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Chapter 5 Modeling with Higher-Order Differential Equations

Chapter 5 Modeling with Higher-Order Differential Equations. Overview. 5.1 Linear Equation : Initial-Value Problems 5.1.1 Spring/Mass System : Free Undamped Motion 5.1.2 Spring/Mass System : Free Damped Motion 5.1.1 Spring/Mass System : Driven Motion 5.1.4 Series Circuit Analogue

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Chapter 5 Modeling with Higher-Order Differential Equations

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  1. Chapter 5 Modeling with Higher-Order Differential Equations 1

  2. Overview • 5.1 Linear Equation : Initial-Value Problems • 5.1.1 Spring/Mass System : Free Undamped Motion • 5.1.2 Spring/Mass System : Free Damped Motion • 5.1.1 Spring/Mass System : Driven Motion • 5.1.4 Series Circuit Analogue • 5.2 Linear Equations : Boundary-Value Problems • 5.3 Nonlinear Equations 2

  3. INTRODUCTION • In section 5.1, we deal exclusively with initial-value problems. • In section 5.2, we examine applications described by boundary-value problem. In this section, we also see how some boundary-value problems lead to the concepts of eigenvalues and eigenfunctions . • Section 5.3, beings with a discussion of the differences between linear and nonlinear springs. 3

  4. 5.1 Linear Equation : Initial-Value Problems • In this section we consider several linear dynamical systems in which each mathematical model is a second-order differential equation with constant coefficients. • Recall that the function g is the input or forcing function of the system. A solution y(t) of the differential equation on an interval containing t0 and satisfies prescribed initial condition y(t0) = y0 , is the output or response of the system. 4

  5. Hook’s Law • Suppose that a flexible spring is suspended vertically from a rigid support and then a mass m is attached to its free end. The spring it self exerts a restoring force F opposite to the direction of elongation and proportional to the amount of elongation s. where k is a constant of proportionality called the spring constant. • The spring is essentially characterized by the number k. 5

  6. Newton’s Second Law • After a mass m is attached to a spring , it stretches the spring by an amount s and attains a position of equilibrium at which its weight W is balanced by the restoring force ks. W = mg g = 32 ft/s2, 9.8m/s2, or 980 cm/s2 • We can equate Newton’s Second Law with the net, or resultant, force of the restoring force and the weight : (1) 6

  7. DE of Free Undamped Motion • By dividing (1) by the mass m we obtain the second-order differential equation (2) where . Equation (2) is said to describe simple harmonic motion or free undamped motion. • Two obvious initial conditions associated with (2) are x(0) = x0 , the amount of initial displacement, and , the initial velocity of the mass. 7

  8. Solution and Equation of Motion • To solve equation (2) we note that the solutions of the auxiliary equation are the complex numbers , . (3) The period of free vibrations described by (3) is , and the frequency is . • A maximum of x(t) is a positive displacement corresponding to the mass’s attaining a maximum height above the equilibrium position. • When the initial conditions are used to determine the constants and in (3), we say that the resulting particular solution or response is the equation of motion. 8

  9. Alternative Form of x(t) • When and , the actual amplitude A of free vibrations is not obvious from inspection of equation (3). Hence it is often convenient to convert a solution of form (3) to the simpler form (6) where and is a phase angle defined by • To verify this we expand (6) by the addition formula for the sine function : (8) then (9) 9

  10. System with Variable Spring Constants • In one model for the aging spring the spring constant k in (1) is replaced by the decreasing function The linear differential equation cannot be solved by the methods considered in Chapter 4. • When a spring/mass system is subjected to an environment in which the temperature is rapidly decreasing, it might make sense to replace the constant k with a function that increases with time. • The resulting model is a form of Airy’s differential equation. 10

  11. 5.1.2 Spring/Mass Systems : Free Damped Motion • When no other external forces are impressed on the system, it follows from Newton’s second law that (10) • The differential equation of free damped motion is or (11) where • The symbol is used only for algebraic convenience since the auxiliary equation is and the corresponding roots are then 11

  12. DE of Free Damped Motion • CASE I : In this situation the system is said to be overdamped since the damping coefficient is large when compared to the spring constant k . • The corresponding solution of (11) is or (13) 12

