MATHEMATICS_NUMERICAL-GRAPHICAL SOLUTIONS OF THE NON-LINEAR VIBRATION MODEL

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NUMERICAL - GRAPHICAL SOLUTIONS OF THE NON-LINEAR VIBRATION MODEL with discrete data input

CO HONG by CO.H . TRAN University of Natural Sciences, TRAN

Digitally signed by CO HONG TRAN DN: cn=CO HONG TRAN, c=VN, o=VNU-HCM, ou=MMI, email=cohtran@math.com Date: 2008.01.01 19:16:58 +07'00'

HCMC Vietnam coth123@math.com & coth123@yahoo.com Copyright 2007 May 06 2007 Abstract : The system of non-linear differential quations with discrete input function is solved by Runge-Kutta method . Subjects: Vibration Mechanics , The Differential equations . NOTE: This worksheet demonstrates Maple's capabilities in the design and finding the numerical solution of the non-linear vibration system .

All rights reserved. Copying or transmitting of this material without the permission of the authors is not allowed .

LOI GIAI SO VA DO THI CUA MAU DAO DONG PHI TUYEN voi so lieu roi rac TRAN HONG CO Dai hoc Khoa hoc tu nhien tp HCM Vietnam coth123@math.com & coth123@yahoo.com

Step 1 : System Definition. > restart: eq1:=(m1+m2)*Diff(y,t$2)*l*cos(phi)+(m1*l^2+J)*Diff(phi,t$2)+m1*g*l*cos(phi)=f(t);eq2:=(m1+m2)*Diff(y,t$2)+m1*l*cos(phi) *Diff(phi,t$2)-m1*l*Diff(phi,t)^2*cos(phi)+b*Diff(y,t)+c1*y+c3*y^3 =h(t);

(1.1)


(1.1)

> with(plots): readlib(spline): with(inttrans):

Warning, the name changecoords has been redefined

Step 2 : Fitting the experimental data by Spline function. > eq1:=(m1+m2)*Diff(y,t$2)*l*cos(phi)+(m1*l^2+J)*Diff(phi,t$2)+m1*g*l*cos(phi)=f(t);eq2:=(m1+m2)*Diff(y,t$2)+m1*l*cos(phi) *Diff(phi,t$2)-m1*l*Diff(phi,t)^2*cos(phi)+b*Diff(y,t)+c1*y+c3*y^3 =h(t);

(2.1)

(2.1)

> datax1:=[0,0.5,1,1.5,2,2.5,3,3.5,4]; datay1:=[0.2,0.5,0.7,0.4,0.65,1.2,2.4,0.9,1.1]; pldataf:= zip((x,y)->[x,y], datax1, datay1): dataplot1 := pointplot(pldataf, symbol=diamond);

(2.2)

(2.2)

(2.2)

> Ft:=spline(datax1, datay1, w, cubic);


(2.3)

> dothif:=plot(Ft, w=0..5, color=red): display(dataplot1,dothif, axes=frame);

By CO HONG TRAN at 7:19 pm, Jan 01, 2008

> fnum:=subs(w=t,Ft);eq1:=subs(f(t)=fnum,eq1);


(2.4)

> datax2:=[0,0.5,1,1.5,2,2.5,3,3.5,4]; datay2:=[0.3,0.5,0.58,0.4,0.85,1.2,1.4,0.9,1.55]; Ht := zip((x,y)->[x,y], datax2, datay2): dataplot2 := pointplot(Ht, symbol=cross);

(2.5)

(2.5)

(2.5)

> Ht:=spline(datax2, datay2, w, cubic);


(2.6)

> dothih:=plot(Ht, w=0..5, color=blue): display(dataplot2,dothih, axes=frame);

> h1:=subs(w=t,Ht);eq2:=subs(h(t)=h1,eq2);


(2.7)

Step 3 : The non-linear vibration system with discrete data input . > T:=5;m1:=1; m2:=1; b:=5; c1:= 1;c3:=1 ; l:= 0.05 ; J:= 0.5 ; g:=9.8;n:=2;

> with(DEtools):with(plots): alias(y=y(t), phi=phi(t), y0=y(0),p0=phi(0), yp0=D(y)(0),pp0=D(phi)(0));

