Nonlinear Dynamic Systems Homework 1
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1 Nonlinear Dynamic Systems Homework 1 1. A particle of mass m is constrained to travel along the path shown in Figure 1, which is described by the following function yx = 5x + 1x 4, 1 where x is defined as the horizontal location of the particle. Develop the equation of motion for the system, find the fixed points for the system, classify the stability of each fixed point, and sketch the neighboring trajectories. Explain how the stable and unstable fixed points match or do not match your intuition? 5 Track Shape For Multiple Potential Wells 4 3 yx [m x [m Figure 1: Graph of the curved path yx in a vertical gravitational field. Prob. 1 Soln. r cm = xî + yxĵ, 1
2 r cm = ẋî + yxĵ,... but... dy dt = y xẋ = y xẋ, 3 T = 1 m r cm r cm = 1 m ẋ + y x ẋ, 4 V = mgyx = mg 5x + 1x 4, 5 Using Lagrange s equation.. the resulting equation of motion becomes... d T T dt ẋ x + V x = 6 mẍ1 + y x + my x y xx ẋ + mgy x =, 7 Fixed points are found from the state space form of the above equation f x =, which results in f x = mgy x = mg 1x f + 4x 3 f = mgx f 1 + 4x f = 8 with fixed points at x f =,,, ± 1. Stability of each fixed point is found from eigenvalues of Jacobian matrix evaluated at the fixed point. Df x x= xf = [ f1 x 1 f f 1 x 1 f x x x= x f 9 Results are that, is an unstable fixed point i.e. at lease one Rλ s >. Stable fixed points are found at, ± 1 since all eigenvalues satisfy Rλ s <. The phase space portriat is shown in Figure.
3 Phase Space: Particle on a Double Potential Well Curve 3 1 x x 1 Figure : Phase space portrait of a particle following a curve with a double potential well. 3
4 . Figure 3 is a schematic of a double pendulum undergoing parametric excitation i.e. time periodic movement of the pivot point. The first pendulum is subjected to a harmonic displacement, of amplitude A and frequency Ω, oriented along an angle α which is at an incline with the horizontal. Angular oscillations for the first pendulum, with a mass of m 1 and length L 1, are described by θ 1. The corresponding angular motions of the second pendulum, with a mass of m and effective length of L, are described by θ. Derive the equations of motion for this system. Pivot Point Motion m1 m Figure 3: Schematic of a double pendulum subject to parametric excitation at a tilted angle. The following is the derivation for the double pendulum equation of motion. Start by considering the position vectors r cm1, to the center of mass for each object and its time derivative. The corresponding equations for m 1 are given by r cm1 = A sin Ωt cos α + L 1 sin θ 1 Î + A sin Ωt sin α L 1 cos θ 1 Ĵ r cm1 = AΩ cos Ωt cos α + L 1 θ1 cos θ 1 Î + AΩ cos Ωt sin α + L 1 θ 1 sin θ 1 Ĵ 1a 1b where Î and Ĵ are the fixed-frame unit vectors in horizontal and vertical directions, respectively. The equations for the second mass are 4
5 r cm = r cm1 + L sin θ Î L cos θ Ĵ, r cm = r cm1 + L θ cos θ Î + L θ sin θ Ĵ, 11a 11b With the position vectors to the center of mass defined, the kinetic energy can be found from T = T 1 + T = 1 m 1 r cm1 r cm1 + 1 m r cm r cm, 1 where the expanded kinetic energy expression is written as T = 1 m 1 [ AΩ cos Ωt cos α + L 1 θ1 cos θ 1 + AΩ cos Ωt sin α + L 1 θ1 sin θ 1 [ AΩ cos Ωt cos α + L 1 θ1 cos θ 1 + L θ cos θ + 1 m + AΩ cos Ωt sin α + L 1 θ1 sin θ 1 + L θ sin θ. 13 For those less familiar with Lagrange s method, it insightful to note that it is more convenient to not write out the square of the terms inside Eq. 13. Therefore, we now write the potential energy of the system as [ [ V = m 1 g A sin Ωt sin α+l 1 1 cosθ 1 +m g A sin Ωt sin α+l 1 1 cosθ 1 +L 1 cosθ, 14 The next step is to apply Lagrange s equation, for θ 1 d T dt θ T + V =, 15 1 θ 1 θ 1 which requires several partial derivatives that are given in Eqns d T dt θ 1 = [ m 1 + m L θ 1 1 AL 1 Ω sin Ωt cos θ 1 cos α + sin θ 1 sin α + AL 1 Ω θ 1 cos Ωt cos θ 1 sin α sin θ 1 cos α + m L 1 L [ θ cosθ 1 θ + θ 1 θ sinθ θ 1 + θ sinθ 1 θ, 16 5
