maple非线性振动

1、1、已知：l, m, c, k, F=sin(0Jt)求：系统的振动微分方程,口时质点的振幅解：建模：质点作圆周运动，杆作定轴转动。 Maple> restart:FmcJ0:=m*1A2:Fe:=-k*3*l*theta(t): >Fd:=-c*2*l*diff(theta(t),t):> F:=F0*sin(omega*t): >eq:=J0*diff(theta(t),t$2)=Fd*2*l+Fe*3*l+F*3*l:>eq:=subs(diff(theta(t),t$2)=DDtheta, diff(theta(t),t)=Dtheta, theta(t)

2、=theta,eq):>eq:=m*lA2*DDtheta+4*c*lA2*Dtheta+9*k*lA2*theta= 3*F0*sin(omega*t)*l:> eq:=expand(eq/(m*lA2);eq := DDtheta4 c Dtheta 9 k 3F0 sin( t)m lb:=h/sqrt(omega0A2-omegaA2)A2+4*deltaA2*omegaA2): b1:=subs(omega=omega0,b);> omega0:=sqrt(9*k/m);1 b1 := 4I ,o k :=3 . 一mdelta:=(2*c)/m:h:=(3*f0)

3、/(m*l):B:=simplify(l*b1,symbolic);B := 1 f0 m:4 c k答：系统振动微分方程为占+丝+ &=3in一 2 %3。一 3 mB=2、已知：m=0.5kg, h=0.1m, k=0.8kN/m ,户=30求：系统的固有频率和振幅，物块的运动方程。 Maple程序> restart:> delta0:=m*g*sin(beta)/k:>eq:=m*diff(x(t),t$2)= m*g*sin(beta)- k*(delta0+x):> eq:=lhs(eq)-rhs(eq)=0:>eq:=subs(diff(x(t

4、),t$2)=DDx, eq):> eq:=simplify(eq);eq := m DDx x k> X:=A*sin(omega0*t+theta):> omega0:=sqrt(k/m):> x0:=-delta0:> v0:=sqrt(2*g*h):> A:=sqrt(x0A2+(v0/omega0)A2):> theta:=arctan(omega0*x0/v0):> m:=0.5:> h:=0.1:> k:=0.8e3:> beta:=Pi/6:0 := 40.00> g:=9.8:> omega0:=e

5、valf(omega0,4);> A:=evalf(A,4);A := .03513> theta:=evalf(theta,4);:=-.08724> X:=eval(X);X := .03513 sin( 40.001 .08724)答： =A=35.1mm,物块的运动方程为 x=35.1sin(40t-0.087)mm。+ax+ b x3 =Acos t3、吸引子的仿真。以杜芬方程为例，杜芬方程表示如下+c Maple程序> restart:> with(plots):> de1:=diff(x(t),t)=y(t):>de2:=diff(y(t

6、),t)=-a*x(t)-b*x(t)A3-c*y(t)+A*cos(Omeg a*t):> a:=-1:b:=1:c:=0,15:A:=0.3:Omega:=1:>duffing:=dsolve(de1,de2,y(0)=-0.5,x(0)=-1,x(t),y(t ),type=numeric,method=lsode): >duffplot:=odeplot(duffing,x(t),y(t),0.200,numpoint s=4000):> duffplot;答：杜芬方程相图如图所示4、已知：l°比求：摆的运动方程解：建模：小球作平面运动 自由度f=1

7、,取广义坐标 Maple程序> restart:> xrho:=l:> xphi:=l*phi:> xrho:=subs(l=l(t),xrho):> xphi:=subs(phi=phi(t),xphi):> vrho:=diff(xrho,t):> vphi:=diff(xphi,t):> V:=vector(vrho,vphi):> vA:=sqrt(vrhoA2+vphiA2):> T:=1/2*m*vAA2;2212T ：= 2m tl(t) l t (t)> T:=subs(diff(phi(t),t)=Dphi,p

