2011《非线性振动》试题解答解析

华中科技大学研究生课程考试试卷课程名称： 非线性振动与控制 课程类别 □公共课专业课 考核形式开卷□闭卷学生类别 考试日期 2011. 12. 21 学生所在院系 学号 姓名 任课教师题1：（20分）Consider the motion of a particle of mass m sliding freely on a wire described by parabola2z px = which rotates about the z -axis as shown in Figure 1. We assume that the wire is weightless and that its angular velocity Ωis changing with the position of the mass along the wire. There is no outside influence acting on the wire. (a) Show that the equations of motion are 20x x Ω+Ω=and 22222(14)4(2)0p x x p x x p g x +++-Ω=2x Ω=wherea constant of integration (essentiallythis is a statement of conservation of angularmomentum) and that the governing equation for x can be written in the form22224(14)4(2)0Hp x x p x x pg x x+++-= (c) Discuss the motion of the mass along parabola. Show that the motion is always bounded in this system.(d) For 1, 32.2, 1000p g h ===and 12H =, plot the trajectories in the phase plane.【注：这里g 为重力加速度，32.2g =这一值的单位为2ft /s 。

】题2：（20分）Determine the singular points and their types for the system2252x x y y xy =+-=-Figure 1 Particle on a rotating parabolaSketch the trajectories and the separatrices in the state plane.题3：（20分）Consider the motion of a system governed by230123(2||)0u u u u u u ωεμμμ++++=where 1ε<<.(a) Show that0cos()()u a t O ωβε=++where2231203043()38a a a a εμμωμωπ=-++ (Note that 3μ must be positive for a realistic system.)(b) Determine the stationary motions and their stability as a function of the magnitudes andthe signs of1μand 2μ.题4：（20分）Consider the system governed by2sin 2cos K t θθμθθ++=Ω(a) When Ω is near unity, show that for small but finite amplitudes of the response3202()cos[()]()a T T T O θεβε=++where 31sin 4a a k μγ'=-+21cos 16ka aγσγ'=++232,1and 2T k K γσβεσε=-=Ω-=Here ε is a measure of the amplitude of the response. Obtain the frequency-response equation.Show that 1/3max (4/)a k μ=. How does this value of max a compare with that the case of linear viscous damping? Plot a versus σ and k . Is there a jump phenomenon?(b) When Ω is near one third (superharmonic response), show that320209()cos[()]cos()()8a T T T K T O θεβε=++Ω+where223212()(cos sin )836a a a μμγγ'=-Λ+-Λ-232121()(sin cos )2836a a γσμγγΛ'=+Λ+++229,31,and 16T K k k γσβεσε=-=Ω-=Λ=Obtain the frequency-response equation. Plot a versus σ and Λ. Is there a jumpphenomenon?(c) When Ω is near 3 (subharmonic response), show that320201()cos[()]cos()()8a T T T K T O θεβε=+-Ω+where2221112()(sin cos )882a a a a μγμγ'=-Λ+-Λ-223111()3(cos sin )2882a a γσγμγ'=+Λ+-Λ+2213,3,and 16T K k k γσβεσε=-=Ω-=Λ= Obtain the frequency-response equation. Plot a versus σ and Λ. Is there a jumpphenomenon?题5：（20分）Consider the system shown in Figure 5 when the tension 0(1sin )T T t εω=+.(a) Show that the governing equation is221/202(1sin )()0mx T t x l x εω-+++=(b) Linearize the governing equation to obtain22002(1sin )0,T x t x mlωεωω++==(c) Determine second-order expansions for the transition curves separating stability frominstability when00,2ωωω≈(d) If 1/2()x O ε=, determine the influence of the nonlinear terms to first order when02ωω≈.Figure 5 Particle attached to stretched string注意：所有的题目并没有给出完整的解答，以此作为提供一个解题思路，希望自己推导一遍（使用自己习惯的一套符号），修改和完善其中的不妥之处，然后补全没有给出解答的部分即可。

切勿雷同！！！题一解：这题关键算Jacobi 积分，可以参考Nayfeh 的《非线性振动》第二章，或用Mathematica 软件计算。

本题有的地方推导过于简单，有些地方没有必要，希望稍作修改。

第三问的分析可能不太恰当！！！ （a ）系统动能为22222222222111()222111(2)()22211(41)22T mz mx m x m pxx mx m x m p x x m x =++Ω=++Ω=++Ω (1.1)系统的势能为2V mgz mgpx ==(1.2)代入Lagrange 方程d T T Vdt q q q∂∂∂-=-∂∂∂ (1.3)这里取广义坐标为ϕ和x ，其中ϕ是金属丝旋转过的角度，有关系ϕ=Ω，由此得到系统的运动微分方程20x x Ω+Ω=(1.4)22222(14)4(2)0p x x p x x pg x +++-Ω=(1.5)（b ）积分式(1.4)得到2x Ω=(1.6)把式(1.6)代入式(1.5)并整理得到22224(14)4(2)0Hp x x p x x pg x x +++-= (1.7)（c ）下面来求出描述相平面上的运动方程。

设42222(2)4,14gp H x x p xv x v v p x -+==-+ (1.8)从方程(1.8)中消去t ，我们得到42222(2)4(14)dv gp H x x p xv dx p x v-+=-+ (1.9)此式可以改写为2224221(14)()(2)402p x d v gp H x xdx p xv dx ++-+= (1.10)方程(1.10)积分有22222(14)2Hp x v gpx h x +++= (1.11)式中h 是常数。

方程(1.11)表明，此系统的T V +不是一个常数。

积分(1.11)称为Jacobi 积分。

改写(1.11)可以得到22222214h gpx H x v p x --=+ (1.12)并由此可以得出42222122232(2)48(2)(14)dv gp H x x hp x Hp xdx h gpx H x p x -+-=±--+ (1.13)注意到0,0H p >>，所以222h gpx H x h --≤-当2x = 式(1.12)右边分子必须半正定，即 220020h pgx H x ---≥(1.14)解得2x ≤≤(1.15)因此运动是有界的，它用围绕原点的一些闭轨线来表示，而原点是一个中心。

（d ）编程的方法课上老师已经交给大家了，自己编写一小段程序即可。

下面的程序仅为示例，不是最终结果。

勿用此程序画的图。

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %%%%% 题一：画轨线图 %%%%%画一条曲线，先确定参数x 范围%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% clear all; clc; p=1.0; g=32.2; h=1000.0; H=12.0; dt=0.0001; x0=0.5;v0=0.5;II=6740;X(1:II)=0.0;V(1:II)=0.0;X(1)=x0;V(1)=v0;for i=2:IIx1=x0+v0*dt;v1=v0-((2*g*p-H/(x1^4))*x1+4*p^2*x1*v0^2)/(1+4*p^2*x1^2)*dt;x0=x1;v0=v1;X(i)=x0;V(i)=v0;endfigure;plot(X,V,'r');hold on;on;题二解： 因为2252x x y y xy =+-=- (2.1)所以系统的奇点满足225020x y xy +-=-= (2.2)由此解得奇点为1234(1,2),(1,2),(2,1),(2,1)s s s s ---- （1）对原方程在奇点1(1,2)s 附近线性化，得111111224x x y y x y =+=+ (2.3)系统矩阵的特征方程为 2360λλ--=(2.4)特征值为123322λλ+== (2.5)由于1λ和2λ异号，所以奇点1(1,2)s 为鞍点，它是一个不稳定奇点。