By COHONGTRAN at 10:53 am, Jun 21, 2009
THE NON-LINEAR VIBRATION MODEL by CO.H . TRAN - University of Natural Sciences , HCMC Vietnam coth123@math.com & coth123@yahoo.com 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 coth123@math.com & coth123@yahoo.com
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 ]
⎛ v2 ⎞ ⎛ v2 ⎞ 2 eq1 := (m1 C m2 ) ⎜ 2 y⎟ l cos(f) C 0 m1 l C J1 ⎜ 2 f⎟ C m1 g l cos(f) = 0 ⎝ vt ⎠ ⎝ vt ⎠
⎛ v2 ⎞ ⎛ v2 ⎞ eq2 := (m1 C m2 ) ⎜ 2 y⎟ C m1 l cos( f ) ⎜ 2 f⎟ K m1 l ⎝ vt ⎠ ⎝ vt ⎠ v 3 y C c1 y C c3 y = 0 Cb vt
0 1 v f vt
2
cos( f )
0 1
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
By COHONGTRAN at 10:53 am, Jun 21, 2009
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],'numeri c'): 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,ti tle=`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
⎛ v2 ⎞ ⎛ v2 ⎞ 0.10 ⎜ 2 y⎟ cos(f) C 0.5025 ⎜ 2 f⎟ C 0.490 cos(f) = 0 ⎝ vt ⎠ ⎝ vt ⎠ ⎛ v2 2⎜ 2 ⎝ vt
⎞ ⎛ v2 ⎞ y⎟ C 0.05 cos(f) ⎜ 2 f⎟ K 0.05 ⎠ ⎝ vt ⎠
0 vtv f1
2
cos(f) C 5
0 vtv y1 C y C y
" Loi giai so bang phuong phap RUNGE - KUTTA "
v v ⎤ ⎡ y = 0., f = 0., f = 0.10000000000000 ⎥ ⎢ t = 0., y = 0., ⎦ ⎣ vt vt
3
=0
⎡ v y = 0.0111722381727157004, ⎢ t = 1., y = 0.00675589433093451692, vt ⎣ v ⎤ f = K.387008927232790512, f = K.864467159209224345 ⎥ vt ⎦ ⎡ v y = 0.00495727202327608722, ⎢ t = 2., y = 0.0176180540970506220, vt ⎣ v ⎤ f = K1.59839728564964444, f = K1.40007678027131388 ⎥ vt ⎦ ⎡ v y = K0.00455738471153587736, ⎢ t = 3., y = 0.0146475314199627280, vt ⎣ v ⎤ f = K2.78819743056849978, f = K.829006565942576712 ⎥ vt ⎦ ⎡ v y = 0.00508244661411777228, ⎢ t = 4., y = 0.0155584290750635548, vt ⎣ v ⎤ f = K3.13736547759258189, f = 0.137786285991567892 ⎥ vt ⎦ ⎡ v y = K0.00100009422503042136, ⎢ t = 5., y = 0.0191731202686913176, vt ⎣ v ⎤ f = K2.52031087032214485, f = 1.06960026754938830 ⎥ vt ⎦
> 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,thic kness=2);od; " Do thi dao ham ",
d y(t), " voi :", t = 1, s dt
" Do thi dao ham ",
d y(t), " voi :", t = 2, s dt
" Do thi dao ham ",
d y(t), " voi :", t = 3, s dt
" Do thi dao ham ",
d y(t), " voi :", t = 4, s dt
" Do thi dao ham ",
d y(t), " voi :", t = 5, s dt
> 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',co lor=black,thickness=2,title='daohamhamphi');od; " Do thi dao ham ",
d f(t), " voi :", t = 1, s dt
" Do thi dao ham ",
d f(t), " voi :", t = 2, s dt
" Do thi dao ham ",
d f(t), " voi :", t = 3, s dt
" Do thi dao ham ",
d f(t), " voi :", t = 4, s dt
" Do thi dao ham ",
d f(t), " voi :", t = 5, s dt
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Activate the following procedure twice to obtain the result completely . ( in Maple 9.5 & 10 ) Animation Code > mohinh(3,5);
By COHONGTRAN at 10:59 am, Jun 21, 2009
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.
Digitally signed by COHONGTRAN DN: cn=COHONGTRAN, c=VN, o=MMI, ou=NCU HUI, email=cohtran@math.com Reason: I am the author of this document Location: HCMC Date: 2009.06.21 10:52:02 +07'00'