maple非线性振动

1、已知：l，m，c，k，F=F0sin(t)求：系统的振动微分方程，0，=0时质点的振幅解：建模：质点作圆周运动，杆作定轴转动。Maple restart: J0:=m*l2: 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)=theta,eq):eq:=m*l2*DDtheta+4*c*l2*Dtheta+9*k*l2*theta=3*F0*sin(omega*t)*l: eq:=expand(eq/(m*l2); b:=h/sqrt(omega02-omega2)2+4*delta2*omega2): b1:=subs(omega=omega0,b); omega0:=sqrt(9*k/m); delta:=(2*c)/m: h:=(3*f0)/(m*l): B:=simplify(l*b1,symbolic);答：系统振动微分方程为+4cm+9km=3F0sin(t)ml， 0=3km，B=f0m4ck。2、已知：m=0.5kg，h=0.1m，k=0.8kN/m，=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),t$2)=DDx,eq): eq:=simplify(eq); X:=A*sin(omega0*t+theta): omega0:=sqrt(k/m): x0:=-delta0: v0:=sqrt(2*g*h): A:=sqrt(x02+(v0/omega0)2): theta:=arctan(omega0*x0/v0): m:=0.5: h:=0.1: k:=0.8e3: beta:=Pi/6: g:=9.8: omega0:=evalf(omega0,4); A:=evalf(A,4); theta:=evalf(theta,4); X:=eval(X);答：0=40rad/s，A=35.1mm，物块的运动方程为x=35.1sin(40t-0.087)mm。3、吸引子的仿真。以杜芬方程为例，杜芬方程表示如下x+cx+ax+=AcostMaple程序 restart: with(plots): de1:=diff(x(t),t)=y(t):de2:=diff(y(t),t)=-a*x(t)-b*x(t)3-c*y(t)+A*cos(Omega*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,numpoints=4000): duffplot;答：杜芬方程相图如图所示。 4、已知：l=l0-vt求：摆的运动方程解：建模：小球作平面运动 自由度f=1， 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(vrho2+vphi2): T:=1/2*m*vA2; T:=subs(diff(phi(t),t)=Dphi,phi(t)=phi,T): T:=collect(T,Dphi): TDphi:=diff(T,Dphi): Tphi:=diff(T,Dphi); TDphi:=subs(l=l0-v*t,Dphi=Dphi(t),TDphi); V:=-m*g*(l0-v*t)*cos(phi): Qphi:=-diff(V,phi); eq:=diff(TDphi,t)-Tphi-Qphi=0; eq:=subs(diff(Dphi(t),t)=DDphi,Dphi(t)=Dphi,eq): eq:=(l0-v*t)*DDphi-2*v*Dphi+g*sin(phi)=0;答：摆的运动微分方程为(l0-vt)-2v+gsin=0。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）， 质点作曲线运动 自由度f=1，取广义坐标。Maple程序 restart: J0:=m*r2: x0:=r*phi: x0:=subs(phi=phi(t),x0): v0:=diff(x0,t):vA:=sqrt(v02+v02-2*v02* cos(phi):T:=1/2*J0*diff(phi(t),t)2+1/2*m*v02+1/2*m*vA2: T:=simplify(T): T:=factor(T); V:=-m*g*r*cos(phi); L:=T-V; L:=subs(diff(phi(t),t)=Dphi,phi(t)=phi,L): L:=collect(L,Dphi): LDphi:=diff(L,Dphi); Lphi:=m*r2*Dphi*sin(phi)-m*g*r*sin(phi): LDphi:=subs(Dphi=diff(phi(t),t),LDphi): eq:=diff(LDphi,t)-Lphi=0:eq:=subs(diff(phi(t),t)=Dphi(t),diff(Dphi(t),t)=DDphi,eq);答：系统的微分方程为2(2-cos)+2+grsin=0。6、已知：R，r，m，J0，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，Maple程序 restart: xP:=r*theta:xP:=subs(theta=theta(t),xP): vP:=diff(xP,t); vOP:=r*omega; l:=phi: l:=subs(phi=phi(t),l): omega1:=diff(l,t): vO:=omega1*(R-r); eq:=vO-vP-vOP=0: SOL:=solve(eq,omega): omega:=subs(SOL,omega); JO:=2/5*m*r2:T:=1/2*JC*omega12+1/2*JO*omega2+1/2*m*vO: V:=-m*g*(R-r)*cos(phi); L:=T-V; 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): 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); Lphi:=diff(L,phi): LDphi:=diff(L,Dphi):LDphi:=subs(Dtheta=Dtheta(t),theta=theta(t),Dphi=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25mR(R-r)-J0+25mR2=0 75mR(R-r)-25R+gsin=07、已知：m1，m2，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Fs解：建模：轮A作平面运动，杆C作平行移动。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: SOL2:=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*R2,a): a:=simplify(a); FA:=subs(phi=Pi/6,l=2*R,J=1/2*m1*R2,FA): FA:=simplify(FA); FN:=subs(phi=Pi/6,l=2*R,J=1/2*m1*R2,FN); Fs:=subs(phi=Pi/6,l=2*R,J=1/2*m1*R2,Fs): Fs:=simplify(Fs); fs:=FN/Fs: fs:=simplify(fs);答：使B端刚好离开地面的F=3g(3m1+2m2)2保证纯滚动时，Fs3m12(3m1+m2)8、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。Maple程序 restart: J0:=1/3*m*(2*a)2: T1:=0: T2:=1/2*J0*omega2: 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,omega2,omega1): Y:=subs(SOL1,Y); omega1:=subs(SOL1,omega1); omega:=subs(SOL1,omega2): 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); YOy:=subs(SOL2,YOy); eq6:=YOx=0: SOL3:=solve(eq6,l): l:=subs(SOL3,l);答：Iox=m1+k(4a-3l)6l6ag，Ioy=0，碰撞中心位置l=4a/3。9、已知：最高摆角，m1，m2，J0，d，h求：子弹的速度v。解：建模：碰撞过程中，系统满足动量守恒，上摆过程中，系统满足机械能守恒。Maple程序 restart: LO1:=m2*v*d:LO2:=J0*omega+m2*d2*omega: eq1:=LO1=LO2:T1:=1/2*J0*omega2+1/2*m2*d2*omega2: T2: