Applications of Second-Order Differential Equations

Transcription

1 Applications of Second-Order Differential Equations ymy/013

2 Building Intuition Even though there are an infinite number of differential equations, they all share common characteristics that allow intuition to be developed: Particular and complementary solutions Effects of initial conditions Lect1 EEE 0

3 Mechanical Vibrations The study of motion of objects and the effect of forces acting on those objects. Example mass spring system, cantilever, pendulum, Electric Circuits Spring: Support Structure Mass: Mass of the bridge structure

4 Spring-Mass Systems We consider the motion of an object with mass at the end of a spring that is either vertical (as in Figure 1) or horizontal on a level surface (as in Figure ). Figure 1 Figure

5 An object of mass m is suspended from the spring and stretches it a length s when the spring comes to rest in an equilibrium position. According to Hooke s Law, the tension force in the spring is ks, where k is the spring constant. The force due to gravity pulling down on the spring is mg, and equilibrium requires that ks mg (1) Hooke s Law: The spring exerts a restoring force F opposite to the direction of elongation and proportional to the amount of elongation

6 Forces acting on the object : Fp mg F k( s y) s F r dy d dt the propulsion force due to gravity the restoring force of the spring s tension a frictional force assumed proportional to velocity The propulsion force (weight), F p pulls the mass downward, but the spring restoring force F s and frictional force F r pull the mass upward. The motion starts at y y 0 with the mass vibrating up and down.

7 The frictional force tends to retard the motion of the object. The resultant of these forces is and by Newton s second law F Fp Fs Fr F ma d y m mg ks ky dt, we then have dy d dt By Equation (1), mg ks 0, so this last equation becomes subject to the initial conditions d y dy m d ky 0 () dt dt y(0) y0, y (0) 0

8 Remark The nature of the vibrations/oscillations described by the differential equation depends on the constants m, d, and k. d is known as the damping constant. The damping may be due to a damper such as a dashpot, internal damping such as friction within the spring, or external damping such as aerodynamic drag. The displacement, y, at any time, t, of the mass is the output from the system.

9 Free, Undamped Vibrations : Simple Harmonic Motion Suppose there is no retarding frictional force. Then d 0 and there is no damping. In this case, once motion has started it will continue for ever. The motion is governed by Second order linear homogeneous constant-coefficient equation If k w m dy m ky 0 (3) dt, then the second-order equation (3) becomes w y 0 with y(0) y 0, y (0) 0 dy dt

10 The auxiliary equation is roots r r w wi 0 having the imaginary The general solution to the differential equation in (3) is y( t) C cos( wt) C sin( wt) 1 Substituting the conditions yields C1 y0 and C 0 y( t) y cos( wt) (4) describes the motion of the object. Equation (4) represents simple harmonic motion of amplitude and period 0 T p w y 0

11 Note The general solution to the differential equation (3), that is can also be written as where w k m y( t) C cos( wt) C sin( wt) 1 y( t) Acos( wt q) frequency A C1 C amplitude frequency at which the system wants to oscillate without external interference. 1 cos q C C sin phase angle A q A q

12 Example 1 A spring with a mass of kg has natural length 0.5 m. A force of 5.6 N is required to maintain it stretched to a length of 0.7 m. If the spring is stretched to a length of 0.7 m and then released with initial velocity 0, find the position of the mass at any time. Solution The motion is governed by From Hooke s Law, the force required to stretch the spring is dy m ky 0 (3) dt so the spring constant, k(0.) k 18 0.

13 Using this value of k together with m = in Equation (3), we have dy dt 18y 0 The solution of this equation is y( t) C cos 8t C sin 8 t (*) Given the initial condition y(0) 0.. But, from Equation (*) Differentiating Equation (*), we get 1 y(0) C1 C1 0. y ( t) 8C 1sin 8t 8C cos 8t Since the initial velocity is y (0) 0, we have C 0 and so the solution is 1 y( t) cos 8t 5

14 Damped Motion The motion of a spring is subject to a frictional force (in the case of the horizontal spring of Figure ) or a damping force (in the case where a vertical spring moves through a fluid as in Figure 3). Figure Coil spring Dashpot for damping Figure 3 Example: the damping force supplied by coilovers in motorcycle suspension

15 So d 0 in Equation (), If we substitute then the differential equation () is d y dy m d ky 0 () dt dt w k m and b d m d y dy b w y 0 (5) dt dt The auxiliary equation is r br w 0 with roots r b b w

16 We now have 3 cases depending upon the relative sizes of b and Case 1: b w The double root of the auxiliary equation is real and equals The general solution to Equation (5) is y( t) ( C1 Ct) e wt This situation of motion is called critical damping and is not oscillatory. r w w

17 Case : b w The roots of the auxiliary equation are real and unequal, given by 1 w and w r b b r b b The general solution to Equation (5) is b b w t b b w t y() t C e C e 1 Here again the motion is not oscillatory and both r 1 and r are negative. Thus y approaches zero as time goes on. This motion is referred to as overdamping.

18 Case 3: b w The roots of the auxiliary equation are complex and given by r b i w b The general solution to Equation (5) is bt 1 w w y( t) e C cos b t C sin b t This situation, called underdamping, represents damped oscillatory motion. p It is analogous to simple harmonic motion of period T but the amplitude is not constant but damped by the factor e. Therefore, the motion tends to zero as t increases, so the vibrations tend to die out as time goes on. w bt b

19 Three examples of damped vibratory motion for a spring system with friction, d 0 (a) damping is just sufficient to suppress vibrations. (b) strong damping force (high-viscosity oil or grease) compared with a weak spring or small mass (c) the motion decays to 0 as time increases.

20 Example Suppose that the spring of Example 1 is immersed in a fluid with damping constant d 40. Find the position of the mass at any time t if it starts from the equilibrium position and is given a push to start it with an initial velocity of 0.6 m/s. Solution From Example 1 the mass is m = and the spring constant is k = 18, so the differential equation (5) becomes d y dy 0 64y 0 dt dt

21 The auxiliary equation is having roots r r 0r 64 ( r 4)( r 16) 0 4 and r 16 so the motion is overdamped and the solution is We are given that we get 4t 16t 1 y() t C e C e y(0) 0 so C1 C 0 4t 16t 1 y ( t) 4C e 16C e so y (0) 4C 16C Differentiating, Since C C1, this gives 1C1 0.6 C Therefore 4t 16t y( t) 0.05( e e )

22 Exercise 1. Find the damping parameter and natural frequency of the system governed by d y dy 6 dt dt 9y 0. A spring with a mass of kg has damping constant 14, and a force of 6 N is required to keep the spring stretched 0.5 m beyond its natural length. The spring is stretched 1 m beyond its natural length and then released with zero velocity. (i) Find the position of the mass at any time t. (ii) Find the mass that would produce critical damping. d y dy 14 1y 0, y(0) 1, y (0) 0 dt dt d d 4mk 0 m kg 4k 4(1) 1

23 3. A spring has a mass of 1 kg and its spring constant is k = 100. The spring is released at a point 0.1 m above its equilibrium position. Graph the position function for the following values of the damping constant d : 10, 15, 0, 5, 30. What type of damping occurs in each case? d d d 10 & 15, underdamped 0, critically damped 5 & 30, overdamping

24 Forced Vibrations Suppose that, in addition to the restoring force and the damping force, the motion of the spring is affected by an external force F(t). Then Newton s Second Law gives dy m restoring force + damping force + external force dt dy = ky d F( t) dt Thus, the motion of the spring is now governed by the following non-homogeneous differential equation: d y dy m d ky F( t) (6) dt dt

25 The general solution to Equation (6) is y( t) y ( t) y ( t) p where the complementary solution will be the solution to the free, damped case and the particular solution will be found using undetermined coefficients or variation of parameters. Note The complementary solution will approach zero as t increases. Because of this the complementary solution is often called the transient solution. The particular solution is often called the steady state solution or forced response. c

26 Remark A commonly occurring type of external force is a periodic force function F( t) F cos w t where w w In the absence of a damping force ( d 0 ) F y( t) C cos t C sin t cos t 0 1 w w 0 m( w w ) w 0 If w0 w, then the applied frequency reinforces the natural frequency and the result is vibrations of large amplitude. This is the phenomenon of resonance. k m

27 Example Suppose a machine is attached to a 1 kilogram mass and the machine exerts a force of sin t newtons on the mass at time t. In addition the mass is attached to a spring having spring constant 4. The mass slides along a frictionless horizontal surface. The equation of motion is dx 4 x( t) sint dt for t > 0. Solution Here x(t) is the displacement from the equilibrium position (which is taken to be 0) at time t.