Digitally signed by COHONGTRAN DN: cn=COHONGTRAN, c=US, o=MMI, ou=NCU HUI, email=cohtran@math.com Reason: I am the author of this document Location: HCMC Date: 2009.06.21 10:40:44 +07'00'

eq1 := .10*Diff(y,`$`(t,2))*cos(phi)+.5025*Diff(phi,`$`(t,2))+.490*cos(phi) = PIECEWISE([.2000000000+.559628129599999968*t+.161487481600000010*t^3, t < .5],[.1596281296+.680743740799999997*t+.242231222385861644*(t-.5)^2-1.60743740800000001*(t-.5)^3, t < 1],[.9826030927.282603092700000002*t-2.16892488954344652*(t-1)^2+3.06826215099999988*(t-1)^3, t < 1.5],[.6254970546.150331369699999995*t+2.43346833578792321*(t-1.5)^2-2.26561119299999980*(t-1.5)^3, t < 2],[.5178571430+.583928571299999977*t-.964948453608247214*(t-2)^2+3.99418262299999994*(t-2)^3, t < 2.5],[5.336542708+2.61461708300000017*t+5.02632547864506574*(t-2.5)^2-10.9111192900000002*(t-2.5)^3, t < 3],[4.027190721.542396906999999984*t-11.3403534609720165*(t-3)^2+12.8502945499999992*(t-3)^3, t < 3.5],[8.7576030922.24502945500000006*t+7.93508836524300420*(t-3.5)^2-5.29005890999999994*(t-3.5)^3, otherwise]); eq2 := 2*Diff(y,`$`(t,2))+.5e-1*cos(phi)*Diff(phi,`$`(t,2))-.5e-1*Diff(phi,t)^2*cos(phi)+5*Diff(y,t)+y+y^3 = PIECEWISE([.3000000000+.401389911599999982*t-.555964654000000014e-2*t^3, t < .5],[.3013899116+.397220176799999990*t.833946980854194387e-2*(t-.5)^2-.932201767300000039*(t-.5)^3, t < 1],[.8902706185-.310270618499999984*t1.40664212076583217*(t-1)^2+2.61436671600000015*(t-1)^3, t < 1.5],[.342065540e1+.243862297300000003*t+2.51490795287187030*(t-1.5)^2-2.40526509599999994*(t-1.5)^3, t < 2],[1.059642857+.954821428699999974*t-1.09298969072164964*(t-2)^2+1.16669366700000010*(t-2)^3, t < 2.5],[.642129970+.736851987999999958*t+.657050810014727760*(t-2.5)^2-2.66150957300000002*(t-2.5)^3, t < 3],[3.206688144-


.602229381300000033*t-3.33521354933726099*(t-3)^2+5.07934462299999990*(t-3)^3, t < 3.5],[1.347770617.127934461999999998*t+4.28380338733431554*(t-3.5)^2-2.85586892500000022*(t-3.5)^3, otherwise]); G:=dsolve({eq1,eq2,y0=0,p0=0,yp0=0,pp0=0.1},[y,phi],'numeric'): print(" Loi giai so bang phuong phap RUNGE - KUTTA "); for i from 0 to T do print(G(i)); od; yy:=t-> rhs(G(t)[2]): pp:=t-> rhs(G(t)[4]): yyp:=t->rhs(G(t)[3]):ppp:=t->rhs(G(t)[5]):plot(yy,0..n*T,color=red,thickness=3,title=`tung do y(t)`); plot(pp,0..n*T,color=blue,thickness=3,title=`goc phi phi(t)`); plot(yyp,0..n*T,color=green,title=`daohamtungdo y'(t)`); plot(ppp,0..n*T,color=black,title=`daohamgocphi phi'(t)`);

By CO HONG TRAN at 7:19 pm, Jan 01, 2008



CO HONG TRAN

Digitally signed by CO HONG TRAN DN: cn=CO HONG TRAN, c=VN, o=VNUHCM, ou=MMI, email=cohtran@math.com Date: 2008.01.01 19:20:03 +07'00'


By CO HONG TRAN at 7:20 pm, Jan 01, 2008


By CO HONG TRAN at 7:20 pm, Jan 01, 2008


Animation Code

Activate the following procedure twice to obtain the result completely . (use Maple 9.5 & 10 ) Mohinh procedure

> mohinh:=proc(M,lan)

> mohinh(0.75,3);

By CO HONG TRAN at 7:21 pm, Jan 01, 2008


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CO HONG TRAN

Digitally signed by CO HONG TRAN DN: cn=CO HONG TRAN, c=VN, o=VNUHCM, ou=MMI, email=cohtran@math.com Reason: I am the author of this document Date: 2008.01.01 19:21:38 +07'00'


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