Unformatted text preview: S. Widnall 16.07 Dynamics Fall 2009 Version 1.0 Lecture L19  Vibration, Normal Modes, Natural Frequencies, Instability
Vibration, Instability
An important class of problems in dynamics concerns the free vibrations of systems. (The concept of free vibrations is important; this means that although an outside agent may have participated in causing an initial displacement or velocityor both of the system, the outside agent plays no further role, and the subsequent motion depends only upon the inherent properties of the system. This is in contrast to "forced" motion in which the system is continually driven by an external force.) We shall consider only undamped systems for which the total energy is conserved and for which the frequencies of oscillation are real. This forms the basis of the approach to more complex studies for forced motion of damped systems. We saw in Lecture 13, that the free vibration of a massspring system could be described as an oscillatory interchange between the kinetic and potential energy, and that we could determine the natural frequency of oscillation by equating the maximum value of these two quantities. (The natural frequency is the frequency at which the system will oscillate unaffected by outside forces. When we consider the oscillation of a pendulum, the gravitational force is considered to be an inherent part of the system.) The general behavior of a massspring system can be extended to elastic structures and systems experiencing gravitational forces, such as a pendulum. These systems can be combined to produce complex results, even for onedegree of freedom systems. We begin our discussion with the solution of a simple massspring system, recognizing that this is a model for more complex systems as well. 1 In the figure, a) depicts the simple mass spring system: a mass M, sliding on a frictionless plane, restrained by a spring of spring constant k such that a force F (x) = kx opposes the displacement x. (In a particular problem, the linear dependence of the force on x may be an approximation for small x.) In order to get a solution, the initial displacement and initial velocity must be specified. Common formulations are: x(0) = 0, and
dx dt (0) = V0 (The mass responds to an initial impulse.); or x(0) = X0 and dx dt (0) = 0 (The mass is given an initial displacement.). The general formulation is some combination of these initial conditions. From Newton's law, we obtain the governing differential equation m with x(0) = X0 , and
dx dt (0) d2 x = kx dt2 (1) = V0 . The solution is of the general form, x(t) = Re(Aeit ), where, at this point in the analysis, both A and are unknown. That is, we assume a solution in which both A and are unknown, and later when the solution is found and boundary conditions are considered, we will end up taking the real part of the expression. Depending upon whether A is purely real, purely imaginary, or some combination, we will in general get oscillatory behavior involving sin s and cos s since eit = cos t + i sin t. For a system without damping, 2 = Real(K) (K being some combination of system parameters.), so that = K. For undamped systems, these two values of , are redundant; only one need be taken. To solve the differential equation, equation(1) is rewritten m d2 x + kx = 0. dt2 (2) With the assumed form of solution, this becomes  2 mx(t) + kx(t) = 0 = x(t) ( 2 m + k) Since x = x(t), for a valid solution, we require = k m. (3) This approach works as well for systems b) and c). These two systems are pendulums restrained by torsion springs, which for small angles ( or ) produce a restoring torque proportional to the angular departure from equilibrium. Consider system b). Its equilibrium position is = 0. The restoring torque from the Tg = mgL. Writing Newton's law in a form appropriate for pendular motion, we obtain mL d2 1 = (Tg + Ts ) dt2 L spring is Ts = . The restoring torque from gravity is Tg = mgLsin which for small angles becomes (4) We assume a form of solution, (t) = Ae(it) , and rewrite the equation as before, moving all terms to the lefthand side. mL 2 + 1 (mgL + ) (t) = 0. L mgL + mL2 2 (5) Therefore for a solution we require = (6) Examining this result, we see that the combination of the spring and gravity acts to increase the natural frequency of the oscillation. Also if there is no spring, = 0, and the result becomes just the frequency of a g pendulum = L . System c) is perhaps a bit more interesting. In this case, we use the small angle . We take the equilibrium position of the spring to be = 0 so that the restoring torque due to the spring is again Ts = . But now in this case, we are expanding the gravitational potential about the point = 0. Since this is an unstable equilibrium point, this gives the restoring (it doesn't restore, it keeps going!) torque due to gravity as Tg = mgLsin or for small , Tg = mgL. Writing the governing equation for this case, we obtain 1 mL 2 + (mgL + ) (t) = 0. L Therefore for a solution we require = mgL + mL2 (7) (8) mgL mL2 , In this case, there is a critical value of for which = 0. On either side of this point, we have = which gives  real for > mgL, and  imag for < mgL. Since we have assumed (t) = eit , a real will produce oscillatory motion; an imaginary will produce exponentially diverging, or unstable, motions. We say that the pendulum for less than the critical value, = mgL, is unstable. Vibration of MultiDegree of Freedom Systems
We begin our treatment of systems with multiple degrees of freedom, by considering a two degree of freedom system. This system contains the essential features of multidegree of freedom systems. Consider the two twomass, twospring systems shown in the figure. 3 In this case, there are two independent variables, x1 (t) and x2 (t); their motion is not independent, but is coupled by their attachments to the springs k1 , k2 and for system b), k3 . The sketch shows the forces Fi acting on the masses as a result of the extension of the spring; these of are equal and opposite at the ends of the springs. We consider both system a) and b). System b) is actually the simpler of the two systems because of its inherent symmetry. The governing equations can be written as for system a) d2 x1 dt2 d2 x2 m2 2 dt m1 for system b) d2 x1 dt2 d2 x2 m2 2 dt m1 = k1 x1 + k2 (x2  x1 ) = k2 (x2  x1 )  k3 x2 (11) (12) = k1 x1 + k2 (x2  x1 ) = k2 (x2  x1 ) (9) (10) In both cases, as before, we assume a solution of the form x1 (t) = X1 eit and x2 (t) = X2 eit . However, as we will see, in this case, we will obtain two possible values for 2 ; both will be real; we will take only the positive value of itself. These will be the two vibration modes of this two degree of freedom system. These results extend to N 2 's for an N degree of freedom system. Again, we will take only the positive value of . Consider first system b). With the assumed form of solution, and rewriting all terms on the lefthand side, we obtain  2 m1 X1 + k1 X1  k2 (X2  X1 )  2 m2 X2 + k2 (X2  X1 ) + k3 X2 This equation can be written in matric form as k1 + k2 k2 m 2  1 k2 k2 + k3 0 or k1 + k2  m1 2 k2 k2 =0 = 0 (13) (14) 0 m2 2 X1 X2 X1 X2 This equation makes a very powerful statement. Since the righthand side of both equations is zero, a condition for a solution is that the determinant of the matrix equals zero. This will give an algebraic equation with two solutions for : 1 and 2 . These are the "natural" frequencies of the two degree of freedom system. In the general case, they are not equal; and both x1 and x2 participate in the oscillation 4 k2 + k3  m2 2 = 0 0 0 0 = . (15) . (16) at each frequency i . Also, as in the single degree of freedom system, the actual values of x1 (t) and x2 (t) are determined by initial conditions; in this case 4 initial conditions are required: x1 (0), x2 (0),
dx2 dt (0). dx1 dt (0),and The actual values of X1 and X2 are of less interest than the relationships between them and the structure of the problem. If the two masses are equal, a particularly simple form of a more general result follows from equation(16). We consider this as an introduction to the more general case. The more general case will be considered shortly. k1 /m + k2 /m  2 k2 /m k2 /m X1 X2 0 0 k2 /m + k3 /m  2 We determine the two values of i (1 and 2 ) by setting the determinant equal to zero. We then substitute each value of i in turn into the matrix equation and determine for each i the coefficients X1i and X2i ; only their ratio can be determined. We write the coefficients X1i and X2i as vectors, X1 = (X11 , X21 ) and X2 = (X12 , X22 ), where the subscript 1 refers to the mode associated with 1 , and 2 refers to the mode associate with 2 . It is a remarkable property of the solution to the governing equations that these vectors are orthogonal: the Consider the simplest case of system b) with both masses equal to m and all springs of stiffness k. In this case we have 2k/m  2 k/m k/m X1 X2 0 dot product of X1 X2 = 0 (We will follow with an example to amplify and clarify this.) = . (17) components of the x1 and x2 motion are: X1 = (1, 1) and X2 = (1, 1). This simple example gives great are called "normal modes". . (18) 0 Setting the