28 General solution of the corresponding homogeneous equation: Particular integral of the non-homogeneous equation: 1 yp( t) sin t 3 The general solution of the original non-homogeneous equation: 1 y( t) sin t C1 cos t C sin t 3 Determine the constants C 1 and C from the initial conditions:

29 Simple Electric Circuits The LCR circuit contains a resistor R, an inductor L, and a capacitor C, connected in series with a switch and battery (d.c. voltage source). V IR potential difference across resistor Q V C potential difference across capacitor V di L potential difference across inductor dt The total potential difference around the circuit must be zero and we have di Q L RI 0 dt C

30 The principle of conservation of charge tells us that the current is equal to the rate of change of charge, that is we have : dq I dt We use this to eliminate Q, we obtain di 1 L RI I dt dt C or to eliminate I, we obtain d Q dq 1 L R Q dt dt C Differentiating either equations, we obtain d I di 1 L R I dt dt C (7)

31 Note We have differential equations for one of the variables: either the charge on the capacitor in the circuit or the current in the circuit. The equations are equivalent, but equation (7) is the most usual and occurs widely in engineering applications.

32 Example A circuit contains a 1 henry inductor, a 000 ohm resistor, and a 1 microfarad capacitor. Initially there is no charge in the circuit and the initial current is 1 ampere. Find the charge in the circuit as a function of time. Solution The equation governing the charge is d Q dt dq Q( t) 0, Q(0) 0, Q (0) 1 dt The characteristic equation has a repeated real root at So the general solution is Q( t) ( C C t) e t

33 Since Q(0) 0, Q (0) 1, 0 = C 1 and 1 = C. Hence Q() t te 1000t Exercise Suppose in the previous example the resistance of the resistor is (i) 1990 ohms (ii) 010 ohms Find the charge in the circuit. (i) (ii) a periodic solution with exponentially decaying amplitude distinct real roots that leads to rapidly decaying charge

34 Oscillations in Electrical Circuits If an alternating voltage signal is applied to a simple LCR electrical circuit, the equation governing the resulting oscillations is also a second-order linear ODE. Suppose the circuit contains an electromotive force (supplied by a battery or generator), a resistor, an inductor, and a capacitor, in series. V IR voltage drop across resistor V I dt C V di L dt voltage drop across inductor voltage drop across capacitor

35 Kirchhoff s voltage law says that the sum of these voltage drops is equal to the supplied voltage: Since I di 1 L RI dt C I dt E() t dq, this equation becomes dt d Q dq 1 L R Q dt dt C E( t) (8) which is a second-order linear differential equation with constant coefficients. If the charge Q 0 and the current I 0 are known at time 0, then we have the initial conditions Q(0) Q0 Q (0) I(0) I0

36 Note the similarity between the differential equations of the two quite different systems given above. As L, R, and C are all positive, this system behaves just like the mass and spring system. The position of the mass is replaced by the current. Mass is replaced by the inductance, damping is replaced by resistance and the spring constant is replaced by one over the capacitance. The change in voltage becomes the forcing function. Hence for constant voltage this is an unforced motion.

37 This similarity is used in analogue systems in which a mechanical system can be simulated by the equivalent electrical system.

38 Example Find the charge and current at time in the circuit if R 40, 4 L 1H, C F, E( t) 100 cos10t charge and current are both 0. and the initial Solution With the given values of L, R, C, and E(t), Equation (8) becomes d Q dq 40 dt dt 65Q 100 cos 10t The characteristic equation have complex roots 0 15i. So the complementary function is 0t Q ( t) e ( C cos15t C sin15 t) c 1

39 Using the method of undetermined coefficients we try the trial solution Substituting into the DE and equating coefficients, we have (DIY) the particular integral and the general solution is Qp( t) Acos10t B sin10t 4 Qp( t) (1cos10t 16 sin10 t) t Q( t) (1cos10 t 16sin10 t) e ( C1cos15t Csin15 t) 697

40 Imposing the initial conditions formula for the charge is Q(0) 0, I(0) 0, we get the 4 e Q( t) (1cos10t 16sin10 t) ( 63cos15t 116sin15 t) t and the expression for the current is 1 0t I( t) 10( 1sin10t 16cos10 t) e ( 190cos15t 13060sin15 t) 091

41 Exercise The equation for an LCR circuit with applied voltage E is di 1 L RI Q E dt C By differentiating this equation find the solution for Q(t) and I(t) if L = 1, R = 100, C = 10 4 and E = 1000 given that Q = 0 and I = 0 at t = 0.