Math 2214 Sol Test 2B Spring 2015

Transcription

1 Mah 14 Sol Tes B Sring 015 roblem 1: An objec weighing ounds sreches a verical sring 8 fee beond i naural lengh before coming o res a equilibrium The objec is ushed u 6 fee from i s equilibrium osiion wih an iniial veloci of f/sec and hen released a) (18s) Se u he iniial value differenial equaion ha models his sring moion Find he soluion and hen conver o he form S Re Cos( ) Find and describe he amliude Using he sandard form of he equaion, we have / ms as ks F There is no resisance and i is homogeneous so ms ks 0 So we need o find mass and he resoring consan k Since = mg where g = f/sec so m = 1 and = ks where s = 8f so k = 4 This gives us he equaion s 4s 0 B solving his DE we ge C1cos( ) Csin( ) Now using he iniial values s(0) = -6 and s / (0) = - find he unknown coefficiens () C cos( ) C sin( ) 6 C C1 6 and C 1 1 / C1 sin( ) C cos( ) C 6cos( ) C sin( ) R[cos( )] where R ( 6) ( 1) 7 Therefore he Amliude = R and is consan Noe ha an 6 b) (4s) Se u a new model for his sring ssem if here is a daming force of 4 lbs when veloci is f/sec and an exernal force of F() = Sin() Do no solve In his ssem we do have resisance and an exernal force Consider he resisance r = as /, so as / = 4 where s / = f/sec and a will = The DE is / s s s 4 sin(4 ) 1 1

2 roblem : Below are soluions for a aricular Sring Mass ssems Analze and describe he soluions in erms of heir comonens (homogeneous and aricular soluions) Indicae simle harmonic, criicall damed, underdamed, over-damed, ransien and/or resonance roeries Remember o invesigae when ) a) (4) [045cos( ) 5sin( )] cos( ) Noe ha 045cos( ) 5sin( ) is simle Harmonic and C cos( ) becomes infiniel large as The increasing oscillaion becomes resonan b) () 004 e 067e Noe ha we have reeaed roos which indicae ha he ssem is criicall damed Therefore he oscillaions become zero as 0, bu he ssem is ver fragil c) () e [05cos( ) 04sin( )] This ssem is underdamed I aroaches zero as, bu he oscillaions never come o a comlee so roblem : Given he second order linear differenial equaion e and e / 0 and soluions a) (7s) Use he Wronskian o deermine ha 1 and form a fundamenal se e e 8 W e ( e 4 e ) (4 e )( e ) e 0 e Therefore e and e do form a fundemenal se (You should alwas be clear in summar of work) b) (7s) Find he values of and Since he soluions involve reeaed roos, we know ha he facors o he characerisic equaion would be homogeneous DE is ( m 4)( m 4) m 8m 16 0 This indicaes ha our original where 8 and 16 (When ou resen / an mahemaical calculaion, here alwas need o be some readable exlanaions)

3 c) (7s) If given iniial values (0) = 1 and / (0) = 6, solve for unknown consans in he general soluion Given he general soluion: he unknown consans C e C e, creae he following ssem o solve for C1e Ce / 4C1e 4Ce Ce Using he iniial values we have: 1 C1 6 4C C e o give ha C 1, C So he aricular soluion is e e roblem 4:( 0s) Use he mehod of variaion of arameers o find he general soluion of he following differenial equaion / / u 1v 0 Use he ssem and / / / / u 1 v g() Firs solve he homogeneous form Now choose he aricular soluion o be Using he given ssem we have ha : / e 4 (no iniial values) 1 A o find o find he general soluion C 0 o ge = C e C e / = u() e ( ) v C e / / u e v e 0 / / u e v e We will now use he inverse of A o solve his ssem where e e 1 1 e e A and A e e e e e

4 / u 1 e e 0 / v e e e 4 and le k 0 / u u d e d k1 1 / v v d e d e k 4 and le k 0 The general soluion is now given o be: = C ( ) = C 1e Ce e e e 1e Ce e roblem 5: Given he following differenial equaion given) g 6 ( ) (no iniial values a) (7s) Find he general soluion C for he homogeneous form Given 6 0 The characerisic equaion is m 6m 0 m 0 6 i, his gives he general soluion = C cos(6 ) C sin(6 ) b) (7s) Find he aricular soluion when g( ) Using mehod of undeermined coefficiens we have he following choice Ae Ae / 9Ae Now subsiue ino he original equaion (9 Ae ) 6( Ae ) 4 e and solve for A o ge A = 4/45 so 45 6 o ge he following c) (1) Sae he final general soluion = C1cos(6 ) Csin(6 ) 45

5 roblem 6: Given he following differenial equaion soluions for he given g() ( do no solve) g 6 ( ), se u he aricular Firs noe he soluion for he homogeneous form is = C1cos(6 ) Csin(6 ) and all choices for he aricular soluion mus be comared o he homogeneous soluion a) (s) Le g( ) ( ) e and se u Using Undeermined Coefficiens we have ( A A A A ) e 4 b) (6s) Le g( ) sin(6 ) and se u Using undeermined coefficiens, choose Acos(6 ) Bsin(6 ) B comaring his choice o he homogeneous form we see ha we acuall have he same soluion and our choice mus be modified Now choose A cos(6 ) B sin(6 ) and his choice will work 5 c) (4s) Le g( ) e cos(6 ) and se u Using undeermined coefficiens choose ( A A ) e cos(6 ) ( B B )sin(6 )