THE EXAM

1. THE EXAM There are four questions of equal value You may use a calculator You may use one 8.5 x 11” poop sheet, which you are to turn in with the exam

2. FAQs

3. What’s the difference between a particular solution and a homogeneous solution? The particular solution satisfies the whole differential equation including the forcing term It is proportional to the amplitude of the the forcing term If there is no forcing term, there is no particular solution

4. What’s the difference between a particular solution and a homogeneous solution? The homogeneous solution satisfies the differential equation with the forcing term crossed out It has no determined amplitude

5. What good are the particular and homogeneous solutions? What do they mean? The particular solution accounts for the forcing. In a damped system, it is the entire solution after some time has passed The homogeneous solution 1. tells you what the system will do if it is disturbed 2. allows you to satisfy initial conditions Repeat the example from last time

6. Simple example of how to solve once you have the equations The initial value problem The homogeneous equation Its solution where This is a constant

7. (remember P0 is constant) Substitute The particular equation Its solution

8. These are connected by the initial conditions

9. Sometimes you use exp(st) and sometimes you use exp(jwt) What’s that all about? For homogeneous problems where I am trying to find an eigenvalue exp(st) is the general expression: s is, in general, complex I use exp(jwt) when there is no damping and I know that s will be purely imaginary

10. If I use exp(st), then I replace dots by s, (double dots by s2, etc.) If I use exp(jwt), then I replace double dots by -w2, etc. (there are no terms with an odd number of dots without damping)

11. How do you invert those 2 x 2 matrices again? Swap diagonal elements Change the sign of the off-diagonal elements Divide by the determinant Remember: this only works for 2 x 2s

12. When is gravity important? All pendulum problems Problems where it can act to restore an equilibrium, such as

13. MORE QUESTIONS?

14. VIBRATIONS’ GREATEST HITS THE DEN HARTOG AWARDS

15. j

16. Exponential and trigonometric functions

17. HARMONIC FUNCTIONS

18. Harmonic functions are periodic at a single frequency The general function may be written in several ways A, B, and C are real numbers — D is complex

19. The forms are equivalent, as we can see from the following The complex part takes a little more work — next slide

21. NATURAL FREQUENCIES

22. 1. The mass-spring system ky y mass mass

23. Admits harmonic solutions

24. We can do the same thing with the pendulum T l q mg m

25. We can make this simpler by a change of variable

26. We’re going to work with linear pendulum equations The mass-spring system and the linear pendulum are essentially the same thing

27. They both admit harmonic motion Natural frequencies: mass-spring: linear pendulum:

28. FORCING AND DAMPING

29. mass-spring-damper system m c f k

30. Here we need to deal with both forcing and damping We split y into a homogeneous and particular solution and we wed them through the initial conditions To find the homogeneous solution, we drop f and seek an exponential solution

32. can be real or imaginary note that we can factor these and write them in a different form

33. Of course, we can do this directly in the equations, and we did Let’s look at solving this for a particular solution for harmonic forcing

34. We will be happier using the exponential approach Remembering to take the real part when we are done

36. Initial conditions

37. From which we need to find A1 and A2

38. That was all for one DOF The processes are the same for two (or more) DOF The only new issues revolve around matrix analysis Recall how this went when we did the automobile suspension system

39. From Gillispie, TD Fundamentals of Vehicle Dynamics (1992) We can fit this to our vertical model

40. Rotate our picture to the vertical c2 k2 z2 f2 m2 c3 k3 f1 z1 m1 c1 k1

41. Gillespie Us z2 m2 k3 c3 z1 m1 k1 zG

42. Free Body Diagrams m2 m1

43. The differential equations, supposing z positive up We want to know both the (damped) natural frequencies and the response to rough roads This means a homogeneous and a particular solution But first let’s look at the Lagrangian method

44. The Lagrangian method: z1 and z2 are the generalized coordinates No generalized forces

45. combine all this and we get which we can see is the same thing as the result from the free body diagrams

46. Start with the homogeneous solution by setting zG to zero Seek exponential solutions: replace each dot by an s and the variable by a constant Z I’m using s because there is damping

47. rearrange put in matrix form

48. Now we have the matrix eigenvalue problem again This is quite messy in general, so let us put in some realistic numbers the same ones we had from Gillespie

49. Put these numbers into our picture z2 m2 = 2.479 lb-sec2/in m2 k3 c3 k3 = 143 lb/in c3 = 15.06 lb-sec/in z1 m1 = 0.194 lb-sec2/in m1 k1 = 1198 lb/in k1 zG

50. The determinant with the parameters substituted in is expanding It has four roots, which come in complex conjugate pairs: -39.3025 ± j70.4975, -2.5494 ± j6.9412 rad/sec