Thompson - Computing for Scientists and Engineers (523188), страница 49
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For simplicity, the bumping forces as a function of time, t, are approximatedas having a single angular frequencythus varying as F0cosThe inertialmass of the system we denote by M, and the system is modeled as having compressional forces from a Hooke’s law spring (restoring force proportional to displacement) having spring constant K and damping forces proportional to speed with proportionality constant B.Exercise 8.2First write down Newton’s force equation, (8.1), for the mechanical system justdescribed, then rearrange it so that all the derivatives are on the left-hand side, inorder to obtain the differential equation(8.4)for the mechanical model of the automobile suspension, where the primes denotederivatives with respect to time.
nNow consider an AC series circuit for which we will describe the free charge inthe circuit at time t,responding to an external EMF of E0cosThe inductive inertia of the system we denote by L, the system has a capacitance of C, andthere is an Ohm’s law resistance (damping) with value R. The electrical analog ofrelation (8.3) between mechanical displacement and momentum is given by260SECOND-ORDER DIFFERENTIAL EQUATIONS(8.5)relating the charge to the time-dependent current, I. Recall that the voltage dropacross an inductor is L dI /dt, that the drop across a resistor is RI, while the dropacross a capacitor is Q/C. This information can be used to derive a differential equation for the AC circuit containing the three elements in series.Exercise 8.3Write down Kirchhoff’s second law (conservation of energy) to equate the sumof voltage drops across the three circuit elements to the source EMF, in order toobtain the differential equation for the electrical model of the series AC circuit,(8.6)with the primes denoting derivatives with respect to time.
nWe see immediately that the mechanical and electrical systems are completely analogous and can be represented by a single second-order differential equation, namely(8.7)in which the derivatives are with respect to control variable x. The correspondencebetween the variables in the last three differential equations is given in Table 8.1.TABLE 8.1 Correspondence of mechanical, electrical, and general differential equation quantities.To understand how these analogies arise, apart from the correctness of the formal equations, try working the following exercise.8.1FORCES, SECOND-ORDER EQUATIONS, RESONANCES261Exercise 8.4In Table 8.1 of the analogies between mechanical and electrical systems, consider the intuitive explanations of the correspondences.
To do this, discuss the frequency response of the amplitude, damping, and restoring terms in the differential equations (8.6) and (8.7). nWith this preparation on setting up and interpreting the differential equations, we areprepared to solve and interpret the linear equations that arise when the source term iszero.Solving and interpreting free-motion equationsThe analysis of the general differential equation (8.7) is easiest to approach by firstconsidering the situation with no driving term, S0 = 0. Then we have a linear second-order equation.
After solving this, it will be straightforward to solve the forcedmotion equation in the next subsection. The differential equation to be solved isnow(8.8)We know from common experience (in the context that x represents time) thatsprings undulate and circuits oscillate, but that their motion eventually damps out ifthere is no driving term. This suggests that we look for solutions of the form(8.9)where m is complex in order to allow both damping and oscillation. By substitutinginto the differential equation (8.8) one obtains directly that(8.10)One solution of this equation is that y(x) = 0 for all x. In our examples the controlling variable x is the time, so this solution states that if the system is initially undisturbed it will remain undisturbed if there are no driving terms.
A solution of(8.10) that is more interesting is to imagine that driving terms were present for someearlier x but were removed, so that the system then evolved according to (8.8).Then the appropriate solution of (8.10) is that the parentheses containing the mterms must be zero.Exercise 8.5(a) Equate the m-dependent expression in (8.10) to zero in order to find the twosolutions for m, namely(8.11)262SECOND-ORDER DIFFERENTIAL, EQUATIONSwhere the discriminant, D, is given by(8.12)(b) Show that the most general solution of (8.8) can therefore be expressed interms of dimensionless variables as any linear superposition of the two basicsolutions(8.13)in which the natural frequency is(8.14)the dimensionless damping parameter is(8.15)0and the dimensionless discriminant is(8.16)so that the corresponding dimensionless solutions are(8.17)(c) Show that the differential equation corresponding to (8.8) is(8.18)in which derivatives are taken with respect to dimensionless variablenThe expressions for the natural angular frequency of oscillation,and the damping parameter, for the mechanical and electrical systems are given in Table 8.1.Plots of the relation (8.17) as a trajectory in the complex plane are explored in Section 2.3.The solution (8.11) does not cover all bases, because we know that any secondorder differential equation has two linearly-independent solutions, whereas (8.11)collapses to a single solution whenthat is, whenThis case requires special treatment.Exercise 8.6Verify by substitution into (8.18) that whenthe expression(8.19)solves the differential equation for any choice of the constants y+ and y-.
n8.1FORCES, SECOND-ORDER EQUATIONS, RESONANCES263FIGURE 8.1 Amplitude of free motion, y. as a function of damping, using dimensionless variables. The solid curveand the dashed curveis for damped oscillation, the dotted curve is for critical dampingis for overdamped motion.The degenerate case (8.19) is called critical damping because it is sensitive to theamount of damping, as indicated by b in the original differential equation (8.8).Critical damping provides a dividing line between the oscillatory and nonoscillatorysolutions of (8.8) that we now consider.Suppose that the modified discriminant, in (8.16), is positive, so that thesquare root in (8.17) is real, then the solutions are clearly oscillatory, but damped.If is negative, the solutions are always exponentially-damped as a function of x,since v± is then real and negative.