  13. DE of Free Damped Motion • CASE II : The system is said to be critically damped since any slight decrease in the damping force would result in oscillatory motion. • The general solution of (11) is or (14) 13

  14. DE of Free Damped Motion • CASE III : In this case the system is said to be underdamped since the damping coefficient is small compared to the spring constant.The roots and are now complex : • Thus the general solution of equation (11) is 14

  15. 5.1.3 Spring/Mass System : Driven Motion • Suppose we now take into consideration an external force acting on a vibrating mass on a spring. • The inclusion of in the formulation of Newton’s second law gives the differential equation of driven or forced motion : (24) • Dividing (24) by m gives (25) where F(t) = f(t)/m and, as in the preceding section, . 15

  16. Pure Resonance • We expect that over a length of time we should be able to substantially increase the amplitudes of vibration. • For we define the solution to be (31) • When the displacements become large; in fact when The phenomenon we have just described is known as pure resonance. 16

  17. 5.1.4 Series Circuit Analogue • LRC Series Circuits (32) • By Kirchhoff’s second law,the sum of the these voltages equals the voltage impressed on the circuit; (33) • The charge on the capacitor is related to the current by then (34) 17

  18. LRC Series Circuits • If E(t) = 0, the electrical vibrations of the circuit are said to be free. • Since the auxiliary equation for (34) , there will be tree forms of the solution with depending on the value of the discriminant In each of these three cases the general solution of (34) contains the factor ,and . 18

  19. 5.2 Linear Equations : Boundary-Value Problems • Assume a beam of length L is homogeneous and has uniform cross-sections along its length. In the absence of any load on the beam (including its weight), a curve joining the centroids of all its cross-section is a straight line called the axis of symmetry. • If a load is applied to the beam in a vertical plane containing the axis of symmetry, the beam. • Undergoes a distortion, and the curve connecting the centroids of all cross-sections is called the defection curve or elastic curve. 19

  20. Deflection of a Beam • In the theory of elasticity it is shown that the bending moment M(x) at a point x analog the beam is related to the load per unit length w(x) by the equation (1) • In addition, the bending moment M(x) is proportional to the curvature k of the elastic curve (2) where E and I are constant. • The product EI is called the flexural rigidity of the beam. 20

  21. When the deflection y(x)is small, the slopeand so If we let , equation (2) become . The second derivative of this last expression is (3) • Using the given result in (1) to replace in (3), we see that the deflection y(x) satisfies the first-order differential equation (4) 21

  22. For a cantilever beam the deflection y(x) must satisfy the following two conditions at the embedded end x = 0 : • = 0 since there is no deflection,and • = 0 since the deflection curve is tangent to the x-axis (in other words, the slope of the deflection curve is zero at this point). • At x = L the free-end conditions are • = 0 since the bending moment is zero, and • = 0 since the shear force is zero. 22

  23. The function is called the shear force. • If an end of a beam is simply supported (also called pin supported, fulcrum supported, and hinged), then we must have y = 0 and = 0 at that end. 23

  24. Example 2 Nontrivial Solution of BVP • Solve the Boundary-value problem , • CASE I. For = 0 the solution of is the condition y(0) = 0 and y(L) = 0 The only solution of the boundary-value problem for = 0 is the trivial solution y = 0. 24

  25. CASE II. For < 0 y(0) = 0 so y(L) = 0 since , so y = 0 • CASE III. For > 0 y(0) = 0 y(L) = 0 25

  26. Ify = 0 , • The last condition implies that the argument of the sine function must be an integer multiple of : • Therefore for any real nonzero , is a solution of the problem for each n. 26

  27. Buckling of a Thin Vertical Column • Let y(x) denote the deflection of the column when a constant vertical compressive force, or load, P is applied to its top. • By comparing bending moments at any point along the column we obtain (5) where E is Young’s modulus of elasticity and I is the moment of inertia of a cross- section about a vertical line through its centroid. 27

  28. Rotating String • The simple linear second-order differential equation (6) • As a model that defines the deflection curve or the shape y(x) assumed by a rotating string. 28

