3 General forced response

Transcription

1 3 General forced response So far, all of he driving forces have been sine or cosine exciaions In his chaper we examine he response o any form of exciaion such as Impulse Sums of sines and cosines Any inegrable funcion

2 Linear Superposiion allows us o break up complicaed forces ino sums of simpler forces, compue he response and add o ge he oal soluion If x, x are soluions of a linear homogeneous 1 2 equaion, hen If and x a x a x is also a soluion. x is he paricular sol of x x f 2 1 n 1 x he paricular sol of x x f 2 2 n 2 ax bx solves x x af bf n 1 2

3 3.1 Impulse Response Funcion Fˆ 2 F() F() Impulse exciaion Fˆ 2 is a small posiive number Figure 3.1

4 From sophomore dynamics The impulse impared o an objec is equal o he change in he objecs momenum Fˆ 2 F() i.e. area under pulse impulse force = F( ) d I( ) F( ) d F( ) d N s Fˆ 2 2 Fˆ F

5 We use he properies of impulse o define he impulse funcion: F() Dirac Dela funcion Equal impulses F( ), If Fˆ F( ) d Fˆ 1, his is he Dirac Dela ()

6 The effec of an impulse on a spring-mass-damper is relaed o is change in momenum. impulse=momenum change F mv m[ v( ) v( )] ˆ ˆ F F mv v m Jus afer impulse F m Jus before impulse Thus he response o impulse wih zero IC is equal o he free response wih IC: x = and v =FΔ/m

7 Recall ha he free response o jus non zero iniial condiions is: The soluion of: mx cx kx x() x x() v in underdamped case: 2 2 v x x x x( ) e sin( an ) n d -n 1 d d d v nx For x his becomes: ve -n ( ) sin x d d

8 Nex compue he response o x()= and v() =FΔ/m The soluion of: mx cx kx x() x x() F / m m in underdamped case from he previous slide is: ˆ -n Fe x( ) sin d m d Fˆ Response o an impulse a =, and zero iniial condiions

9 So for an underdamped sysem he impulse response is (x = ) ˆ n Fe x( ) sin (response o ˆ d F) (3.6) m d n ( ) ˆ e x Fh( ), where h( ) sin d (3.8) m d Fˆ uni impulse response funcion x() M.5 h( ) k c Time Response o an impulse a =, and zero iniial condiions

10 The response o an impulse is hus defined in erms of he impulse response funcion, h(). So, he response o ( ) is given by h( ). n e h( ) sin d (3.8) m Wha is he response o a uni impulse applied a a ime differen from zero? The response o ( - ) is h( - ). This is given on he following slide d

11 ( ) ( ) n h e sin d ( ) m d for he case ha he impulse occurs a noe ha he effecs of non-zero iniial condiions and oher forcing erms mus be super imposed on his soluion For example: If wo pulses occur a wo differen imes hen heir impulse responses will superimpose (see Equaion (3.9)) h2 h1 +h = = Time

12 Consider he undamped impulse response Seing in he equaion (3.8) Response o uni impulse applied a, i.e. ( - ) is: 1 h( ) sin n ( ) m n

13 Example Design a camera moun wih a vibraion consrain Consider example of he securiy camera again only his ime wih an impulsive load

14 Using he siffness and mass parameers of Example 2.1.3, does he sysem say wih in vibraion limis if hi by a 1 kg bird raveling a 72 kmh? The naural frequency of he camera sysem is n k m c 3Ebh 12m c 3 3 (7.1 x 1 N/m)(.2 m)(.2 m) 3 4(3 kg)(.55) rad/s From equaions (3.7) and (3.8) wih ζ =, he impulsive response is: F mv b x( ) sinn sinn m m c n c n The magniude of he response due o he impulse is hus X mv c b m n

