaely gives A 0 and B F 0 (v g ) Therefore p () F 0 sin Applying he given iniial condiions o he general soluion () c cos c sin F 0 sin yields c 0 and c gf 0 v(v g ) Thus he soluion is () F 0 ( ) ( sin sin ), (0) FIGURE 54 Pure resonance Pure Resonance Alhough equaion (0) is no defined for g v, i is ineresing o observe ha is limiing value as can be obained by applying L Hôpial s Rule This limiing process is analogous o uning in he frequency of he driving force (g p) o he frequency of free vibraions (v p) Inuiively, we epec ha over a lengh of ime we should be able o subsanially increase he ampliudes of vibraion For g v we define he soluion o b d sin sin () lim F 0 F 0 lim : ( ) : sin cos F 0 lim F 0 sin cos F 0 sin F 0 cos As suspeced, when :, he displacemens become large; in fac, ( n ) B when n np v, n,, The phenomenon ha we have jus described is known as pure resonance The graph given in Figure 54 shows ypical moion in his case In conclusion i should be noed ha here is no acual need o use a limiing process on (0) o obain he soluion for g v Alernaively, equaion () follows by solving he iniial-value problem : : ( d sin sin ) d d F 0 sin, (0) 0, (0) 0 direcly by convenional mehods d d ( ) () Copyrigh 0 Cengage Learning All Righs Reserved May no be copied, scanned, or duplicaed, in whole or in par Due o elecronic righs, some hird pary conen may be suppressed from he ebook and/or echaper(s) Ediorial review has deemed ha any suppressed conen does no maerially affec he overall learning eperience Cengage Learning reserves he righ o remove addiional conen a any ime if subsequen righs resricions require i

11 5 LINEAR MODELS: INITIAL-VALUE PROBLEMS 0 If he displacemens of a spring/mass sysem were acually described by a funcion such as (), he sysem would necessarily fail Large oscillaions of he mass would evenually force he spring beyond is elasic limi One migh argue oo ha he resonaing model presened in Figure 54 is compleely unrealisic because i ignores he rearding effecs of ever-presen damping forces Alhough i is rue ha pure resonance canno occur when he smalles amoun of damping is aken ino consideraion, large and equally desrucive ampliudes of vibraion (alhough bounded as : ) can occur See Problem 4 in Eercises 5 E FIGURE 55 L R C LRC-series circui 54 SERIES CIRCUIT ANALOGUE LRC-Series Circuis As was menioned in he inroducion o his chaper, many differen physical sysems can be described by a linear second-order differenial equaion similar o he differenial equaion of forced moion wih damping: m d () d d k f() d If i() denoes curren in he LRC-series elecrical circui shown in Figure 55, hen he volage drops across he inducor, resisor, and capacior are as shown in Figure 4 By Kirchhoff s second law he sum of hese volages equals he volage E() impressed on he circui; ha is, L di () d Ri q E() C Bu he charge q() on he capacior is relaed o he curren i() by i dq d, so () becomes he linear second-order differenial equaion L d q dq R (4) d d q E() C The nomenclaure used in he analysis of circuis is similar o ha used o describe spring/mass sysems If E() 0, he elecrical vibraions of he circui are said o be free Because he auiliary equaion for (4) is Lm Rm C 0, here will be hree forms of he soluion wih R 0, depending on he value of he discriminan R 4L C We say ha he circui is overdamped if R 4L/C 0, criically damped if R 4L/C 0, and underdamped if R 4L/C 0 In each of hese hree cases he general soluion of (4) conains he facor e R/L, so q() : 0 as : In he underdamped case when q(0) q 0, he charge on he capacior oscillaes as i decays; in oher words, he capacior is charging and discharging as : When E() 0 and R 0, he circui is said o be undamped, and he elecrical vibraions do no approach zero as increases wihou bound; he response of he circui is simple harmonic EXAMPLE 9 Underdamped Series Circui Find he charge q() on he capacior in an LRC-series circui when L 05 henry (h), R 0 ohms ( ), C 000 farad (f), E() 0, q(0) q 0 coulombs (C), and i(0) 0 SOLUTION Since C 000, equaion (4) becomes q 0q 000q 0 or q 40q 4000q 0 4 Copyrigh 0 Cengage Learning All Righs Reserved May no be copied, scanned, or duplicaed, in whole or in par Due o elecronic righs, some hird pary conen may be suppressed from he ebook and/or echaper(s) Ediorial review has deemed ha any suppressed conen does no maerially affec he overall learning eperience Cengage Learning reserves he righ o remove addiional conen a any ime if subsequen righs resricions require i