6 T θ 1 = AL 1 Ω θ 1 cos Ωt m 1 + m cos θ 1 sin α sin θ 1 cos α + m L 1 L θ1 θ sinθ θ 1, 17 V θ 1 = m 1 + m gl1 sin θ 1 18 The final equation of motion for the θ 1 coordinate is θ 1 AΩ L 1 sin Ωt cos θ 1 cos α + sin θ 1 sin α + m L θ cosθ 1 θ + m 1 + m L θ sinθ 1 θ + g sin θ 1 =, 19 1 L 1 A second equation of motion is required to fully describe the system. This equation is found by applying Lagrange s equation to θ d T dt θ T + V = θ θ which again requires several partial derivatives that are given in Eqns. 1 3 d T dt θ = m [L θ AL Ω sin Ωt cos θ cos α + sin θ sin α + AL Ω θ cos Ωt cos θ sin α sin θ cos α + L 1 L θ1 cosθ 1 θ θ 1 θ 1 θ sinθ 1 θ, 1 T [ = m AL Ω θ θ cos Ωt cos θ sin α sin θ cos α + L 1 L θ1 θ sinθ 1 θ, V θ = m gl sin θ 3 The final equation of motion for the θ coordinate is 6
7 θ A Ω sin Ωt cos θ cos α + sin θ sin α + L 1 [ θ1 cosθ 1 θ L L θ 1 sinθ θ 1 + g L sin θ =, 4 3. Use the method of isoclines to construct a phase portrait for the van der Pol equation ẍ + ɛ x 1 ẋ + x =, 5 using the following value ɛ =.5. Plot the vector fields for x vs. ẋ over the range of 5 < x < 5 and 5 < ẋ < 5. Find the fixed points for the system and classify their stability. Van Der Pol Oscillator Phase Space Diagram x x 1 Figure 4: Phase portrait in the vicinity of the fixed point located at the origin. 4. The drag force on a body immersed in a fluid, moving a very low Reynolds numbers, is commonly approximated with proportional damping. The equation of motion for a viscously damped 7
8 pendulum undergoing free vibrations is given by Eq. 6. Plot the phase space diagram, using any method you would like, over the range of 3π < θ < π and 4 < θ < 4. Label the separatrix and the different basins of attraction. Does the system have fixed point attractors, periodic attractors, chaotic attractors or some combination of each type of attractor? θ + ζ θ + ω sin θ =. 6 with parameters ζ =.5 and ω = 1 [rad/s. Pivot Point Pendulum Figure 5: Schematic of a pendulum undergoing free vibrations. 4 Phase Space: Pendulum Oscillations With No Forcing 3 1 x x 1 Figure 6: Phase portrait for a viscously damped pendulum free oscillations with fixed points at ±nπ, where n =, ±1,... All attractors for this example are fixed point attractors. Separatix is the line of separation between final solutions and the basins of attraction are the regions of the phase space that that eventually go towards a particular attractor. 8
9 5. The contact of a rigid sphere with a perfectly elastic surface is often referred to as Hertzian contact [1 3. The typical model for the motion of a vibration system in intimate contact with the surface is given by mẍ + cẋ + kx E Rx 3/ = F s + F o cos Ωt, 7 Using the parameters listed below, simulate the response of the system and plot: 1 displacement vs. time; a power spectrum of the displacement signal PSD; 3 the phase space of the system i.e. displacement vs. velocity; and 4 a Poincaré section for the system i.e. periodic samples of the displacement vs. velocity. m = [Kg, c =.47 [Ns/m, k = [N/m, E = [N/m R = [µm, F s = 1 [µn, F o = [µn, Ω = 1 [rad/s Figure 7: Schematic of a spherical indenter undergoing oscillatory motion while in contact with a flat material surface. References [1 S. Hu, S. Howell, A. Raman, R. Reifenberger, and M. Franchek, Frequency domain identification of tip-sample van der waals interactions in resonant atomic force microcantilevers, Journal of Vibration and Acoustics, vol. 16, pp , 4. [ K. L. Johnson, Contact Mechanics. Cambridge, UK: Cambridge University Press, [3 G. M. Pharr, W. C. Oliver, and F. R. Brotzen, On the generality of the relationship among contact stiffness, contact area, and elastic modulus during indentation, Journal of Material Research, vol. 7, no. 3, pp ,
10 Displacement [nm Time [s 4 x Phase Space x Poincare Section 1 7 Velocity [m V n [m Displacement [m x X [m n x 1 7 Figure 8: Displacement vs. time, the phase space, and the Poincaré section 1 5 Power Spectral Density Pwr frequency [Hz Figure 9: Power Spectrum of Displacement Signal 1
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