TREATING MULTI-DIMENSIONAL SYSTEMS CHAPTER 10 TREATING MULTI-DIMENSIONAL SYSTEMS Chapter 1 through Chapter 9 developed basic principles of dynamics and control of single degree of freedom systems. The interacting parameters were composed of physical parameters, dynamic performance parameters (both natural and controlled), and control parameters. Since we were concerned with single degreeof-freedom systems, the relationships that were developed were temporal in nature, that is, associated with how quantities change in time. In this chapter, we’ll develop basic principles of dynamics and control of two degree-of-freedom systems. While doing this we’ll develop new relationships that are spatial in nature. We will be able to answers questions like where the actuators should be located and how many actuators to use. Perhaps the first thing to realize about multi-dimensional systems is that they’re more complex than single degree-of-freedom systems and that a way of handling this added complexity is necessary. We handle this added complexity through another separation principle. This separation principle enables us to treat the multidimensional system as separate single degree-of-freedom systems. Each of the single degree-of-freedom systems is a “mode” of the system. In Chapter 11 we’ll see under certain circumstances that a multi-dimensional system can be controlled by controlling its individual modes. 1. Equations Consider the undamped two degree of freedom system shown in Fig. 10 - 1. Applying Newton’s second law of motion to each mass, we get the two equations that govern the dynamics of the system (10 – 1) m1 x1 k 2 ( x 2 x1 ) k1 x1 f1 m2 x2 k 2 ( x 2 x1 ) k 3 x 2 f 2 Fig. 10 - 1 CONTROL OF DYNAMICAL SYSTEMS: AN INTRODUCTORY APPROACH TREATING MULTI-DIMENSIONAL SYSTEMS Equation (10 – 1) can be rewritten in the matrix-vector form (10 – 2) m1 0 x1 k1 k 2 0 m x k 2 2 2 k 2 x1 f1 k 2 k3 x2 f 2 or more compactly as Mx Kx F (10 – 3) where x is called the position vector, F is called the force vector, M is called the mass matrix, and K is called the stiffness matrix. Of course, systems with more than two degrees-of-freedom can be written in this form, too. 2. The Eigenvalue Problem Let’s study freely vibrating two degree-of-freedom systems. We let F = 0 in Eq. (10 – 3) and try solutions in the form (10 – 4) x e st Substitute Eq. (10 – 4) into Eq. (10 – 3) and divide the result by e st to get (10 – 5) [ s 2 M K ]φ 0 Equation (10 – 5) admits a nontrivial solution only if the matrix in brackets is singular, in which case (10 – 6) det[ s 2 M K ] 0 Equation (10 – 6) is called the characteristic equation of the system. It yields the values of s and corresponding in Eq. (10 – 4) that satisfy the free vibration problem. For the mass matrix and stiffness matrix given in Eq. (10 – 2), Eq. (10 – 6) becomes (10 – 7) s 2 m1 k1 k 2 k2 0 det 2 k2 s m2 k 2 k 3 ( s 2 m1 k1 k 2 )( s 2 m2 k 2 k 3 ) k 22 m1 m2 s 4 [m1 (k 2 k 3 ) m2 (k1 k 2 )]s 2 (k1 k 2 k 2 k 3 k 3 k1 ) CONTROL OF DYNAMICAL SYSTEMS: AN INTRODUCTORY APPROACH TREATING MULTI-DIMENSIONAL SYSTEMS Equation (10 – 7) is a quadratic equation in terms of s2, admitting two solutions s12 and s 22 . Each solution is substituted back into Eq. (10 – 5) to obtain the associated vectors φ1 and φ 2 . Equation (10 – 5) is called the eigenvalue problem associated with the equations of motion, Eq. (10 – 3). The complex numbers s are called eigenvalues and the corresponding vectors, φ 1 and φ 2 , are called eigenvectors. The eigenvectors are also called natural modes of vibration. Since the eigenvalue problem is a homogeneous equation, the eigenvectors are unique up to a multiplicative constant. In other words, any multiple of an eigenvector is itself an eigenvector. Therefore, without loss of generality, when finding the eigenvectors we can let the first entry in each eigenvector be equal to one, that is, (10 – 8) 1 1 φ 1 , φ 2 1 2 The eigenvectors are found by substituting Eq. (10 – 8) back into Eq. (10 – 5). The eigenvectors in Eq. (10 - 8) is said to be normalized. Note that we’ll normalize the eigenvectors in a different way in Section 3. Example Consider the two degree-of-freedom system given in which m1 = m2 = m, k1 = k3 = k, and k2 = k in Eq. (10 – 2). The two masses are the same and the middle spring is times stiffer than the springs on the ends. The system mass matrix and stiffness matrix are 1 0 1 M m K k 1 0 1 The eigenvalue problem, Eq. (10 – 6) with Eq. (10 – 8), is s 2 m k (1 ) 1 0 k k s 2 m k (1 ) 0 The characteristic equation, Eq. (10 – 7), reduces to 0 m 2 s 4 2mk (1 ) s 2 k 2 (1 2 ) The solution is s12 k m s 22 k (1 2 ) m CONTROL OF DYNAMICAL SYSTEMS: AN INTRODUCTORY APPROACH TREATING MULTI-DIMENSIONAL SYSTEMS Substituting these solutions into Eq. (10 – 6) yields 1 = 1 and 2 = –1 so the corresponding eigenvectors are 1 1 1 2 1 1 3. Properties of Eigenvalues and Eigenvectors The two degree of freedom systems considered above are composed of mass and stiffness. In Eq. (10 – 2) the mass matrix and the stiffness matrix are both symmetric matrices. This was not coincidental. In fact, in any two degree-offreedom system composed of mass and stiffness, the equations can be derived so that the mass and stiffness matrices are symmetric. Furthermore, the mass and stiffness matrices have another property; the property is analogous to the sign property of numbers. The mass matrix can be said to be positive much like the mass in a single degree of freedom system is positive, and the stiffness matrix can be said to be nonnegative, much like the stiffness in a single degree of freedom system is nonnegative. Of course, M and K are matrices, so we’ll need to define what we precisely mean by positive and nonnegative within the context of matrices. A symmetric matrix A is defined to be positive if the number y T Ay 0 for any nonzero y. Similarly, a symmetric matrix A is defined to be nonnegative if the number y T Ay 0 for any nonzero y. Utilizing these definitions, the mass and stiffness matrices satisfy the following symmetry and definiteness properties: M0 (10 – 9) M MT K0 K KT The eigenvalues and eigenvectors of systems that satisfy Eq. (10 – 9) are said to be normal mode systems. Normal mode systems satisfy three important properties. (1) The first normal mode property states that the eigenvalues are pure imaginary. Pre-multiplying Eq. (10 – 5) by φ T yields (10 – 10) 0 φ T [ s 2 M K ]φ s 2 (φ T Mφ ) (φ T Kφ ) From Eq. (10 – 9) (10 – 11) s2 φ T Mφ 0, φ T Kφ CONTROL OF DYNAMICAL SYSTEMS: AN INTRODUCTORY APPROACH TREATING MULTI-DIMENSIONAL SYSTEMS so s i 0 is pure imaginary, in which 0 is called a natural frequency of oscillation. The eigenvalue problem, Eq. (10 – 5), can now be written as (10 – 12) 02 Mφ Kφ This is the form of the eigenvalue problem that is usually encountered. (2) The second normal mode property states that the eigenvectors are real up to a multiplicative constant. To show this, write the eigenvector in terms of its real and imaginary components as (10 – 13) φ φ R φ Ii Substitute Eq. (10 – 13) into Eq. (10 – 12) and separate the real and imaginary parts to get (10 – 14) 02 Mφ R Kφ R , 02 Mφ I Kφ I Notice in Eq. (10 – 14) that the real part of the eigenvector and the imaginary part of the eigenvector satisfy the same eigenvalue problem. Therefore, the eigenvectors must be real – up to a multiplicative constant. (3) The third normal mode property is the orthogonality property of the modes. To show this property, pre-multiply the eigenvalue problem, Eq. (10 – 12), by the transpose of the eigenvectors, as shown below: 2 φ T2 : 01 Mφ 1 Kφ 1 (10 – 15) 2 φ 1T : 02 Mφ 2 Kφ 2 Next, subtract the two equations, recognizing the properties given in Eq. (10 – 9), to get (10 – 16) 2 2 ( 01 02 )φ T2 Mφ 1 0 Equation (10 – 16) reveals that eigenvectors having distinct eigenvalues are orthogonal with respect to the mass matrix. Substituting Eq. (10 – 15) into Eq. (10 – 16) further reveals that eigenvectors having distinct eigenvalues are orthogonal with respect to the stiffness matrix, too. Indeed, (10 – 17) φ 1T Mφ 2 0, φ 1T Kφ 2 0 Next, let’s normalize the eigenvectors, not in the manner given in Eq. (10 – 8), but rather according to the following: CONTROL OF DYNAMICAL SYSTEMS: AN INTRODUCTORY APPROACH TREATING MULTI-DIMENSIONAL SYSTEMS (10 – 18) φ 1T Mφ 1 1, φ T2 Mφ 2 1 Substituting Eq. (10 – 15) into Eq. (10 – 18) further reveals that (10 – 19) 2 2 φ 1T Kφ 1 01 , φ 1T Kφ 2 02 Finally, the orthogonality property and the normalization in Eq. (10 – 16) through Eq. (10 – 19) are written together in the more compact form as (10 – 20) φ Tr Mφ s rs , φ Tr Kφ s 02r rs where rs is the Kronecker-delta symbol in which rs = 0 when r and s are different and rs =1 when r and s are equal. Orthogonal vectors that have been normalized this way are called orthonormal. In summary, the eigenvalues and eigenvectors of two degree-of-freedom systems composed of mass and stiffness are normal in the sense that the eigenvalues are pure imaginary (the system oscillates), the eigenvectors are real, and they satisfy orthogonality properties with respect to the mass and stiffness matrices. As it turns out, the third property is precisely what enables us to separate a multi-degree-offreedom system into an independent set of single degree-of-freedom systems. 4. Modal Equations of Motion In the previous section, the eigenvalue problem was developed from the free vibration problem. However, since the eigenvectors form a basis set, any vector x can be expressed as a linear combination of the eigenvectors, whether x is associated with the freely vibrating system or not. Indeed, we’ll now see the merit of representing the general response of a two degree-of-freedom system in terms of the eigenvectors. We begin by representing x as (10 – 21) x(t ) φ 1q1 (t ) φ 2q2 (t ) where q1 (t ) and q2 (t ) are called modal displacements. In Eq. (10 – 21) we see that the r-th modal displacement represents the level of participation of the r-th mode of vibration in the overall response. Pre-multiplying Eq. (10 – 21) by 1T M and 2T M , we get (10 – 22) q1 (t ) φ 1T Mx(t ), q2 (t ) φ T2 Mx(t ) Equation (10 – 21) and Eq. (10 – 22) are transformations between the physical displacements x and the modal displacements q1 (t ) and q2 (t ) . CONTROL OF DYNAMICAL SYSTEMS: AN INTRODUCTORY APPROACH TREATING MULTI-DIMENSIONAL SYSTEMS Let’s now replace the physical coordinates in Eq. (10 – 4) with modal coordinates. Substituting Eq. (10 – 21) into Eq. (10 – 4) yields (10 – 23) M[φ 1q1 (t ) φ 2 q2 (t )] K[φ 1q1 (t ) φ 2 q2 (t )] F Pre-multiplying Eq. (10 – 23) by 1T and 2T , and invoking the orthogonality properties, Eq.(10 – 20), we get the equations that govern the modal motions of the two degree-of-freedom system 2 q1 (t ) 01 q1 (t ) Q1 (t ) (10 – 24) 2 q2 (t ) 02 q2 (t ) Q2 (t ) where (10 – 25) Q1 (t ) φ 1T F, Q2 (t ) φ T2 F are called modal forces. Equations (10 – 24) are two independent modal equations of motion; each has the form of an undamped single degree-of-freedom system subject to a modal force. The difference is that the coordinates are not physical – they’re more abstract. The coordinates are amplitudes of components of the motion. Likewise the model forces are not physical forces. They’re amplitudes of the components of force, as you’ll see momentarily. The independence of the modal equations implies that Q1 (t ) can excite or control only the first mode and that Q2 (t ) can excite or control only the second mode. But, again, what are the modal forces Q1 (t ) and Q2 (t ) ? How are they related to the physical forces? The physical forces are expressed in terms of the modal forces as follows: (10 – 26) F Mφ 1Q1 (t ) Mφ 2 Q2 (t ) Equation (10 – 26) can be verified by substituting Eq. (10 – 25) into Eq. (10 – 26) to render an identity. Thus, just as the modal displacements represent the degree to which a mode of vibration φ participates in the overall response, so too