12 04 CHAPTER 5 MODELING WITH HIGHER-ORDER DIFFERENTIAL EQUATIONS Solving his homogeneous equaion in he usual manner, we find ha he circui is underdamped and q() e 0 (c cos 60 c sin 60) Applying he iniial condiions, we find c q 0 and c q 0 Thus q() q 0 e 0 cos 60 sin 60 Using (), we can wrie he foregoing soluion as q() q 00 e 0 sin(60 49) When here is an impressed volage E() on he circui, he elecrical vibraions are said o be forced In he case when R 0, he complemenary funcion q c () of (4) is called a ransien soluion If E() is periodic or a consan, hen he paricular soluion q p () of (4) is a seady-sae soluion EXAMPLE 0 Seady-Sae Curren Find he seady-sae soluion q p () and he seady-sae curren in an LRC-series circui when he impressed volage is E() E 0 sin g SOLUTION equaion The seady-sae soluion q p () is a paricular soluion of he differenial L d q dq R d d C q E 0 sin Using he mehod of undeermined coefficiens, we assume a paricular soluion of he form q p () A sin g B cos g Subsiuing his epression ino he differenial equaion, simplifying, and equaing coefficiens give A E 0 L C L L C C R, B I is convenien o epress A and B in erms of some new symbols E 0 R L L C C R If If X L, hen X L L C C C Z X R, hen Z L L C C R Therefore A E 0 X ( gz ) and B E 0 R ( gz ), so he seady-sae charge is q p () E 0X Z sin E 0R Z cos Now he seady-sae curren is given by i p () q p (): i p () E 0 Z R Z sin X Z cos (5) The quaniies X Lg Cg and Z X R defined in Eample 0 are called he reacance and impedance, respecively, of he circui Boh he reacance and he impedance are measured in ohms Copyrigh 0 Cengage Learning All Righs Reserved May no be copied, scanned, or duplicaed, in whole or in par Due o elecronic righs, some hird pary conen may be suppressed from he ebook and/or echaper(s) Ediorial review has deemed ha any suppressed conen does no maerially affec he overall learning eperience Cengage Learning reserves he righ o remove addiional conen a any ime if subsequen righs resricions require i

13 5 LINEAR MODELS: INITIAL-VALUE PROBLEMS 05 EXERCISES 5 5 SPRING/MASS SYSTEMS: FREE UNDAMPED MOTION A mass weighing 4 pounds is aached o a spring whose spring consan is 6 lb/f Wha is he period of simple harmonic moion? A 0-kilogram mass is aached o a spring If he frequency of simple harmonic moion is p cycles/s, wha is he spring consan k? Wha is he frequency of simple harmonic moion if he original mass is replaced wih an 80-kilogram mass? A mass weighing 4 pounds, aached o he end of a spring, sreches i 4 inches Iniially, he mass is released from res from a poin inches above he equilibrium posiion Find he equaion of moion 4 Deermine he equaion of moion if he mass in Problem is iniially released from he equilibrium posiion wih a downward velociy of f/s 5 A mass weighing 0 pounds sreches a spring 6 inches The mass is iniially released from res from a poin 6 inches below he equilibrium posiion (a) Find he posiion of he mass a he imes p, p 8, p 6, p 4, and 9p s (b) Wha is he velociy of he mass when p 6 s? In which direcion is he mass heading a his insan? (c) A wha imes does he mass pass hrough he equilibrium posiion? 6 A force of 400 newons sreches a spring meers A mass of 50 kilograms is aached o he end of he spring and is iniially released from he equilibrium posiion wih an upward velociy of 0 m/s Find he equaion of moion 7 Anoher spring whose consan is 0 N/m is suspended from he same rigid suppor bu parallel o he spring/mass sysem in Problem 6 A mass of 0 kilograms is aached o he second spring, and boh masses are iniially released from he equilibrium posiion wih an upward velociy of 0 m/s (a) Which mass ehibis he greaer ampliude of moion? (b) Which mass is moving faser a p 4 s? A p s? (c) A wha imes are he wo masses in he same posiion? Where are he masses a hese imes? In which direcions are he masses moving? 8 A mass weighing pounds sreches a spring fee Deermine he ampliude and period of moion if he mass is iniially released from a poin foo above he equilibrium posiion wih an upward velociy of f/s Answers o seleced odd-numbered problems begin on page ANS-7 How many complee cycles will he mass have compleed a he end of 4p seconds? 