Thompson - Computing for Scientists and Engineers (523188), страница 50
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Notice that the amplitude,Ais asymmetric about= 1, with the asymmetry decreasing as thedamping decreases and the response becomes more resonant. The phase,isnearly symmetric about= 1, and it varies more rapidly in the neighborhoodof the resonance point as the damping decreases.Resonant oscillations are important in practical applications; for example, oneusually wants to avoid resonant oscillations of mechanical structures, but one oftenwants to enhance resonant behavior in an electrical circuit. It is therefore importantto know for what driving frequency,the amplitude is a maximum.Exercise 8.10Differentiate expression (8.25) for the amplitude with respect to and therebyshow that the only physically meaningful solution for a maximum of Ais atgiven by(8.29)assuming that the quantity under the square root is positive, otherwise there is norelevant maximum.
nNotice that the amplitude has a maximum at a frequency slightly below the naturaloscillation frequency for free motion, . The maximum rate of change of thephase, however, occurs right at this frequency for any value of the damping parameter , as is clear from Figure 8.4.268SECOND-ORDER DIFFERENTIAL EQUATIONSFor the typical resonances that are of interest to scientists and engineers thedamping is relatively weak, that is,The amplitude formulas can then, withonly mild approximations, be made analytically much more tractable.
One sets(8.30)and approximates the amplitude expression (8.25) by(8.3 1)and the phase expression (8.26) by(8.32)Exercise 8.11Show the steps in the approximations that are necessary to produce (8.31) and(8.32) from the original expressions (8.27) and (8.28). nThe function AL will be revisited in Chapter 10. Within overall normalization,it is the square root of the Lorentzian function, with the damping parameter b beingproportional to the width, of the Lorentzian.
We show in Figures 8.5 and 8.6comparisons of the exact and approximate amplitude and phase expressions forFIGURE 8.5 Resonance amplitude,and its Lorentzian approximation,as a function of the ratio of the frequency to the resonance frequency. Shown for damping parameter8.2 CATENARIES, CATHEDRALS, AND NUPTIAL ARCHES269FIGURE 8.6 Resonance phase,and its Lorentzian approximation,as a function ofthe ratio of the frequency to the resonance frequency. Shown for damping parameterNotice that the Lorentzian function gives distributions that are symmetric aboutthe resonance frequency, and that the approximation is particularly good for thephases.
In applications of resonance, a value of of order l0-6 is typical, so thenthe Lorentzian approximations (8.31) and (8.32) become very accurate. We exploreLorentzian functions and their properties extensively in Sections 10.2 - 10.4, wheretheir relations to Fourier transforms and convolutions are emphasized. The relationsbetween resonances, response, stability, and chaos are discussed extensively in Pippard’s text on the physics of vibration.The study of vibrations and waves, which we introduced in the context of complex variables in Chapter 2.4, is discussed extensively in the context of differentialequations in the texts by Braun et al. and by Ingard.
The musical connections arewell described in the book by Backus and that by White and White.8.2 CATENARIES, CATHEDRALS, AND NUPTIAL ARCHESWe now turn to the problem of modeling a static mechanical system, hanging chainsof various density distributions. This provides us opportunities to develop our understanding of applications of differential equations.
We also bring together manyinteresting and insightful results on general catenaries that have appeared in scatteredform in the research and teaching literature of the past three centuries but which arenot much appreciated by modern pedagogues.The famous problem of the equilibrium shape of a cable or chain hanging undergravity provides insight into setting up and solving differential equations of secondorder. Also, the differential equation is nonlinear in that the quantity of interest, theheight, y, as a function of horizontal distance from the center of the cable, x, is notlinearly related to its derivatives.
Once the equilibrium shapes have been determined — they’re called “catenaries” from Latin for “chains”- the same equations270SECOND-ORDER DIFFERENTIAL EQUATIONScan be applied with minor changes to the optimal shapes of cathedral archways andto the shape assumed by an arch of string uniformly levitated by balloons.The equation of the catenaryThe classical mechanics problem of finding the equilibrium shape of a flexible chainprovides an introduction to the mechanics of continuous media (“continuum mechanics”) and to nonlinear second-order differential equations.
If we observe a chainhanging in a curve of given shape, what is its associated density distribution? Also,from an engineering viewpoint, what density distribution provides a constantstrength (density-to-tension ratio), and what is the equilibrium shape of the chain?In the following we derive a general formula for the density distribution in termsof first and second derivatives of the shape, then we introduce dimensionless variables (scaled units) to allow general characterization of catenaries.
Examples forvarious shapes are then given and we examine the system under the condition of uniform strength.In setting up the equations for the equilibrium of a chain, we allow the weightper unit length of the chain, w(X), to depend on position along the chain, (X,Y).We assume that w is an even function of X, so that the chain hangs symmetricallyabout its midpoint X = 0.