does a modal force represent the degree to which a mode of force Mφ participates in the overall force. In summary, the modal transformations of the displacements and of the forces are x(t ) φ 1q1 (t ) φ 2 q2 (t ), q1 (t ) φ 1T Mx(t ), q2 (t ) φ T2 Mx(t ) (10 – 27) F Mφ 1Q1 (t ) Mφ 2 Q2 (t ), Q1 (t ) φ 1T F, Q2 (t ) φ T2 F The modal displacements act as coordinates of single degree-of-freedom systems, each governed by a modal equation of motion, Eq. (10 – 24). Modal displacements CONTROL OF DYNAMICAL SYSTEMS: AN INTRODUCTORY APPROACH TREATING MULTI-DIMENSIONAL SYSTEMS are excited by modal forces. The spatial relationship between the modal components of force and motion are indicated symbolically below. (10 – 28) x(t ) φ 1 q1 (t ) F Mφ 1Q1 (t ) x(t ) φ 2 q 2 (t ) F Mφ 2 Q2 (t ) 5. The Double Pendulum As an illustration, consider the double pendulum shown in Fig. 10 – 2. Summing moments about each of the pivot points yields the two equations governing the motion of the double pendulum. m1 ga 1 M 1 2 m ga I 22 ka 2 ( 2 1 ) 2 2 M 2 2 I 11 ka 2 ( 2 1 ) (10 – 29) Fig. 10 - 2 Equation (10 – 29) can be written in the compact matrix-vector form of Eq. (10 – F 4) by letting x 1 and F 1 . The mass matrix and the stiffness matrix are 2 F2 0 m1 a 2 0 , I2 0 m2 a 2 2 m1 ga ka 2 ka 2 K m ga ka 2 ka 2 2 2 I M 1 0 (10 – 30) Substituting Eq. (10 – 30) into Eq. (10 – 6) yields CONTROL OF DYNAMICAL SYSTEMS: AN INTRODUCTORY APPROACH TREATING MULTI-DIMENSIONAL SYSTEMS (10 - 31) m ga 2 0 I1 ka 2 1 ka 2 2 0 det[ 02 M K ] det m ga ka 2 02 I 2 ka 2 2 2 2 3k 3k 3g 3g 3k 3k 3g 04 02 a 2a m1 m2 2a m1 m2 Equation (10 - 31) is a quadratic equation in 02 . Solving Eq. (10 - 31) (10 - 32) 1/ 2 3k 3k 3 g 2 a 1 3k 3k 3 g m1 m2 012 ,02 2 m1 m2 a 3 g 3k 3k 3g 2 4 2a m m 2a 2 1 3g 1 3k 3k 3 g 3k 3k 2a 2 m1 m2 a m1 m2 3 g 3k 3k 2a m1 m2 Assuming eigenvectors in the form of Eq. (10 - 5) and substituting Eq. (10 - 32) back into Eq. (10 - 5), we obtain the eigenvectors of the double pendulum (10 - 33) 3k m 1 3 k m2 3k m 2 3k m2 3k 1 m1 1 0 φ1 3k 1 0 1 m2 3k m m1 1 0 φ 2 2 3k 2 0 m1 m1 Finally, let’s normalize the modes according to Eq. (10 - 20). Doing this yields the normalized modes of vibration (10 - 34) CONTROL OF DYNAMICAL SYSTEMS: AN INTRODUCTORY APPROACH TREATING MULTI-DIMENSIONAL SYSTEMS 3 φ 1 2 (m1 m2 )a 1/ 2 1 3 ,φ 2 2 1 m1 m2 (m1 m2 )a 1/ 2 m2 m1 The normalized modes of force are (10 - 35) a2 Mφ 1 3(m1 m2 ) 1/ 2 a 2 m1 m2 m1 , M φ 2 m 2 3(m1 m2 ) 1/ 2 1 1 The modes of vibration and the modes of force are shown in Fig. 10 – 3. Fig. 10 - 3 Equation (10 - 32) provides us with temporal information concerning the double pendulum and Eq. (10 - 34) and (10 - 35) provide us with spatial information concerning the double pendulum. We see that the first mode is a “pendulum” mode wherein the system moves as a rigid body pendulum. The second mode is a “pseudo-elastic” mode associated with the spring and gravity. Equation (10 - 34) and Eq. (10 - 350 further reveal that applied moments in the proportions of m1 to m2 produce displacements in the proportions of 1 to 1, and that applied moments in the proportions of 1 to –1 produce displacements in the proportions of m2 to - m1 . Information like this is very useful to know when setting out to control the motion of a system. CONTROL OF DYNAMICAL SYSTEMS: AN INTRODUCTORY APPROACH TREATING MULTI-DIMENSIONAL SYSTEMS 6. Actuator Dynamics The transient response of multi-degree-of-freedom systems was treated in the previous sections. Let’s now look at the steady-state response of multi-degree-offreedom systems. We will do this in this section while examining actuator dynamics. As mentioned earlier, the dynamics of an actuator can interact with a system, producing a coupled system comprised of the mechanical part and the actuator. Of course, such interactions are undesirable in the sense that they complicate the design process – violating the goal to separate the components of a system into independent parts. Furthermore, the manner in which an actuator is mounted to a system is important. The method of mounting an actuator to a system affects the coupling of the actuator to the system. Even if the dynamics of an actuator and a system are uncoupled, the manner in which the actuator is mounted to the system dictates whether the actuator should be regarded as prescribing a force on the system or whether the actuator should be regarded as prescribing a displacement on the system. To sort this out, consider the two degree of freedom system given earlier in Eq. (10 - 2) and Eq. (10 - 3). Assume that the right mass and the right force represent an actuator, that the left mass and the left spring represent a mechanical part, and that the center spring is a mount between the actuator and the mechanical part (See Fig. 10 – 4). Fig. 10 - 4 Furthermore, assume that the actuator harmonically excites the mechanical part through the applied force F2. Let’s conduct a steady-state analysis to find the response of this harmonically-excited system. The mechanical force and the steady-state response are of the form (10 - 36) 0 0 X F it e it F0 e it , x p 1 e it Xe it X 2 F0 e F0 in which denotes excitation frequency. Substituting Eq. (10 - 36) into Eq. (10 3) and dividing by eit yields CONTROL OF DYNAMICAL SYSTEMS: AN INTRODUCTORY APPROACH TREATING MULTI-DIMENSIONAL SYSTEMS [ 2 M K ] X F0 (10 - 37) Let’s now recall the explicit form of the mass and stiffness matrices (10 - 38) m M 1 0 0 k k 2 , K 1 m2 k2 k2 k 2 and let’s express the physical parameters in these matrices in terms of the quantities m k k (10 - 39) r 2 , 12 1 , 22 2 m1 m1 m2 where r is the mass ratio, 1 is the frequency of the mechanical part (when isolated from the actuator), 2 is the frequency of the mount, and is the frequency of the excitation (as stated earlier). Substituting Eq. (10 - 39) into Eq. (10 - 37), dividing the first equations by m1 , and dividing the second equation by m2 , yields the mass and stiffness matrices (10 - 40) 2 r 2 1 0 1 2 M , K 2 0 1 2 r 22 22 From Eq. (10 - 38) we get (10 - 41) 2 12 r 22 22 r 22 1 X F0 2 2 m2 2 Solving Eq. (10 - 41) yields the amplitude of the steady-state response of the two degree-of-freedom system F r 22 X 0 m2 2 12 r 22 (10 - 42) 4 2 [12 22 r 22 ] 12 22 Also notice that the amplitude of the mount force between the masses is (10 - 43) FM k 2 ( X 2 X 1 ) F0 22 ( 2 12 ) The actuator and the system are uncoupled only if the excitation frequency has no affect on X1. From Eq. (10 - 42), it follows that the actuator and the sensor can be CONTROL OF DYNAMICAL SYSTEMS: AN INTRODUCTORY APPROACH TREATING MULTI-DIMENSIONAL SYSTEMS uncoupled only if the excitation frequency is significantly lower than the frequency of the mechanical part; so we impose the necessary condition (10 - 44) 1 The actuator dynamics and the dynamics of the mechanical part are uncoupled in the following two situations. In the first situation, the actuator prescribes a force and in the second situation the actuator prescribes a displacement. (a) Prescribed Force (Hard Mount) Let the actuator be significantly less massive than the mechanical part and let it be “hard mounted,” that is let r << 1 and 2 >> From Eq. (10 - 43) (10 - 45) FM = F0 The mount force is equal to the excitation force and independent of the excitation frequency. (b) Prescribed Displacement (Soft Mount) Next, let the actuator be significantly less massive than the mechanical part and let it be “soft mounted,” that is let r << 1 and 2 << From Eq. (10 - 42) F (10 - 46) X2 0 k2 We see that the displacement of the actuator is proportional to the excitation force and independent of the excitation frequency. In summary, an actuator that acts on a single degree of freedom system, if it produces a force according to Eq. (10 - 36), can be uncoupled from the system if the excitation frequency is much lower than the frequency of the mechanical part. Under these conditions, when the actuator is hard mounted to the