8、hi(t)=phi,T):> T:=collect(T,Dphi):> TDphi:=diff(T,Dphi):> Tphi:=diff(T,Dphi);T := 0> TDphi:=subs(l=l0-v*t,Dphi=Dphi(t),TDphi);2TDphi := m(lo vt)Dphi > V:=-m*g*(l0-v*t)*cos(phi):> Qphi:=-diff(V,phi);Q := m g (l0 v t) sin()> eq:=diff(TDphi,t)-Tphi-Qphi=0;2eq := 2 m (l0 v t) Dphi (

9、t) v m (l0 v t) t Dphi (t) m g (l0v t) sin( ) 0> eq:=subs(diff(Dphi(t),t)=DDphi,Dphi(t)=Dphi,eq):> eq:=(l0-v*t)*DDphi-2*v*Dphi+g*sin(phi)=0;eq := (l0 vt) DDphi 2 v Dphi g sin( ) 0答：摆的运动微分方程为0-讥痴-2丽+妙” =°。5、已知：m, r求：系统的微分方程解：建模：圆环作平面运动(纯滚动)，二质点作曲线运动./ 自由度f=1 ,取广义坐标中。:1r'1IK Maple 程序

10、9;、上.，> restart:Tmg > J0:=m*rA2:一二> x0:=r*phi:mg> x0:=subs(phi=phi(t),x0):> v0:=diff(x0,t):>vA:=sqrt(v0A2+v0A2-2*v0A2* cos(phi):>T:=1/2*J0*diff(phi(t),t)A2+1/2*m*v0A2+1/2*m*vAA2:> T:=simplify(T):> T:=factor(T);2_2T := m r (t)( 2 cos()> V:=-m*g*r*cos(phi);V := mgr cos()&

11、gt; L:=T-V;2,2L := m r (t)( 2 cos( ) mgr cos()> L:=subs(diff(phi(t),t)=Dphi,phi(t)=phi,L):> L:=collect(L,Dphi):> LDphi:=diff(L,Dphi);一 2 一一LDphi := 2 m r Dphi ( 2 cos()> Lphi:=m*rA2*Dphi*sin(phi)-m*g*r*sin(phi):> LDphi:=subs(Dphi=diff(phi(t),t),LDphi):> eq:=diff(LDphi,t)-Lphi=0:>

12、;eq:=subs(diff(phi(t),t)=Dphi(t),diff(Dphi(t),t)=DDphi,eq);22eq := 2 m r DDphi ( 2 cos( ) m r Dphi sin( ) mgr sin( ) 0答：系统的微分方程为2(2-ms0广十例n3 二。6、已知：R, r, m,圆球C作纯滚动求：系统的运动微分方程解：建模：圆槽绕O作定轴转动, 小球C作平面运动，自由度f=2,取广义坐标”，隼 Maple程序> restart:> xP:=r*theta:>xP:=subs(theta=theta(t), xP):> vP:=diff(x

13、P,t);vp:= r t (t)> vOP:=r*omega;Vop：=t (t) R t (t) r r t (t)> l:=phi:> l:=subs(phi=phi(t),l):> omega1:=diff(l,t):> vO:=omega1*(R-r);Vo :=t (t) (R r)> eq:=vO-vP-vOP=0:> SOL:=solve(eq,omega):> omega:=subs(SOL,omega);t (t) R t (t) r r t (t)r> JO:=2/5*m*rA2:>T:=1/2*JC*omega

14、1A2+1/2*JO*omegaA2+1/2*m*vOA:> V:=-m*g*(R-r)*cos(phi);V := m g (R r) cos()> L:=T-V;2211L := 2Jc t (t)5m t (t) R t (t) r r t (t)21 22 m t (t) (R r) mg(R r) cos()> L:=simplify(L):>L:=subs(diff(theta(t),t)=Dtheta,theta(t)=theta, diff(phi(t),t)=Dphi,phi(t)=phi,L):> L:=collect(L,Dtheta):&g