For the vertical coordinate, we choose Y = 0 at the midpoint. Referring to Fig. 8.7, we see that at P = (X,Y) the conditions for equilibrium are that the tension T (tangential to the chain since the chain is flexible), theweight W, and the horizontal tension H (the same constant value on each side of theorigin) are related byFIGURE 8.7Forces acting at the point P = (X, Y) of a chain hanging under gravity.8.2CATENARIES, CATHEDRALS, AND NUPTIAL ARCHES271(8.33)(8.34)In terms of the arc length along the chain, S, we have(8.35)From geometry we know that(8.36)and also that(8.37)Our aim is to find an expression for the weight distribution, w(X), given the shape,Y(X). By eliminating T between (8.33) and (8.34) then differentiating the resultingexpression with respect to X and using (8.35), we find immediately that(8.38)Use of (8.35) and (8.36) now gives an expression for the weight distribution interms of the shape(8.39)We may verify this result for the uniform-density chain by substituting the standard equation for the catenary(8.40)to verify directly by substituting in (8.39) that(8.41)where the dimensionless variable p is a constant for the uniform-density catenary.Exercise 8.12Verify the result for the uniform-density chain by taking the first and secondderivatives of (8.40) for the catenary then substituting into (8.39) to produce(8.41).
n272SECOND-ORDER DIFFERENTIAL EQUATIONSDimensionless variables are helpful when comparing catenaries with variousweight distributions, because even for a uniform chain the shape depends upon thehorizontal force (H) exerted. For the remainder of our analysis we therefore changeto the following dimensionless variables in terms of an appropriate characteristiclength, L, and the horizontal force, H :in which the weight-to-tension ratio,provides in dimensionless units the tensilestrength required for the chain at each point.The scaled density distribution is immediately found by inserting (8.39) for winto the definition in (8.42):(8.43)The tension at any point in the chain can readily be calculated by solving for T from(8.31), then using (8.36) to eliminate the angle variable.
We thus find the tension inscaled units(8.44)The weight of chain below x is then, from (8.42) with similar angle elimination, inscaled units,(8.45)The strength of the chain is obtained by taking the ratio of p to(8.44), in scaled units this isFrom (8.43) and(8.46)Exercise 8.13Solve for T and W as suggested, then transform to the dimensionless variablesof (8.42), to verify equations (8.43) through (8.46). ■8.2CATENARIES. CATHEDRALS, AND NUPTIAL ARCHES273With the above scaling, the characteristic dimensions (such as L for a uniformdensity catenary) can be chosen such that all flexible chains will hang with the sameshape near the bottom, where the slope of y on x is zero. Their scaled density, tension, weight below the bottom, and strength will satisfy(8.47)provided that at the origin d2y /dx2 = 1.
In order to achieve this invariance, a givenshape choice y(x) should be expanded in a Maclaurin series in x, then constants inits formula adjusted so that the constant and first-derivative terms vanish and thesecond-derivative term has unity coefficient in order to make p(0) = 1. This procedure can be verified for the uniform-density catenary, (8.40), and is demonstratedfor the parabolic shape in the next subsection.Thus, we have solved a mechanics problem and have been able to express thesignificant quantities, (8.42), purely in terms of the geometry of the curve shape,y(x). We now illustrate these scaled curves and corresponding scaled densities andforces.Catenaries of various shapes and strengthsIn the statics problem solved above we obtained the distribution of density, tension,weight below a point, and strength of a chain, given its shape, with the assumptionsbeing that the shape, y(x), is an even function of x and that its second derivative ispositive, as required in (8.38).
With these mild restrictions, a large variety of realistic catenaries can be investigated.The parabolic catenary is the first shape that we examine. We write the shapeequation as Y (X) = X2/L, where L is some characteristic length. This examplewill explain why Galileo was nearly correct in his assertion in his Two New Sciences that a uniform-density chain hangs in a parabolic shape, and it will clarify useof the dimensionless variables introduced at the end of the preceding subsection.
Interms of scaled lengths, we have y(x) =where the dimensionless quantityis determined by the condition p(0) = 1, that is, unity second derivative of y atx = 0, according to (8.43). Therefore = 1/2, and the scaled parabolic shape is(8.48)It is now simple to take the first and second derivatives of y and by using (8.43)through (8.46) to compute the mechanical quantities of interest. Their formulas aregiven in Table 8.2, and the corresponding distributions with respect to x are shownas curves (1) in Figures 8.8 -8.11.274SECOND-ORDER DIFFERENTIAL EQUATIONSTABLE 8.2 Examples of the relations between the shape of the equilibrium curve, the weight ofchain below each point, the local linear density of the chain, the tension at a point (x,y), and thestrengthof the chainExercise 8.14For the parabolic catenary calculate the first and second derivatives, then substitute into (8.43) through (8.46) for the scaled density, tension, weight, andstrength, thus verifying the results for line (1) in Table 8.2.
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