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THE NON-LINEAR VIBRATION MODEL
by CO.H . TRAN - University of Natural Sciences , HCMC Vietnam -
[email protected] & [email protected]
Copyright 2006 Feb 06 2006 -------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------
** Abstract : The system of non-linear differential quations 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 TRAN HONG CO - Dai hoc Khoa hoc tu nhien - tp HCM Vietnam [email protected] & [email protected]
A . Xac dinh he thong . [ System Definition ]
B. Mo hinh dao dong . [ Vibration Model ]Khao sat mau vat the dang giai tich co hinh mo phong nhu tren [ Consider an analytical model which has the simulation figure above ] He phuong trinh vi phan chuyen dong : [ System of differential equations ]
eq1 := (m1 Cm2 ) ⎛⎜⎝v2
vt2 y⎞⎟⎠
l cos(f) C 0m1 l2 C J1 ⎛⎜⎝v2
vt2 f⎞⎟⎠Cm1 g l cos(f) = 0
eq2 := (m1 Cm2 ) ⎛⎜⎝v
2
vt2 y⎞⎟⎠Cm1 l cos(f)
⎛⎜⎝v
2
vt2 f⎞⎟⎠Km1 l 0 v
vt f1
2 cos(f)
C b 0 vvt
y1 C c1 yC c3 y3 = 0
Xac dinh cac dieu kien dau . [ Define initial conditions ] : y0, p0, yp0, pp0
Thay cac gia tri cua tham so m1,m2 , b , c1 , c3 , l , g , J . [ Substitute the parameter values m , b , c1 , c3 , l ,g , J ] . > interface(warnlevel=0):
C . Chuong trinh tinh toan mo hinh dao dong phi tuyen . [ Calculation procedures for the vibration model ]
Bang hoat trinh Maple nay gom 2 phan . Ví du ve cach su dung trong cac bai toan thuc te , hay xem phan tiep sau . * (khoi luong m1 , m2 , hang so can nhot b, he so lo xo c1, he so lo xo c3, solan T,chdai l ,moment J , giatoc trong truong g) This Maple worksheet contains 2 parts. For examples of applying them to real problems, see the following action . * (mass m1 , m2 ,viscous damping constant b,spring constant c1, spring constant c3,number of points , moment J ,acceleration of gravity g )
> restart;T:=5;m1:=1; m2:=1; b:=5; c1:= 1;c3:=1 ; l:= 0.05 ; J:= 0.5 ; g:=9.8;
5
1
1
5
1
1
0.05
0.5
9.8
> 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)); eq1:=(m1+m2)*Diff(y,t$2)*l*cos(phi)+(m1*l^2+J)*Diff(phi,t$2)+m1*g*l*cos(phi)=0;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 =0; 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..T,0..0.05,color=red,thickness=3,title=`tung do y(t)`); plot(pp,0..T,-3.5..0.5,color=blue,thickness=3,title=`goc phi phi(t)`);plot(yyp,0..T,color=green,title=`daohamtungdo y'(t)`); plot(ppp,0..T,color=black,title=`daohamgocphi phi'(t)`); ;;;;;;;;;;;;;;;;;;;;;;
y, f, y0, p0, yp0, pp0
0.10 ⎛⎜⎝v
2
vt2 y⎞⎟⎠
cos(f) C 0.5025 ⎛⎜⎝v
2
vt2 f⎞⎟⎠C 0.490 cos(f) = 0
2 ⎛⎜⎝v
2
vt2 y⎞⎟⎠C 0.05 cos(f)
⎛⎜⎝v
2
vt2 f⎞⎟⎠K 0.05 0 vvt
f12 cos(f) C 5 0 vvt
y1 C yC y3 = 0
" Loi giai so bang phuong phap RUNGE - KUTTA "
⎡⎢⎣
t = 0., y = 0., v
vt y = 0., f = 0., v
vt f = 0.10000000000000⎤⎥⎦
⎡⎢⎣
t = 1., y = 0.00675589433093451692, v
vt y = 0.0111722381727157004,
f = K.387008927232790512, v
vt f = K.864467159209224345 ⎤⎥⎦
⎡⎢⎣
t = 2., y = 0.0176180540970506220, v
vt y = 0.00495727202327608722,
f = K1.59839728564964444, v
vt f = K1.40007678027131388 ⎤⎥⎦
⎡⎢⎣
t = 3., y = 0.0146475314199627280, v
vt y = K0.00455738471153587736,
f = K2.78819743056849978, v
vt f = K.829006565942576712 ⎤⎥⎦
⎡⎢⎣
t = 4., y = 0.0155584290750635548, v
vt y = 0.00508244661411777228,
f = K3.13736547759258189, v
vt f = 0.137786285991567892 ⎤⎥⎦
⎡⎢⎣
t = 5., y = 0.0191731202686913176, v
vt y = K0.00100009422503042136,
f = K2.52031087032214485, v
vt f = 1.06960026754938830 ⎤⎥⎦
> interface(warnlevel=0):> ;for k from 1 to T do print(" Do thi ham ",y(t)," voi :",t=k,s);plot(yy,0..k,thickness=4,title='hamy');od;
" Do thi ham ", y( t) , " voi :", t = 1, s
" Do thi ham ", y( t) , " voi :", t = 2, s
" Do thi ham ", y( t) , " voi :", t = 3, s
" Do thi ham ", y( t) , " voi :", t = 4, s
" Do thi ham ", y( t) , " voi :", t = 5, s
> for k from 1 to T do print(" Do thi dao ham ",diff(y(t),t)," voi :",t=k,s);plot(yyp,0..k,title='daohamhamy',color=green,thickness=2);od;
" Do thi dao ham ", ddt
y( t), " voi :", t = 1, s
" Do thi dao ham ", ddt
y( t), " voi :", t = 2, s
" Do thi dao ham ", ddt
y( t), " voi :", t = 3, s
" Do thi dao ham ", ddt
y( t), " voi :", t = 4, s
" Do thi dao ham ", ddt
y( t), " voi :", t = 5, s
> for k from 1 to T do print(" Do thi ham ",phi(t)," voi :",t=k,s);plot(pp,0..k,thickness=4,color=blue,title='hamphi');od;
" Do thi ham ", f( t) , " voi :", t = 1, s
" Do thi ham ", f( t) , " voi :", t = 2, s
" Do thi ham ", f( t) , " voi :", t = 3, s
" Do thi ham ", f( t) , " voi :", t = 4, s
" Do thi ham ", f( t) , " voi :", t = 5, s
> for k from 1 to T do print(" Do thi dao ham ",diff(phi(t),t)," voi :",t=k,s);plot(ppp,0..k,color=black,title='daohamhamphi',color=black,thickness=2,title='daohamhamphi');od;
" Do thi dao ham ", ddt
f( t), " voi :", t = 1, s
" Do thi dao ham ", ddt
f( t), " voi :", t = 2, s
" Do thi dao ham ", ddt
f( t), " voi :", t = 3, s
" Do thi dao ham ", ddt
f( t), " voi :", t = 4, s
" Do thi dao ham ", ddt
f( t), " voi :", t = 5, s
------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------ Activate the following procedure twice to obtain the result completely . ( in Maple 9.5 & 10 )Animation Code
> mohinh(3,5);
Legal Notice: The copyright for this application is owned by Maplesoft. The application is intended to demonstrate the use of Maple to solve a particular problem. It has been made available for product evaluation purposes only and may not be used in any other context without the express permission of Maplesoft.