9 A mass weighing 8 pounds is aached o a spring When se in moion, he spring/mass sysem ehibis simple harmonic moion (a) Deermine he equaion of moion if he spring consan is lb/f and he mass is iniially released from a poin 6 inches below he equilibrium posiion wih a downward velociy of f/s (b) Epress he equaion of moion in he form given in (6) (c) Epress he equaion of moion in he form given in (6 ) 0 A mass weighing 0 pounds sreches a spring 4 foo This mass is removed and replaced wih a mass of 6 slugs, which is iniially released from a poin foo above he equilibrium posiion wih a downward velociy of 5 4 f/s (a) Epress he equaion of moion in he form given in (6) (b) Epress he equaion of moion in he form given in (6 ) (c) Use one of he soluions obained in pars (a) and (b) o deermine he imes he mass aains a displacemen below he equilibrium posiion numerically equal o he ampliude of moion A mass weighing 64 pounds sreches a spring 0 foo The mass is iniially released from a poin 8 inches above he equilibrium posiion wih a downward velociy of 5 f/s (a) Find he equaion of moion (b) Wha are he ampliude and period of moion? (c) How many complee cycles will he mass have compleed a he end of p seconds? (d) A wha ime does he mass pass hrough he equilibrium posiion heading downward for he second ime? (e) A wha imes does he mass aain is ereme displacemens on eiher side of he equilibrium posiion? (f) Wha is he posiion of he mass a s? (g) Wha is he insananeous velociy a s? (h) Wha is he acceleraion a s? (i) Wha is he insananeous velociy a he imes when he mass passes hrough he equilibrium posiion? (j) A wha imes is he mass 5 inches below he equilibrium posiion? (k) A wha imes is he mass 5 inches below he equilibrium posiion heading in he upward direcion? Copyrigh 0 Cengage Learning All Righs Reserved May no be copied, scanned, or duplicaed, in whole or in par Due o elecronic righs, some hird pary conen may be suppressed from he ebook and/or echaper(s) Ediorial review has deemed ha any suppressed conen does no maerially affec he overall learning eperience Cengage Learning reserves he righ o remove addiional conen a any ime if subsequen righs resricions require i

14 06 CHAPTER 5 MODELING WITH HIGHER-ORDER DIFFERENTIAL EQUATIONS A mass of slug is suspended from a spring whose spring consan is 9 lb/f The mass is iniially released from a poin foo above he equilibrium posiion wih an upward velociy of f/s Find he imes a which he mass is heading downward a a velociy of f/s Under some circumsances when wo parallel springs, wih consans k and k, suppor a single mass, he effecive spring consan of he sysem is given by k 4k k (k k ) A mass weighing 0 pounds sreches one spring 6 inches and anoher spring inches The springs are aached o a common rigid suppor and hen o a meal plae As shown in Figure 56, he mass is aached o he cener of he plae in he double-spring arrangemen Deermine he effecive spring consan of his sysem Find he equaion of moion if he mass is iniially released from he equilibrium posiion wih a downward velociy of f/s 7 8 FIGURE 57 Graph for Problem 7 k k FIGURE 58 Graph for Problem 8 0 lb 9 FIGURE 56 Problem Double-spring sysem in 4 A cerain mass sreches one spring foo and anoher spring foo The wo springs are aached o a common rigid suppor in he manner described in Problem and Figure 56 The firs mass is se aside, a mass weighing 8 pounds is aached o he double-spring arrangemen, and he sysem is se in moion If he period of moion is p 5 second, deermine how much he firs mass weighs 5 A model of a spring/mass sysem is 4 e 0 0 By inspecion of he differenial equaion only, discuss he behavior of he sysem over a long period of ime 6 A model of a spring/mass sysem is 4 0 By inspecion of he differenial equaion only, discuss he behavior of he sysem over a long period of ime 0 FIGURE 59 Graph for Problem 9 FIGURE 50 Graph for Problem 0 5 SPRING/MASS SYSTEMS: FREE DAMPED MOTION In Problems 7 0 he given figure represens he graph of an equaion of moion for a damped spring/mass sysem Use he graph o deermine (a) wheher he iniial displacemen is above or below he equilibrium posiion and (b) wheher he mass is iniially released from res, heading downward, or heading upward A mass weighing 4 pounds is aached o a spring whose