15 Nex compue he momenum of he bird o complee he magniude calculaion: km 1 m hour mv b 1 kg 72 2 kg m/s hour km 36 s Nex use his value in he expression for he maximum value: X mv b 2 kg m/s m 3 kg rad/s c n.88 m This max value exceeds he camera olerance

16 Example 3.1.3: wo impacs, zero iniial condiions (double hi). x() x x 1 2 m 1 kg, c.5 kg/s, k 4 N/m Fˆ 2 N s and F( ) 2 ( ) ( ) 2,.125 n 2e x e n.25 1( ) sind 1.8 sin(1.984 ), m d x e.25( ) 2( ).54 sin(1.984( )),.25( ) 1.8 sin(1.984 ) e.25.25( ) 1.8 sin(1.984 ).54 sin(1.984( )) e e

17 Example wo impacs and iniial condiions x 2x 4 x ( ) ( 4), x 1 mm, x 1 mm/s Solve hree simple problems and add he resuls. Homogeneous soluion ( 2rad/s, =.5, = 3 rad/s) v x x e x n n h( ) [ sind cos d ] d 11 e [ sin 3 cos 3 ] e cos 3 3 n d Noe, no need o redo consans of inegraion for impulse exciaion (ohers, yes)

18 Compuaion of he response o firs impulse: Trea ( ) as x and v 1, 4 v 1 n ( ) x sin I e d e sin 3 d 3 4

19 Toal Response for < < 4 x1 ( ) xh ( ) xi ( ) e (cos 3 1 sin 3 ), 3 for 4

20 Nex compue he response o he second impulse: 1 sin 3( 4), x2 e 4 e sin 3( 4) H ( 4) 3 Heaviside Sep funcion Here he Heaviside sep funcion is used o urn on he response o he impulse a = 4 seconds.

21 To ge he oal response add he parial soluions: 4 1 e x( ) e ( sin 3 cos 3 ) sin 3( 4) H( 4) 3 3 iniial condiion fris impulse second impulse 1.5 x

22 3.2 Response o an Arbirary Inpu The response o general force, F(), can be viewed as a series of impulses of magniude F( i )Δ Response a ime due o he i h impulse zero IC x i () = [F( i )D ]. h(- i ) for > i x F() i Impulses h If (he i ime inerval) I F( i ) I x( ) [ F( ) ] h( ) I i i i1, i 1, 2, 3 i x( ) F( ) h( ) d convoluion inegral (3.12)

23 Properies of convoluion inegrals: I is symmeric meaning: Le, fixed so ha = and d d. Also : : x( ) F( ) h( ) d = F( ) h( )( d) = F( ) h( ) d

24 The convoluion inegral, or Duhamel inegral, for underdamped sysems is: 1 n n x( ) e F( ) e sin ( ) d d m d 1 m d n F( ) e sin d (3.13) d The response o any inegrable force can be compued wih eiher of hese forms Which form o use depends on which is easies o compue

25 Example 3.2.1: Sep funcion inpu mx cx kx F x, v, 1 Figure 3.6 Sep funcion To solve apply (3.13): d d m d F e n e n sin d ( ) d d 1 1 n n n n x( ) e () e sin ( ) d e F e sin d( ) d m m

26 Inegraing (use a able, code or calculaor) yields he soluion: F F n x e 2 d k k 1 ( ( ) ) cos ( ), (3.15) 1 an (3.16) 2 1

27 Example: undamped oscillaor under IC and consan force For an undamped sysem: 1 h( ) sinn m n F() F The homogeneous soluion is v x sin x cos, h n n 1 n Good unil he applied force acs a 1, hen: x F( ) h( ) d, F( ) h( ) d F( ) h( ) d 1 x() 1 2 k m F ()

28 Nex compue he soluion beween 1 and 2 For x F sin n ( ) d m F ( 1)( 1) cos n ( ) m F [1 cos n( 1)] m 2 n n n n 1

29 Now compue he soluion for ime greaer han 2 For x 2 F( ) h( ) d F( ) h( ) d F( ) h( ) d 2 n F 1 cos n ( ) mn n 1 F [cos n( 2) cos n( 1)] m