determinant equal to zero gives two solutions for : 1 = k/m and 2 = 3k/m. The 2k/m  2 physical insight to the more general problem. The natural frequency is i ; the components Xi = (Xi1 , Xi2 ) = Normal Modes of MultiDegree of Freedom Systems Examining the first "normal mode", we see an oscillation in which X1 = (1, 1) occurs at an oscillation frequency 1 = k/m. Since X1 = (1, 1), the central spring does not deform, and the two masses oscillate, each on a single spring, thus giving a frequency = k/m. The second "normal mode" has a frequency = 3k/m, with X 2 = (1, 1); thus the masses move in opposite directions, and the frequency of oscillation is increased. It can be seen by inspection that the vector X1 and X2 are orthogonal (their dot product is zero.) If such a system was at rest, and an initial impulse was given to one of the masses, both modes would be excited and a free oscillation would occur with each "mode" oscillating at "its" natural frequency. The equation for general values of k1 , k2 and k3 can be written 5 k1 /m + k2 /m k2 /m k2 /m k2 /m + k3 /m  2 1 0 0 1 X1 X2 = 0 0 . (19) Characteristic Value Problem
This problem is called a characteristicvalue or eigenvalue problem. Formulated in matrix notation it can be written A11 A21 A12 A22 The requirements on the simplest form of the characteristic value problem are that the matrix [A] is symmetric (This will always be true for combinations of masses and springs.), and that the characteristic value multiplies an identity matrix. (An identity matrix has all 1's on the diagonal and 0's off the diagonal; this will be true if all the masses are equal; if not, a more general form must be used yielding analogous results. This more general from will be considered shortly.) For the form of the governing equation shown in equation(20), the characteristic value = 2 . The general will result in 4 's). For each i , a vector Xi will be obtained. For this form of the characteristic value problem, the dot product between any two of these vectors is zero. (Xi,1 , Xi,2 ).(Xj,1 , Xj,2 ) = 0 (21) solution to this problem will yield a set of solutions for equal to the size of the matrix :(i.e a 4 4 matrix  1 0 0 1 X1 X2 = 0 0 . (20) numerical matrix methodsuch as MATLAB will yield both the i 's (called the eigenvalues) and the Xi 's, called the eigenvectors for a particular matrix [A]. A similar result is obtained for the modes of vibration of for i = j. These are the normal modes of the system, and the 's are the natural frequencies. Any a continuous system such as a beam. The displacement of the various mode of vibration of a uniform beam are orthogonal. The general solution for the motion of the masses is then given by an expansion in the normal modes, Xi , x1 (t) X11 X21 = A1 ei1 t + A2 ei2 t , (22) x2 (t) X12 X22 where X1 = (X11 , X21 ) and X2 = (X12 , X22 ) are the eigenvectors or normal modes from the solution of the characteristicvalue problem, obtained by hand or numerically, and i is the natural frequency of that mode. Again, it is a remarkable and extremely useful property that the dot product of Xi and Xj is zero unless i = j. Good form would suggest that we normalize each Xi so that the magnitude of Xi Xi equals 1, but we usually don't and therefore need to define Ci = Xi Xi , the vectormagnitudesquared, for later use. 6 Expansion in Normal Modes; Satisfaction of Initial Conditions
The general form of solution is given by equation (22). All that remains is to determine the coefficients A1 and A2 . This is done by satisfying the initial conditions on displacement and velocity. In the general case, since eit = 1 at t = 0, the initial displacement can be written x1 (0) X11 X21 = A1 + A2 . X12 X22 x2 (0) (23) and for an initial velocity, since ieit = i at t = 0, the initial velocity can be written v1 (0) X11 X21 = iA1 + iA2 . X12 X22 v2 (0) (24) It should be noted that in general A is complex; the real part relates to the initial displacement; the imaginary part to the initial velocity. If we consider Ai to be real, we are automatically assuming no initial velocity. Case 1: initial displacement nonzero; initial velocity zero We first consider the case of an initial condition on the displacement, specifically x1 (0) = x10 and x2 (0) = x20 , with v1 (0) = 0 and v2 (0) = 0. We define the initialcondition vector as X0 = (x1 (0), x2 (0)). To complete the solution, we need to obtain the values of A1 and A2 from equation(23). This is done by taking the dot product of both sides of equation(23) with the first mode X1 . Finally we get our payoff for all our hard work. Since the dot