consan is lb/f The medium offers a damping force ha is numerically equal o he insananeous velociy The mass is iniially released from a poin foo above he equilibrium posiion wih a downward velociy of 8 f/s Deermine he ime a which he mass passes hrough he equilibrium posiion Find he ime a which he mass aains is ereme displacemen from he equilibrium posiion Wha is he posiion of he mass a his insan? Copyrigh 0 Cengage Learning All Righs Reserved May no be copied, scanned, or duplicaed, in whole or in par Due o elecronic righs, some hird pary conen may be suppressed from he ebook and/or echaper(s) Ediorial review has deemed ha any suppressed conen does no maerially affec he overall learning eperience Cengage Learning reserves he righ o remove addiional conen a any ime if subsequen righs resricions require i

15 5 LINEAR MODELS: INITIAL-VALUE PROBLEMS 07 A 4-foo spring measures 8 fee long afer a mass weighing 8 pounds is aached o i The medium hrough which he mass moves offers a damping force numerically equal o imes he insananeous velociy Find he equaion of moion if he mass is iniially released from he equilibrium posiion wih a downward velociy of 5 f/s Find he ime a which he mass aains is ereme displacemen from he equilibrium posiion Wha is he posiion of he mass a his insan? A -kilogram mass is aached o a spring whose consan is 6 N/m, and he enire sysem is hen submerged in a liquid ha impars a damping force numerically equal o 0 imes he insananeous velociy Deermine he equaions of moion if (a) he mass is iniially released from res from a poin meer below he equilibrium posiion, and hen (b) he mass is iniially released from a poin meer below he equilibrium posiion wih an upward velociy of m/s 4 In pars (a) and (b) of Problem deermine wheher he mass passes hrough he equilibrium posiion In each case find he ime a which he mass aains is ereme displacemen from he equilibrium posiion Wha is he posiion of he mass a his insan? 5 A force of pounds sreches a spring foo A mass weighing pounds is aached o he spring, and he sysem is hen immersed in a medium ha offers a damping force ha is numerically equal o 04 imes he insananeous velociy (a) Find he equaion of moion if he mass is iniially released from res from a poin foo above he equilibrium posiion (b) Epress he equaion of moion in he form given in () (c) Find he firs ime a which he mass passes hrough he equilibrium posiion heading upward 6 Afer a mass weighing 0 pounds is aached o a 5-foo spring, he spring measures 7 fee This mass is removed and replaced wih anoher mass ha weighs 8 pounds The enire sysem is placed in a medium ha offers a damping force ha is numerically equal o he insananeous velociy (a) Find he equaion of moion if he mass is iniially released from a poin foo below he equilibrium posiion wih a downward velociy of f/s (b) Epress he equaion of moion in he form given in () (c) Find he imes a which he mass passes hrough he equilibrium posiion heading downward (d) Graph he equaion of moion 7 A mass weighing 0 pounds sreches a spring fee The mass is aached o a dashpo device ha offers a damping force numerically equal o b (b 0) imes he insananeous velociy Deermine he values of he damping consan b so ha he subsequen moion is (a) overdamped, (b) criically damped, and (c) underdamped 8 A mass weighing 4 pounds sreches a spring 4 fee The subsequen moion akes place in medium ha offers a damping force numerically equal o b (b 0) imes he insananeous velociy If he mass is iniially released from he equilibrium posiion wih an upward velociy of f/s, show ha when he equaion of moion is () 5 SPRING/MASS SYSTEMS: DRIVEN MOTION 8 e / sinh 8 9 A mass weighing 6 pounds sreches a spring fee The mass is iniially released from res from a poin fee below he equilibrium posiion, and he subsequen moion akes place in a medium ha offers a damping force ha is numerically equal o he insananeous velociy Find he equaion of moion if he mass is driven by an eernal force equal o f() 0 cos 0 A mass of slug is aached o a spring whose consan is 5 lb/f Iniially, he mass is released foo below he equilibrium posiion wih a downward velociy of 5 f/s, and he subsequen moion akes place in a medium ha offers a damping force ha is numerically equal o imes he insananeous velociy (a) Find he equaion of moion if he mass is driven