  29. Rotating String • The net force is (7) When angles and (measured in radians) are small, we have and . 29

  30. Since and are, in turn, slopes of the lines containing the vectors T1 and T2 , we can also write Thus (7) becomes (8) • The mass of string on the interval is ; the acceleration of a body rotating with angular speed in a circle of radius r is . With small we take r = y . 30

  31. The net vertical force is also approximated by (9) where the minus sign comes from the fact that the acceleration points in the direction opposite to the positive y-direction. • By equating (8) and (9) (10) 31

  32. For close to zero the difference quotient in (10) is approximated by the second derivative . Finally we arrive at the model (11) • Since the string is anchored at its ends x = 0 and y = L , we expect that the solution y(x) of the last equation in (11) should also satisfy the boundary conditions y(0) = 0 and y(L) = 0. 32

  33. 5.3 Nonlinear Equations • The mathematical model in (1) of section 5.1 has the form (1) where F(x) = kx. • A spring acting under a linear restoring force F(x) = kx is naturally referred to as a linear spring. But spring are seldom perfectly linear. • Depending on how it is constructed and the material used, a spring can range from “mushy”,or soft, to”stiff”, or hard,so its restorative force may vary from something below to something above that given by the linear law. 33

  34. Nonlinear Springs • The cube of the displacement x of the mass beyond its equilibrium position or that F(x) is a linear combination of powers of the displacement such as that given by the nonlinear functions . • A spring whose mathematical model incorporates a nonlinear restorative force, such as (2) is called a nonlinear spring. 34

  35. Nonlinear Springs • The nonlinear differential equation (3) is one model of a free spring/mass system with damping proportional to the square of the velocity. • One can then envision other kinds of models : • Linear damping and nonlinear restoring force • Nonlinear damping and nonlinear restoring force • The point is that nonlinear characteristics of a physical system lead to a mathematical model that is nonlinear. 35

  36. Both and are odd functions of x in (2). • Let us express F as a power series centered at the equilibrium position x = 0 : when the displacements x are small, the value of are negligible for n sufficiently large. • If we truncate the power series with, say, the fourth term, then 36

  37. In order for the force at x > 0 and the force at –x < 0 to have the same magnitude but act in opposite directions, we must have F(-x) = -F(x). • Since this means F is an odd function, we must have c0 = 0 and c2 = 0 and so . • A restoring force with mixed powers, such as , and the corresponding vibrations are said to be unsymmetrical. 37

  38. Hard and Soft Springs • The restoring force is given by , k > 0. If k1 > 0, the spring is said to be hard. If k1 < 0, the spring is said to be soft. 38

  39. Nonlinear Pendulum • Any object that swings back and forth is called a physical pendulum. • The displacement angleof the pendulum, measured from the vertical. 39

  40. The arc s of a circle of radius l is related to the central angle by the formula . Hence angular acceleration is • In direction this force is , since it point to the left for > 0 and to the right for < 0. • We equate the two different versions of the tangential force to obtain or (6) 40

  41. Linearization • The Maclaurin series for is given by and if the approximation , equation (6) becomes • If we assume that the displacement are small enough to justify using the replacement , then (6) becomes (7) 41

  42. Suspended Wire • Three forces are acting on the wire : the weight of the segment P1P2 and the tension T1 and T2 in the wire at P1 and P2 , respectively. • The tension T2 resolves into the horizontal and vertical components (scalar quantities) and . • Dividing the last two equations, we find . That is, (8) 42

  43. Since the arc length between point P1 and P2 is given by (9) it follows from the fundamental theorem of calculus that the derivative of (9) is (10) • Differentiating (8) with respect to x and using (10) leads to (11) 43

  44. Rocket Motion • The differential equation of a free-falling body of mass m near the surface of the earth is given by where s represents the distance from the surface of the earth to the object and the positive direction is considered to be upward. • If the positive direction is upward and air resistance is ignored, then the differential equation of motion after fuel burnout is 44

  45. Rocket Motion • To determine the constant k we use the fact that when . Thus the last equation in (12) becomes (13) 45

  46. Variable Mass • The second law of motion, as originally advanced by Newton, state that when a body of mass m moves though a force field with velocity v, the time rate of change of the momentum mv of the body is equal to the applied or net force F acting on the body. (14) • If m is constant, then (14) yields the more familiar form , where a is acceleration. 46

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