product of X1 with X2 is zero; the dot product of X1 with X1 is C1 ; and the dot product of X0 with X1 is some G1 , we obtain G1 A1 = X0 X1 /C1 = C1 and taking the dot product of X0 with X2 (which equals some G2 ), we obtain A2 = X0 X2 /C2 = G2 C2 (26) (25) With A1 and A2 determined, we have a complete solution to the problem. Case 2: initial displacement zero; initial velocity nonzero We now consider the case of an initial condition on the velocity, specifically x1 (0) = v10 and x2 (0) = v20 , with x1 (0) = 0 and x2 (0) = 0. We define the initialcondition vector as V0 = (v10 , v20 ). We use the coefficient Bi to define the solution for this case. The solution is again written as an expansion in normal modes oscillating at their natural frequency i of amplitude Bi , which is unknown at this point. x1 (t) X11 X21 = B1 eit + B2 eit . X12 X22 x2 (t) 7 (27) As previously noted, B is complex; the real part relates to the initial displacement; the imaginary part to the initial velocity. If Bi is imaginary, there is no initial displacement. The velocity at t = 0 is given by v1 (0) X11 X21 = iB1 + iB2 . (28) X12 X22 v2 (0) To complete the solution, we need to obtain the values of B1 and B2 from equation(28). This is again done zero, and the dot product of X1 with X1 is C1 , we obtain B1 =  Q1 i C1 (29) by taking the dot product of equation(28) with the first mode X1 . Since the dot product of X1 with X2 is where Q1 = V0 X1 and taking the dot product of X0 with X2 we obtain B2 =  Q2 i C2 (30) where Q2 = V0 X2 and X2 X2 C2 =. The fact that Bi is purely imaginary confirms our earlier observation that real coefficients imply a nonzero initial displacement while purely imaginary coefficients imply a nonzero initial velocity. A purely imaginary Bi simply implies that the displacement has a sin(t) behavior in contrast to a cos(t) behavior, since the real part of eit = cos(t), while the real part of ieit = sin(t). A solution for general initial conditions on x1 (0), x2 (0), x1 (0) and x2 (0) would be a linear combination of these solutions. Solution for Unequal Masses
If the masses, mi , are not equal we must use a more general form of the eigenvalue problem. Returning to Equation (19) for equal masses. For the case of equal masses, from Equation (19), this can be written k1 /m + k2 /m k2 /m X1 1 = 2 k2 /m k2 /m + k3 /m X2 0 For unequal masses, we rewrite the equation as k1 + k2 k2 X1 m = 2 1 k2 k2 + k3 X2 0 Methods of Applied Mathematics.) a11 a12 b  1 a21 a22 0 0 m2 0 1 (31) This equation is in the form of the generalized characteristic or eigenvalue problem. (See Hildebrand; X1 X2 (32) 0 b2 X1 X2 = 0 0 . (33) 8 where equation (20) has been rewritten as (33) in the extended formulation ([A]  [B])(X) = 0. Following Hildebrand we note that both [A] and [B] are symmetric matrices. Moreover, [B] is a diagonal matrix. Folthe resulting orthogonality condition between the "normal" modes is modified as Xj,1 m1 0 =0 (Xi,1 , Xi,2 ) m2 0 Xj,2 lowing the solution of the generalized characteristic or eigenvalue problem, by numerical or other technique, (34) mass distribution to result in orthogonality. A similar result is obtained for the vibration of a continuous of normal modes goes through as before with the modification of the normality condition. In more general configurations, ([A]  [B])(X) = 0, the matrix [B] may not be diagonal, that is b11 b12 [B] = b21 b22 As long as [B] is symmetric, the orthogonalization goes through as b11 b12 Xj,1 =0 (Xi,1 , Xi,2 ) b21 b22 Xj,2 for i = j. Thus the vectors of displacement for the normal modes of vibration must be multiplied by the system, such as a beam with nonuniform mass distribution. The process of expanding the solution in term (35) (36) Observations
The process we have outlined for finding the solution to the initial value problem to a multidegree of freedom system, outlined from equation(20) on, works for system with degrees of freedom from 2 to 20,000 and beyond. This approach is of fundamental importance in analyzing vibrations in a wide variety of systems. The expansion in normal modes is also useful in more complex problems such as forced motions at frequencies other than i . References
[1] Hildebrand: Methods of Applied Mathematics; for a discussion of the characteristic value problem, matrices and vectors. 9 MIT OpenCourseWare http://ocw.mit.edu 16.07 Dynamics
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This note was uploaded on 05/02/2011 for the course DYNAM 101 taught by Professor Matuka during the Spring '11 term at MIT.
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