Such solutions are called overdamped.Thus, the free-motion solutions to the damped harmonic oscillator equation areof three types: damped oscillatorycritically dampedand overdampedThey are illustrated in Figure 8.1.Figure 8.1 shows that critical damping lies at a boundary between solutions thatoscillate with X, no matter how slowly, and solutions that die away uniformly without oscillation. It might therefore be expected that the transition from oscillatory tooverdamped motion is smooth.
This is not so, as you can show by working thefollowing exercise.Exercise 8.7(a) To investigate the behavior of the amplitude y as a function of parameterin (8.13) choose they+ solution and in (8.19) choose y- = 0, y+ = 1, so thaty values are the same atThus, continuity of values is assured. Next,differentiate the expression for each y with respect to Show that asfrom below (oscillation side) the slope is continuous, but that asfromabove (overdamped side) the slope(b) Make graphs of y against at fixed x values, say atfor in the range 0 to 2.
Show that there is always a cusp at= 1. n264SECOND-ORDER DIFFERENTIAL EQUATIONSFIGURE 8.2 Phase-space plot (Poincaré map) for slightly damped motionin dimen-sionless units).It is interesting to explore how y' = dy /dx, the analog of the speed in the mechanical system or the current in the electrical system, changes with y, the analog ofdisplacement or charge, respectively. The graph of y' against y is called a “phasespace plot” or “Poincaré map.” Figure 8.2 shows the phase-space plot for(dimensionless x units) andwhich is not much damped.Notice in Figure 8.2 how the motion is gradually damped as the “velocity” and“displacement” both tend to zero as x (or “time”) increases. You may improve yourunderstanding of the notion of phase space by making similar graphs yourself.Exercise 8.8(a) Prepare phase-space plots for the free-motion damped oscillator by choosinga value ofthen preparing tables of y (x) against x and of y' (x) against x,with x being in scaled unitsand with y' (0) = -y (0) = 1.
Thenplot y' against y, as in Figure 8.2, which has = 1/4. Other suitable valuesof are 1/8 (very little damping) and = 1/2 (strong damping).(b) Show that forthe phase-space plot is a straight line with slope givenbynExtensive discussions of the phase plane and Poincaré maps are given in Pippard’somnibus text on response and stability, and also in Haberman’s book on mathematical models.Now that we have investigated the solutions for free motion, it is time to investigate the effects of a source term on the motion of an oscillator.8.1FORCES, SECOND-ORDER EQUATIONS, RESONANCES265Forced motion and resonancesWe return to the problem posed near the beginning of this section, namely motion with a source term, as given in general form by (8.7) and exemplified for themechanical and electrical systems by (8.4) and (8.6) through the correspondencesshown in Table 8.1.
The equation with a driving term is(8.20)where the parameter S0 indicates the strength of the source term, while parameter(angular frequency if x is time, but wavenumber, k, if x is distance) indicates thefrequency of the forcing term. We assume, for simplicity, that only a single drivingfrequency is present. By superposing solutions for differentas indicated inChapter 10, one can generalize the treatment given here.It is most convenient to work with a complex-number representation of the variable, y, and the source term, S.
As discussed in Sections 2.1 and 2.4, the modulusof each complex variable indicates its magnitude and the relation between the imaginary and real parts indicates its phase. So we write the source term in (8.20) as(8.21)with the understanding that it is the real part of this expression that counts.From everyday experience, such as pumping a swing, we know that a mechanical system under an oscillatory driving influence will settle into steady-state oscillations having the same frequency as the driver. Therefore, let’s try a steady-state solution of (8.20) of the form(8.22)The algebra and calculus of the solution are straightforward, so work them out foryourself.Exercise 8.9(a) Perform the indicated differentiations of (8.22) that are required for (8.20),substitute them in the latter, and notice that the complex-exponential factor can befactored out for all x, so that (8.22) is a solution of (8.20) under the conditionthat the complex amplitude y0 is related to the source amplitude S0 by(8.23)in which the frequency for free motion without damping appears as(8.24)266SECOND-ORDER DIFFERENTIAL EQUATIONS(b) Convert (8.23) into amplitude and phase form as(8.25)for the amplitude, and(8.26)which gives the phase angle by which the response, y(x), lags the driving term,S(x).(c) Show that the amplitude relation (8.25) can be expressed in terms of the freand the dimensionless damping variable from (8.15), as thequency ratiodimensionless ratio(8.27)while the phase can be expressed in terms of the same variables as(8.28)which is usually displayed in the range 0 tofunction ofnwith a continuous distribution as aFIGURE 8.3 Resonance amplitude (dimensionless units) as a function of frequency relative toresonance frequency for three values of the damping parameter .FORCES, SECOND-ORDER EQUATIONS, RESONANCES267FIGURE 8.4 Resonance phase as a function of the ratio of frequency to resonance frequency forthree values of damping parameterfrom small dampingto moderate dampingThe amplitude and phase are shown in Figures 8.3 and 8.4 for the same valuesof the damping parameteras in Figure 8.2, which shows the response of the system under free motion for different amounts of damping.
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