30 Toal soluion is superposiion: v sinn xcosn 1 n v F x( ) sinn x cos n [1 cos ( 2 n 1)] 1 2 n mn v F sin x cos cos ( ) cos ( ) n n 2 n 2 n 1 2 n mn m F 1, 8, 2, 4, x.1, v n 1 2 Check poins: x increases afer applicaion of F. Undamped response around x.3 Displacemen x() Time (s)

31 Example 3.2.3: Saic versus dynamic load md g mx cx kx This has max value of mg 1 d n x( ) 1 e cos 2 d k 1 mg d x( ) (1 cos d) k x mg max 2 d k, wice he saic load

32 Numerical simulaion and ploing A he end of his chaper, numerical simulaion is used o solve he problems of his secion. Numerical simulaion is ofen easier hen compuing hese inegrals I is wise o check he wo approaches agains each oher by ploing he analyical soluion and numerical soluion on he same graph

33 3.3 Response o an Arbirary Periodic Inpu x x x F F F T 2 2 n n ( ) where ( ) ( ) We have soluions o sine and cosine inpus. Wha abou periodic bu nonharmonic inpus? Displacemen x().5 T We know ha periodic funcions can be represened by a series of sines and cosines (Fourier) Response is superposiion of as many RHS erms as you hink are necessary o represen he forcing funcion accuraely Time (s) Figure 3.11

34 Recall he Fourier Series Definiion: Assume F( ) an cosn bn sin n (3.2) 2 n1 a 2 T 2 T T T F( ) d (3.21) : wice he average a F( ) cos d (3.22) : Oscillaions around average n T a 2n where n n T 2 bn T F( ) sin n d (3. 23) n

35 The erms of he Fourier series saisfy orhogonaliy condiions: T T T m n sin nt sin mtd (3.24) T m n 2 m n cos nt cos mtd (3.25) T m n 2 cos n sin m d (3.26) T T

36 F() Fourier Series Example F 1 2 =T Sep 1: find he F.S. and deermine how many erms you need F () F 2 1,,

37 Fourier Series Example Force.6F() F() 2 coefficiens 1 coefficiens 1 coefficiens Time (s)

38 Having obained he FS of inpu The nex sep is o find responses o each erm of he FS And hen, jus add hem up! Danger!!: Resonance occurs whenever a muliple of exciaion frequency equals he naural frequency. You may excie a 1rad/s and observe resonance while naural frequency is 5rad/s!! Backwards?

39 Soluion as a series of sines and cosines o x x x F 2 2 n n ( ) The soluion can be wrien as a summaion x ( ) x ( ) x ( ) x ( ) p cn sn n1 where x ( ) is a soluion o a x 2 x x x ( ) n 2 n 2 2 2n and x ( ) and x ( ) are a soluions o cn x x x a n sn 2 2n n n cos( T ) x x x b n 2 2n n n sin( T ) a Soluions calculaed from equaions of moion (see secion Example 3.3.2)

40 3.4 Transform Mehods An alernaive o solving he previous problems, similar o secion 2.3

41 Laplace ransformaion Laplace Transform s F( s) f ( ) e d L { f ( )} (3.41) Laplace ransforms are very useful because hey change differenial equaions ino simple algebraic equaions Examples of Laplace ransforms (see page 244) in book) f() Sep funcion, u() e -a sin( ) F(s) 1/s 1/(s+a) / ( s )

42 Laplace Transform Example: Laplace ransform of a sep funcion u() s s e { u( )} e L d s 1 u() s Example: Laplace ransform of e -a a a s ( s a) e e e d e d L{ } e -a ( sa) a e L{ e } 1 ( s a) ( s a)