15、t; LDtheta:=diff(L,Dtheta):> Ltheta:=diff(L,theta):> LDtheta:=subs(Dtheta=Dtheta(t),LDtheta):> eq1:=diff(LDtheta,t)-Ltheta=0:> eq1:=subs(diff(Dtheta(t),t)=DDtheta,eq1);eq1 := m r2 DDtheta05> Lphi:=diff(L,phi):> LDphi:=diff(L,Dphi):>LDphi:=subs(Dtheta=Dtheta(t), theta=theta(t),Dp

16、hi=Dphi(t),LDphi):> eq2:=diff(LDphi,t)-Lphi=0:>eq2:=subs(diff(Dtheta(t),t)=DDtheta, Dtheta(t)=Dtheta, diff(Dphi(t),t)=DDphi, Dphi(t)=Dphi,eq2):> eq2:=simplify(eq2):答：系统运动微分方程为mR(R-r)。一仇十M铲A = ° 72- c/c、；田B + dsin 9 - mR(R-r)W-5" ' =07、已知：叫，R, l=2R,轮A作纯滚动求：F为多大方可使B端离开地面，保证纯滚动时1

17、解：建模：轮A作平面运动，杆C作平行移动Mia' Maple程序> restart:> alpha:=a/R:> FIC:=m2*a:> FIA:=m1*a:> MIA:=J*alpha:>eq1:=FIC*l/2*sin(phi)-m2*g*l/2*cos(phi)=0:> SOL1:=solve(eq1,a):>eq2:=FA*R-FIA*R-MIA-FIC*l/2*sin(phi)- m2*g*l/2*cos(phi):> eq3:=FA-Fs-FIA-FIC=0:> eq4:=FN-m1*g-m2*g=0:> S

18、OL2:=solve(eq2,eq3,eq4,FA,FN,Fs):> a:=subs(SOL1,a):> FA:=subs(SOL2,FA):> FN:=subs(SOL2,FN):> Fs:=subs(SOL2,Fs):> a:=subs(phi=Pi/6,l=2*R,J=1/2*m1*RA2,a):> a:=simplify(a);a := g 3> FA:=subs(phi=Pi/6,l=2*R,J=1/2*m1*RA2,FA):> FA:=simplify(FA);1 一FA := 2 g 3 (2 m23 m1)> FN:=sub

19、s(phi=Pi/6,l=2*R,J=1/2*m1*RA2,FN);Fn := mi g m2 g> Fs:=subs(phi=Pi/6,l=2*R,J=1/2*m1*RA2,Fs):> Fs:=simplify(Fs);L1CFs := 2 mi g 3> fs:=FN/Fs:> fs:=simplify(fs);2 (mm2) 3f :=s 3 m1+ 2m2)答：使B端刚好离开地面的F= 2保证纯滚动时,% -市吗+ mJ8、已知：m, 2a,恢复因数k求：轴承的碰撞冲量，碰撞中心的位置解：建模：杆在碰撞中作定轴转动，满足动量定理和动量矩守恒定理，碰撞前机械能守恒

20、。 Maple程序> restart:> J0:=1/3*m*(2*a)A2:> T1:=0:> T2:=1/2*J0*omegaA2:> V1:=0:> V2:=-m*g*a:> eq1:=T2+V2=T1+V1:> LO1:=-J0*omega:> LO2:=J0*omega1:> eq2:=LO2-LO1=Y*l:> v:=l*omega:> v1:=l*omega1:> eq3:=k=v1/v:> SOL1:=solve(eq1,eq2,eq3,Y,omegaA2,omega1):> Y:=sub

21、s(SOL1,Y);24 m a (k 1) Y:= 3 l> omega1:=subs(SOL1,omega1);:=k> omega:=subs(SOL1,omegaA2):：=26> omega:=sqrt(omega);> Px1:=m*omega*a:> Px2:=-m*omega1*a:> eq4:=Px2-Px1=YOx-Y:> eq5:=0=YOy:> SOL2:=solve(eq4,eq5,YOx,YOy):> YOx:=subs(SOL2,YOx);fA m 6 g a ( 3 kl 3 l 4 a k 4 a)1aYOx := 6l> YOy:=sub