by an eernal force equal o f() cos sin (b) Graph he ransien and seady-sae soluions on he same coordinae aes (c) Graph he equaion of moion A mass of slug, when aached o a spring, sreches i fee and hen comes o res in he equilibrium posiion Saring a 0, an eernal force equal o f() 8 sin 4 is applied o he sysem Find he equaion of moion if he surrounding medium offers a damping force ha is numerically equal o 8 imes he insananeous velociy In Problem deermine he equaion of moion if he eernal force is f() e sin 4 Analyze he displacemens for : When a mass of kilograms is aached o a spring whose consan is N/m, i comes o res in he equilibrium posiion Saring a 0, a force equal o f() 68e cos 4 is applied o he sysem Find he equaion of moion in he absence of damping 4 In Problem wrie he equaion of moion in he form () Asin(v f) Be sin(4 u) Wha is he ampliude of vibraions afer a very long ime? 8 Copyrigh 0 Cengage Learning All Righs Reserved May no be copied, scanned, or duplicaed, in whole or in par Due o elecronic righs, some hird pary conen may be suppressed from he ebook and/or echaper(s) Ediorial review has deemed ha any suppressed conen does no maerially affec he overall learning eperience Cengage Learning reserves he righ o remove addiional conen a any ime if subsequen righs resricions require i

16 08 CHAPTER 5 MODELING WITH HIGHER-ORDER DIFFERENTIAL EQUATIONS 5 A mass m is aached o he end of a spring whose consan is k Afer he mass reaches equilibrium, is suppor begins o oscillae verically abou a horizonal line L according o a formula h() The value of h represens he disance in fee measured from L See Figure 5 (a) Deermine he differenial equaion of moion if he enire sysem moves hrough a medium offering a damping force ha is numerically equal o b(d d) (b) Solve he differenial equaion in par (a) if he spring is sreched 4 fee by a mass weighing 6 pounds and b, h() 5 cos, (0) (0) 0 6 A mass of 00 grams is aached o a spring whose consan is 600 dynes/cm Afer he mass reaches equilibrium, is suppor oscillaes according o he formula h() sin 8, where h represens displacemen from is original posiion See Problem 5 and Figure 5 (a) In he absence of damping, deermine he equaion of moion if he mass sars from res from he equilibrium posiion (b) A wha imes does he mass pass hrough he equilibrium posiion? (c) A wha imes does he mass aain is ereme displacemens? (d) Wha are he maimum and minimum displacemens? (e) Graph he equaion of moion In Problems 7 and 8 solve he given iniial-value problem 7 L d 4 5 sin cos, d (0), (0) d sin, (0), (0) 0 d 9 (a) Show ha he soluion of he iniial-value problem is () F 0 (cos cos ) suppor h() FIGURE 5 Oscillaing suppor in Problem 5 d d F 0 cos, (0) 0, (0) 0 (b) Evaluae lim (cos cos ) : F 0 40 Compare he resul obained in par (b) of Problem 9 wih he soluion obained using variaion of parameers when he eernal force is F 0 cos v 4 (a) Show ha () given in par (a) of Problem 9 can be wrien in he form () F 0 sin ( ) sin ( ) (b) If we define ( ), show ha when is small an approimae soluion is () F 0 sin sin When is small, he frequency g p of he impressed force is close o he frequency v p of free vibraions When his occurs, he moion is as indicaed in Figure 5 Oscillaions of his kind are called beas and are due o he fac ha he frequency of sin is quie small in comparison o he frequency of sin g The dashed curves, or envelope of he graph of (), are obained from he graphs of (F 0 g) sin Use a graphing uiliy wih various values of F 0,, and g o verify he graph in Figure 5 FIGURE 5 Beas phenomenon in Problem 4 Compuer Lab Assignmens 4 Can here be beas when a damping force is added o he model in par (a) of Problem 9? Defend your posiion wih graphs obained eiher from he eplici soluion of he problem d d d d F 0 cos, (0) 0, (0) 0 or from soluion curves obained using a numerical solver 4 (a) Show ha he general soluion of d d d d F 0 sin Copyrigh 0 Cengage Learning All Righs Reserved May no be copied, scanned, or duplicaed, in whole or in par Due o elecronic righs, some hird pary conen may be suppressed from he ebook and/or echaper(s) Ediorial review has deemed ha any suppressed conen does no maerially affec he overall learning eperience Cengage Learning reserves he righ o remove addiional conen a any ime if subsequen righs resricions require i