43 Laplace Transforms of Derivaives Laplace ransform of he derivaive of a funcion df ( ) df ( ) s L e d d d Inegraion by pars gives, df () ( ) s s L f e s f ( ) e d d df () L f () sl f ( ) d

44 Laplace Transform Procedures Laplace ransform of he inegral of a funcion L 1 L f ( ) d f ( ) f ( ) d s Seps in using he Laplace ransformaion o solve DE s Find differenial equaions Find Laplace ransform of equaions Rearrange equaions in erms of variable of ineres Conver back ino ime domain o find resuling response (inverse ransform using ables)

45 Laplace Transform Shif Propery Noe hese shif properies in and s spaces... a e f F s a L f a ( a) e F s hus L as L L as ( ) 1 ( a) e

46 Example 3.4.3: compue he forced response of a spring mass sysem o a sep inpu using LT The equaion of moion is mx( ) kx( ) ( ) Taking he Laplace Transform (zero iniial condiions) 1 1 1/ m s s ms k s s 2 ( ms k) X ( s) X ( s) Taking he inverse Laplace Transform yields: 1/ m 1 x( ) 1- cos n 1- cos n k 2 n ( ) ( n ) Compare his o he soluion given in (3.18)

47 From Fourier series of non-periodic funcions Allow period o go o infiniy Similar o Laplace Transform Useful for random inpus j X ( ) x( ) e d Corresponding inverse ransform ( ) ( ) j x X e d 1 2 Fourier ransform of he uni impulse response is he frequency response funcion j H( ) h( ) e d Fourier Transform h( ) w n =2 and M= Time (s) Normalized H() (db) 1 Fourier Transform Frequency (Hz)

48 3.5 Random Vibraions So far our exciaions have been harmonic, periodic, or a leas known in advance These are examples of deerminisic exciaions, i.e., known in advance for all ime Tha is given we can predic he value of F() exacly Responses are deerminisic as well Many physical exciaions are nondeerminisic, or random, i.e., can wrie explici ime descripions Rockes Earhquakes Aerodynamic forces Rough roads and seas The responses x() are also nondeerminisic

49 Random Vibraions Saionary signals are hose whose saisical properies do no vary over ime Funcions are described in erms of probabiliies Mean values* Sandard deviaions Random oupus relaed o random inpu via sysem ransfer funcion *ie given we do no know x() exacly, bu raher we only know saisical properies of he response such as he average value

50 Auocorrelaion funcion and power specral densiy The auocorrelaion funcion describes how a signal is changing in ime or how correlaed he signal is a wo differen poins in ime. 1 T Rxx( ) lim x( ) x( ) d T T The Power Specral Densiy describes he power in a signal as a funcion of frequency and is he Fourier ransform of he auocorrelaion funcion. 1 j Sxx ( ) Rxx ( ) e d 2

51 Examples of signals A HARMONIC T Signal A rms RANDOM x() ime x() ime -A A 2 /2 T Auocorrelaion A 2 rms R xx () -A 2 /2 ime shif R xx () ime shif

52 Power Specral Densiy S xx (w) S xx (w) 1/T Frequency (Hz) Frequency (Hz)

53 More Definiions x Average : T T 1 lim x( ) d (3.47) T Mean-square: T T T x lim x ( ) d (3.48) rms: T T T xrms x lim x ( ) d (3.49)

54 Expeced Value (or ensemble average) The expeced value = T T x () E[ x( )] lim d x (3.63) T The Probabiliy Densiy Funcion, p(x), is he probabiliy ha x lies in a given inerval (e.g. Gaussian Disribuion) The expeced value is also given by E[ x] xp( x) dx (3.64)

55 Recall he Basic Relaionships for Transforms: Recall for SDOF ransfer funcion : G( s) 2 ms 1 cs k frequency responsefuncion : G( j) 1 H( ) 2 k m cj uni impulseresponse funcion : h( ) 1 m d e n sin d 1 L[ h( )] G( s) 2 ms cs k And he Fourier Transform of h() is H(ω)

