1. Lecture 11: Introduction to Two Degree of Freedom Systems For any system we need to build a model figure out the natural frequencies figure out the homogeneous solution(s) figure out how to get particular solutions start with undamped, unforced 2DOF systems figure out the natural frequencies figure out the homogeneous solution(s)

2. We’ll begin with a simple two DOF system: the masses move independently k1 k3 k2 m1 m2 y1 y2

3. It should be clear that once we know the natural frequencies we know everything there is to know about the (homogeneous, unforced, undamped) system though we do need to think about initial conditions It is clear that the middle spring is what makes this different

4. Without the middle spring, we have two independent 1 DOF systems k1 k3 k2 m1 m2 y1 y2

5. If the middle spring becomes infinitely stiff it turns into a weld k1 k2 m1 m2 y1 y2

6. Let’s do some analysis Free body diagrams for the two masses k3(y2 – y1) k1y1 m1 m2 -k2y2 k3(y2 – y1) Take y positive to the right and write the dynamical equations

7. How do we solve these? Homogeneous ordinary differential equations with constant coefficients have exponential solutions Just like the one degree of freedom problem

8. What do we expect to happen? In the 1DOF case we had two values of s In the undamped case these were We had two values of s and two undetermined constants to satisfy initial conditions. We should expect four values of s, and four undetermined constants for 2DOF The four values of s will become obvious as we go forward In the undamped case, these will be

9. How does this work with the assumed form of the solution? Put in the generality we expect (for the undamped case) EIGHT undetermined constants! Too many. What happens is that the differential equations supply relationships among these coefficients, reducing the number if independent constants to 4

10. We can get at this a number of ways — one is to reframe the problem in matrix form this is easier than working with the equations as they are

11. So we can do the solution in vector terms, writing M is invertible, so we can rewrite this as which looks like the matrix eigenvalue problem for s2 and Y s2 plays the role of the eigenvalue and Y the role of eigenvector

12. We can stick with the original version, without dividing by the ms

13. This set will have a nontrivial solution if which is OR

14. The characteristic equation is a quadratic in s2 Its solution is rather lengthier than it makes sense to write down However both values of s2 are negative: the exponents are purely imaginary we can denote them by ±jw Note that w2 is in general NOT EQUAL TO OR The system is truly coupled and needs to be so treated

15. BTW . . . We could have realized up front that no dissipation/damping implies purely imaginary exponents and have written (4) with jw instead of s I don’t generally recommend making leaps of faith, but if you’re feeling sure, go ahead. Carry on with solving the problem; we have frequencies, what’s next?

16. Here are the algebraic equations again This matrix equation resembles an eigenvalue problem The s2 values we found are a kind of eigenvalue I will call them modal values There are associated modal vectors, or modes and we can understand the problem better if we look at the modes, which are nontrivial vectors that satisfy the system (3) There will be one pair for each modal frequency

17. Why I refuse to call s2 an eigenvalue I want to reserve the word eigenvalue for s and I want to reserve the word eigenvector for vectors in state space I haven’t introduced state space yet, but it will be coming soon For now, please, please, please do not get hung up on language

18. We can do a 2 x 2 system by hand k3 k1 k2 m1 m2 The modal vectors for this problem are y1 y2 where s2 must take on its modal value, -w2 As I show on the next slide

19. rectangle x rectangle plus oval x oval is clearly zero The other product is the characteristic equation, and so it, too, is zero

20. BTW . . . These are unnormalized modal vectors Sometimes it I convenient to normalize the modal vectors such that V.V = 1 I’m not going to do that for today’s work

21. There will be two distinct modal vectors, one for each frequency They are very messy in general, so let’s look at a simple special case mi = 1, ki = 1 The masses move together at the lower frequency and apart at the higher frequency

22. We can write My initial conditions will be

23. Which we can interpret or

24. And we can put all this together

25. I’ve called these modes, and talked about modal vectors which all may be a bit abstract. We can see this better for a triple pendulum (analysis in Chap. 4) w = 0.644806, 1.51469, and 2.50798 times √(g/l)