system, it prescribes a force. When the actuator is soft mounted to the system it prescribes a displacement. These decoupling conditions are used in design to reduce complexity. CONTROL OF DYNAMICAL SYSTEMS: AN INTRODUCTORY APPROACH TREATING MULTI-DIMENSIONAL SYSTEMS PROBLEM SYSTEMS The systems described below go with the problems that are given in this chapter and Chapter 11. System 10 – 1 An elbow-block system is shown. Let mg = 20 lb, L = 4 ft, H = 3 ft, and k = 10 lb/in. The applied moment is M. The elbow is initially held at = 0 with the block hanging from the elbow at rest when the elbow is released. System 1: Elbow-Block System System 10 – 2 A block-blade system is shown. Let m = 12 kg, a = 5 m, M = 30 kg, rg = 2 m, g = 10 m/s2, and k = 36 Nm. Assume that the system lies in the horizontal plane and it is shown in its equilibrium position. The applied force is f. The blade is initially held at = 0 and the spring compressed 0.1 m when released from rest. System 2: Block-Blade System CONTROL OF DYNAMICAL SYSTEMS: AN INTRODUCTORY APPROACH TREATING MULTI-DIMENSIONAL SYSTEMS System 10 – 3 A double pendulum is shown. Let a1 = a2 = 1 m, m1 = m2 = 5 kg, and g = 10 m/s2. Applied moments M1 and M2 act on the respective bars (not shown). M1 is an external moment and M2 is an internal moment. Initially, the system is at rest with 1 = 10° and 2 = 0. System 3: Double Pendulum System System 10 – 4 A hanging two-bar system is shown. Let L = 0.75 m, m = 2 kg, A0 = B0 = 30°, and k = 10 N/m. Applied moments MA and MB act on the respective bars. Initially the system is at rest and A = 40° and B = 30°. System 4: Hanging Two-Bar System System 10 – 5 A bar-disk system is shown. Let mA = 1 kg, mB = 5 kg, L = 0.2 m, R = 0.1 m, and k = 25 N/m. The disk is uniform and rolls without slip. The applied moments MA and MB act on the system. The system is shown in its equilibrium position. Initially, the system is at rest, the bar is vertical and the spring is stretched 0.05 m. System 5: Bar-Disk System CONTROL OF DYNAMICAL SYSTEMS: AN INTRODUCTORY APPROACH TREATING MULTI-DIMENSIONAL SYSTEMS System 10 – 6 A hanging two leg system is shown. Let m = 14 slug, k = 900 lb/ft, and a = 0.5 ft. The equilibrium angle of the system is 0 = 60°. The bars are uniform. The applied moments MA and MB act on the system. Initially the system is at rest, A = 10° andB = 0. System 6: Hanging Two-Leg System PROBLEM STATEMENTS Problem 10 – 1: Deriving Linear Differential Equations (a) Carefully draw a free body diagram of each body. In the diagram, show the system in a displaced position. Remember that the displacement is small. The diagram should show a coordinate system for each body, all of the forces acting on each body, and other dimensions that are needed. The diagram should also list all of the given parameters. Note: In the diagram there are rotated points. Say, point A is rotated about point O. Remember for a small angle of rotation that point A moves perpendicular to line OA. The displaced point A should be shown that way in the diagram. (b) Derive the two linear differential equations that govern the motion of the system. (c) Place the equations in the vector form of Eq. 10 – 3, indicating the displacement vector x and the mass and stiffness matrices M and K. Problem 10 – 2: Natural Frequencies and Natural Modes This problem follows Problem 10 – 1. (a) Find the two natural frequencies of the system. (b) Find the two natural modes of the system. Normalize them so that the first entry is 1. (c) Verify that the modes are orthogonal. CONTROL OF DYNAMICAL SYSTEMS: AN INTRODUCTORY APPROACH TREATING MULTI-DIMENSIONAL SYSTEMS Problem 10 – 3: Free Response This problem follows Problem 10 – 2. Set the applied moments and applied forces to zero. (a) Find and graph the response of each of the degrees of freedom. CONTROL OF DYNAMICAL SYSTEMS: AN INTRODUCTORY APPROACH

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# II Single Degree of Freedom ( 1DOF ) Systems