56 Wha can you predic? The response of SDOF wih f() as inpu: Deerminisic Inpu: X ( s) G( s) F( s) x( ) h( ) f ( ) d Random Inpu: S ( ) H ( ) S ( ) xx 2 2 [ ] ( ) ff ( ) E x H S d 2 ff In a Lab, he PSD funcion of a random inpu and he oupu can be measured simply in one experimen. So he FRF can be compued as heir raio by a single es, insead of performing several ess a various consan frequencies. Here we ge an exac ime record of he oupu given an exac record of he inpu. Here we ge an expeced value of he oupu given a saisical record of he inpu.

57 Example PSD Calculaion Consider mx cx kx F( ), where he PSD of F( ) is consan S The corresponding frequency response funcion is: 1 H ( ) (2.59) 2 k m cj H ( ) k m c j k m c j k m c j ( k m ) c From equaion (3.62) he PSD of he response becomes: S H( ) S xx 2 ff S ( k m ) ( c)

58 Example Mean Square Calculaion Consider he sysem of Example and compue: E x S d 2 k mn cj m S kcm S kc Here he evaluaion of he inegral is from a abulaed value See equaion (3.7).

59 Secion 3.6 Shock Specrum Arbirary forms of shock are probable (earhquakes, ) The specrum of a given shock is a plo of he maximum response quaniy (x) agains he raio of he forcing characerisic (such as rise ime) o he naural period. Maximum response gives maximum sress. x( ) F( ) h( ) d (3.71)

60 Using he convoluion equaion as a ool, compue he maximum value of he response Recall he impulse response funcion undamped sysem: 1 h( ) sin n( ) (3.73) m n 1 x( ) max F( )sin ( ) (3.74) n d m max n Such inegrals usually have o be compued numerically

61 Example Compue he response specrum for gradual applicaion of a consan force F. Assume zero iniial condiions mx( ) kx( ) F( ) 1 =infiniy, means saic loading F() F 1 F( ) F ( ) F ( ) F ( ) 1 2 F 1 1 ime shif and negaive, like half sine problem F2 () 1 ( ) F The characerisic ime of he inpu

62 Spli he soluion ino wo pars and use he convoluion inegral n F F sinn x1 ( ) sin n( ) d k 1 k 1 n1 1 (3.75) For x 2 apply ime shif 1 F sin ( ) x ( ), 1 n k 1 n1 (3.76) F sinn F 1 sin n( 1) x( ) x1 ( ) x2( ) ( 1) k 1 n1 k 1 n1 (3.77)

63 Nex find he maximum value of his response To ge max response, differeniae x(). In he case of a harmonic inpu (Chaper 2) we compued his by looking a he coefficien of he seady sae response, giving rise o he Magniude plos of figures 2.8, 2.9, Need o look a wo cases 1) < 1 and 2) > 1 For case 2) solve: (wha abou case 1? Is max is Xsaic) d F d kn 1 n1 sinn sin n( 1)

64 Solve for a xmax, denoed p cos cos ( ) n n 1 p cos cos cos sin sin n p n p n 1 n p n cosn1 n an p sinn1 sin (1 cos ) 2 2 n 1 n 1 2 1cos n 1 1 cos n 1 n p sin n 1

65 From he riangle: sin n cos n p p 1 (1 cosn 1) 2 sin n 1 2(1 cos ) n 1 Minus sq roo aken as + gives a negaive magniude Subsiue ino x( p ) o ge nondimensional X max : x max F k cos n n 1 1 s erm is saic, 2 nd is dynamic. Plo versus: 1 1 T n 1 2 Inpu characerisic ime Sysem period