26. Low frequencies have simple mode shapes High frequencies have complicated mode shapes

27. QUESTIONS? I took that one a little further than I originally intended so let’s take another look at what I just did

28. Solving the initial value problem using the modes (This may well be new to you all even though we just did it.) Combine the motion of the two masses into an abstract vector This is a modal vector The general vector y has contributions from both modes

29. where the vectors V1 and V2 are the nontrivial solutions to or, in this undamped system

30. means or, putting in the expressions for the modes

31. In component form

32. We need to know how to apply initial conditions if we are to use this method As in the 1 DOF problems, we need conditions on position and velocity

33. Recall that each modal vector goes with one of the natural frequencies so the time behavior of each contribution is going to be harmonic The squares of the natural frequencies are the solutions to a quadratic so there will be a large one and a small one Denote the smaller frequency by w1 and the larger by w2 The combination is then of two harmonic functions, the coefficients of which are the eigenvectors we are back to equation (6)

34. Differentiate and substitute to obtain the velocity condition Working this through in the abstract is much too complicated I’ll do some examples, but I’d like to summarize what we have at this point

35. COMPACT SUMMARY OF THE PROCEDURE The differential equation Solution in terms of constant modal vectors and scalar functions of time these are harmonic functions The initial conditions

36. EXAMPLES I want to consider three cases They are all dissipation –free so I can use the characteristic polynomial in terms of frequency

37. We define the three cases and substitute the values into Eq. (4’) case 1: m1 = 1 = m2, k1 = 100 = k2 case 2: m1 = 1, m2 = 2, k1 = 100, k2 = 200 case 3: m1 = 1 = m2, k1 = 100, k2 = 400 I will look at a specific initial condition for all three of these — I’ll go through the details for case 1, just results for cases 2 and 3

38. The initial condition for all three is a displacement of mass 1 to the right with the second mass held still and initial velocities equal to zero k1 k3 k2 m1 m2 initial offset k1 k3 k2 m1 m1 m2

39. Let’s start with case 1 w1 = 10, w2 = √(100 + 2k3), this is √(k/m) despite what I said we’ll see why shortly We need to see what the motion looks like and to do that we need to understand something about modal vectors

40. For case 1 motion of m1 motion of m2 For w2 = √(100 + 2k3), s2 = -100 – 2k3 and this becomes The masses move equally in opposite directions For w1 = 10, s2 = -100 and this becomes The masses move equally in the same direction

41. We can draw picture of how the frequencies vary with k3

42. If the two masses move in the same direction the spring k3does not flex and that’s why the frequency is the same k/m we have doubled both k and m 100 100 1 1 y1 y2 The effective mass is 1 + 1 and the effective spring constant is 100 + 100

43. Put in the initial conditions Start from rest and offset only m1, the left hand mass These initial conditions are The algebra is not too bad, but I’ll spare you and give you the answer

44. And the answer is The sum of these goes like w2 the difference goes like w1, as I said For small time y2 is essentially zero, but this changes with time The system passes the motion back and forth between the two masses

45. I’ll look at all three cases in Mathematica when we’re done setting them up

46. CASE 2 The two side mechanisms are no longer the same but one of the modes has the same natural frequency w1 = 10, w2 = √(100 + 3/2k3) The low frequency of the coupled system remains the same, and we can explain it in the same way

47. If the two masses move in the same direction the spring k3does not flex and that’s why the frequency is the same k/m 200 100 1 2 y1 y2 The effective mass is 1 + 2 and the effective spring constant is 100 + 200 the ratio remains 100

48. We can draw picture of how the frequencies vary with k3 This is, of course, exactly the same as for Case I

49. The eigenvectors are not the same and the opposite motion is not the same, although the idea of opposite motion continues to hold

50. For case 2 motion of m1 motion of m2 For w2 = √(100 + 3/2k3), s2 = -100 – 3/2k3 and this becomes The masses move in opposite directions but no longer equally This is the only change from case 1 For w1 = 10, s2 = -100 and this becomes The masses move equally in the same direction