66 Response Specrum Indicaes how normalized max oupu changes as he inpu pulse widh increases. Very much like a magniude plo. Shows very small 1 can increase he response significanly: impac, raher han smooh force applicaion The larger he rise ime, he smaller he peaks The maximum displacemen is minimized if rise ime is a muliple of naural period Design by MiniMax idea X xmaxk F

67 Comparison beween impulse and harmonic inpus Impulse Inpu Transien Oupu Max ampliude versus normalized pulse frequency Harmonic Inpu Harmonic Oupu Max ampliude versus normalized driving frequency

68 Review of The Procedure for Shock Specrum 1. Find x() using convoluion inegral 2. Compue is ime derivaive 3. Se i equal o zero 4. Find he corresponding ime 5. Evaluae he max possible value of x (be careful abou poins where he funcion does no have derivaive!!) 6. Plo i for differen inpu shocks

69 3.7 Measuremen via Transfer Funcions Apply a sinusoidal inpu and measure he response Do his a small frequency seps The raio of he Laplace ransform of hese o signals hen gives and experimen ransfer funcion of he sysem

70 Several differen signals can be measured and hese are named X( s) 1 recepance: (3.86) 2 F() s ms cs k sx () s s mobiliy: (3.87) 2 F() s ms cs k 2 2 s X () s s inerance: (3.87) 2 F() s ms cs k

71 The magniude of he compliance ransfer funcion yields informaion abou he sysems parameers 1 H( j) (3.89) 2 2 ( k m) ( c) k 1 H( j ) (3.9) m c 1 H () (3.91) k n

72 3.8 Sabiliy Sabiliy is defined for he soluion of free response case: Sable: x( ) M, Asympoically Sable: Unsable: lim x ( ) if i is no sable or asympoically sable

73 Recall hese sabiliy definiions for he free response Sable Asympoically Sable Divergen insabiliy Fluer insabiliy

74 Sabiliy for he forced response: mx( ) cx( ) kx( ) F( ) Bounded Inpu-Bounded Oupu Sable x() bounded for ANY bounded F() Lagrange Sable wih respec o F() If x() is bounded for THE given F()

75 Relaionship beween sabiliy of he homogeneous sysem and he force response If x homo is Asympoically sable hen he forced response is BIBO sable (Bounded inpu, bounded oupu) If x homo is Sable hen he forced response MAY be Lagrange Sable or Unsable

76 Sabiliy for Harmonic Exciaions The soluion o: mx( ) kx( ) F cos is: v f f x( ) sinn x cos cos 2 2 n 2 2 n n n As long as ω n is no equal o ω his is Lagrange Sable, if he frequencies are equal i is Unsable

77 For underdamped sysems: f x 1 n p ( ) cos( an ) ( n ) (2 n) n X n x( ) Ae sin( ) X cos( ) d homogeneous or ransien soluion paricular or seady sae soluion mx( ) cx( ) kx( ) F cos 2 Add homogeneous and paricular o ge oal soluion: Bounded Inpu-Bounded Oupu Sable

78 Example sin cos sin 2 M m k mg Force from Spring momen arm force momen arm The equaion of moion afer a small angle approximaion is given becomes: 2 2 m k mg ( ) ( ) ( ) 2 2 m k mg ( ) ( ) ( ) This will be sable if and only if he coefficen of is posiive or if k mg The sysem is hus Lagrange sable. Physically his ells us he spring mus be large enough o overcome graviy

79 Find a force of he form F a b o make he sysem asympoically sable (BIBO) 2 2 m ( k mg ) a b 2 2 m b k mg a ( )) Choose b and a mg 2 2 m b k Then he sysem is asympoically sable and BIBO

80 3.9 Numerical Simulaion of he response As before in Secion 2.8 wrie equaions of moion as sae space equaions The Euler inegraion is jus A i i i i ) ( ) ( ) ( ) ( 1 F x x x

81 Example wih delay Le he inpu force be a sep funcion a = F F() F 3 N k 1 N/m n 2 s x() M F ( ) k c

82 Example Analyical versus numerical x e.316( ( ).3.3 ) cos[3.144( ).11] ( ) Response o sep inpu clear all %% Analyical soluion (example 3.2.1) Fo=3; k=1; wn=3.16; zea=.1; o=; hea=aan(zea/(1-zea^2)); wd=wn*sqr(1-zea^2); =:.1:12; x k c F x 2 x 2 ( ) m m m Heaviside=sepfun(,o);% define Heaviside Sep funcion for <<12 x = (Fo/k - Fo/(k*sqr(1-zea^2)) * exp(-zea*wn*(-o)) * cos (wd*(-o)- hea))*heaviside(-o); plo(,x); hold on %% Numerical Soluion xo=[; ]; s=[ 12]; [,x]=ode45('f',s,xo); plo(,x(:,1),'r'); hold off % funcion v=f(,x) Fo=3; k=1; wn=3.16; zea=.1; o=; m=k/wn^2; v=[x(2); x(2).*-2*zea*wn + x(1).*-wn^2 + Fo/m*sepfun(,o)]; x

83 Malab Code x=[;]; s=[ 12]; [,x]=ode45('func',s,x); plo(,x(:,1)).6.5 Displacemen (x) Time (s) funcion v=func(,x) F=3; k=1; wn=3.16; z=.1; =2; m=k/(wn^2); v=[x(2); x(2).*-2*z*wn+x(1).*-wn^2+f/m*sepfun(,)];

84 Problem 3.22 A wave consising of he wake from a passing boa impacs a seawall. I is desired o calculae he resuling vibraion. Figure P3.22 illusraes he siuaion and suggess a model. Calculae he resuling response.

85 Numerical soluion of Problem 3.22 %problem 3.19 m=1; E=3.8e9; A=.3; L=2; k=e*a/l; =.2; F=1; global F k m %numerical soluion x=[;]; s=[.5]; [,x]=ode45('f_3_19',s,x); plo(,x(:,1)) F() F () F (1 ) F funcion v=f_3_19(,x) global F k m A=x(2); F=(((1-./).*sepfun(,))-((1-./).*sepfun(,)))*F/m; B=(-k/m)*x(1)+F; v=[a; B];

86 P3.22

87 3.9 Nonlinear Response Properies Euler inegraion formula: x ( i 1) x( i ) F( x( i )) f( i ) Nonlinear erm Analyical soluions no available so we mus inerrogae he numerical soluion

88 Example cubic spring subjec o pulse inpu mx( ) cx( ) kx( ) k x ( ) 15 ( ) ( ) The sae space form is: x ( ) x ( ) x ( ) 2 x ( ) x ( ) x ( ) 15 ( ) ( ) n 2 n

89 Naure of Response Red (solid) is nonlinear response. Blue (dashed) is linear response Is here any jusificaion? Yes, hardening nonlinear spring. The firs par is due o IC.

90 Malab Code clear all xo=[.1; 1]; s=[ 8]; [,x]=ode45('f',s,xo); plo(,x(:,1)); hold on % The response of nonlinear sysem [,x]=ode45('f1',s,xo); plo(,x(:,1),'--'); hold off % The response of linear sysem % funcion v=f(,x) m=1; k=2; c=2; wn=sqr(k/m); zea=c/2/sqr(m*k); Fo=15; alpha=3; 1=1.5; 2=5; v=[x(2); x(2).*-2*zea*wn + x(1).*-wn^2 - x(1)^3.*alpha+ Fo/m*(sepfun(,1)- sepfun(,2))]; % funcion v=f1(,x) m=1; k=2; c=2; wn=sqr(k/m); zea=c/2/sqr(m*k); Fo=15; alpha=; 1=1; 2=5; v=[x(2); x(2).*-2*zea*wn + x(1).*-wn^2 - x(1)^3.*alpha+ Fo/m*(sepfun(,1)- sepfun(,2))];