chapter 5c

chapter 5c - 4/4 - 3]) in the text take [h - g 129, WWW“...

Info iconThis preview shows pages 1–9. Sign up to view the full content.

View Full Document Right Arrow Icon
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 2
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 4
Background image of page 5

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 6
Background image of page 7

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 8
Background image of page 9
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 4/4 - 3]) in the text take [h - g 129, WWW“ e f“rm will.) J's!) “Mu”: ’ x":- Jug=n€ by x and then adding. We readily Solve first {0 I ' Wank-OD fl. = 36,}, so 4 "1:91 ; m munbcr that —- for variation of parameters j the diffemmial . I be mine“ in standard fem; with leading coefficient 1 i We therefore 11 - 2 equafionwithoomplemcnm function J’c — W “‘6er3 as y” - (my + (6/x1)y = I- :- g a. .F , ') yields I, = —szx’-x-x"a§r+x3 [xi-x-x'qdr = —x‘Idr+x’ I(llx)dr = x3(lnx —l). 6]. 62. 63. 64. fix} =- if”: M's: ' u, = —12x”‘ IS. yp = —72x“315 yl = cosfln x), fix) = (In nix: “I = (In x)cos(ln x) — sin(ln x) "I (In x)sin(ln x) + cosfln 1:) yr, = lnx (i) yl = x! W = x2_19 u; (1+x’m1 —x2), y“ : —x2 +x mu +x)/(1 —x)] +(1/2x1 +£)1n|1 - fl This is simply a matter of solving the equations in (31) rm t h = sinfln x), '— Iii-— y; =1+x2, fir)=1 u; = xl(x2-l) _ y2(x)f(x) Yi(x)f(x) d ’ = W(x) an a, integrating each, and then substituting the results in (32}. Here we have yl(x) = cosx, y2(x) = sin I, W(x) = 1, fix) =2- b, yp(x) = — (cos x] 55in: -25inxdx +(sinx) [mu-2 equation y”+ y = 0. = —xcosx+(sinx)(cos’x+sin2x)‘ Z —(cosx) j(1—coszx)¢tx+(sinx)[2(s' ' V _ (cos x)(x — Sinxcos x) + (sin 1mm}. . —xcosx+sinx Wm ' %- 1 3 Firstweappl _ ' x = AcosSI-I-B‘ solution xp =_ 2 cos 2:. (can mu 3 I no first-derivative term __ mere 1. ‘ 6e general solution is .1113 initial Conditions x(0) = 375' x10) = 0 nowyief. the part ofthc solution with frequency a, = m is _ - x: = 372 cos 10: — 2 sin 10! =J|3a3ss[ 372 10,_ 2 Uisssssm lSSSBEsmm = J138388cos(10: —-a) where o: > 2.7?- tan“(1/136) «.~. 6.2778 isafourth-quadrant figure shows the graph of x0). one: yieldsthepanicularsolurion 1cp — -sin 31. Then we impmelhE l ' 1(0) '= r01) = 0 on the general solution 1 kit/i. 375 ' :6!) = clw321+qsin 2t — sin 3!, . .2. gsian — sin 3!. The following figure ShOWs the graph of Km "375 ' t 4. Noting that there is no firstwderivative term, we try 1; = 4 . H solution xp = 10cos 4:. Then imposition ofthe inil; solution :50) = c1c0551+czsin5r+lilcos4t yields-2'- 350) = (— [0 cos 5: + 18 sin 5!) + 10 cos-41- in' r ,_ 50 120 _- w. 169c0531+fi§sm31 z—tan"(1215) a: 1.9656 F1 L. ‘1‘ 'l ' ‘I -4 i Cm”! gives C " F0!“ - mml}. Then x40) = 0 on the general solution xp = AcosSHBsinSr; —20A+lSB = -4, 15:11-20 16 12 . 4 4 3. ) '- v = —— St—— 5: 2 ~— ~cosSl——sm5! = "fi‘m 125603 1255'" 25(5 5 ;,__;V a T 2H.-mn"(3/4) 2-. 5.6397 (4th quadrantangle) Xsp l . I “ “’0 111° Milking resonance solution ofour initial value =. 2mvll + F3! . Sll'lfllaL solution iii? E‘flvomng undete . _. cgeflicicnts, and finally the - firsuhe general solution x0) invai T“. men the equations that determine these . , so that x0) = xu(:) + xwu) stisfies the a . owsthe graphs ofbolh x0) and xv”). In graph Sh " Ksp - . _ 11. x9 = Acos3I+BSIn3l; -4AHEB =10, 1'2“ I . 1 _ me) = 7 ~cos3t+§sin31= fl(_ ; 4 4 4 10 10 A " ‘v ‘ u = zrfitan ‘0) a 1.3925 (2nd quadrant angle) 1“. _' . i m) 3 L. 1.(c,cosr+czsin1)+xm(f); cl..} = o, -2cl+c1fi_ 5 -, 1 7 . J5 I. 1 I ' x..U) : e1 [icosh—smt) = —e'“[ cost- ' 4 4 4 75-6 75““. - :: ix/Ee'“ cosh — fl) _‘ 122 (-3“: quadnm V I _.noslflt—-lg-sifl10! 10309 X 1 0 SP 171 ‘ , 0f—————-—SH1|0I = — 61‘ J257725 J 793 f “3‘10" W) (HINTS) H 3.485] {3rd quadrant angle) —97A+303 = 8, 30A+978 :76 I} : 21r—tan"(7) 2 4.8543 (4th quadrant angle) 12. xp AcosSHBsinSI; 12,4-303 = G. 30A+llB-¥‘= 25 l . x50“) = —§7c055!-—Efism5( 1469 : +2c _ ’ 50’37 = 2 {113314 f, - 7 2 1469 “(5' fl = 23—m"(421!12395) - 5.250'5‘ 1.1. \‘F : Acosl+Bsinr; 24A+83 = 200, 444.243 = 520 444nm? = 600. 20A+74B = U .rwtrJ 2 cosr+225inr : J485[ l cost+ 22 slut]: 485cc J485 3485 a = [an 1(\22) a 1.5254 (Isl quadrant angle) 300 37 I0 . 300 xi!) : e'“(c,(:053r+clsin3£)+x (I); — cos]0!+ smlm = ~ 5 5“ $454 $469 JI459 J .F459C“5{10‘ a) rl+1= 730, —4c,+3c1+22 = —l() fin" ["3 a 7) H quadrant 1"“) Z 8 4: (_31COS_~“ (szsinm) = \/36659““[— J%Ec0331— J%§sin3r] = 3665 n+tan"(52/3l) :1 4.1748 (3rd quadrant angle) ' '- [I The figure at the top of the next page shows the graph of _ Millie III) -‘= A{w)c°s‘m+ Maggi" M mm the differential . .wimmegimnumerical values of m, c, k. and 14:). we a. thalmun upon collection ofooefficienls of cos a]! and and 3&9) um we gel by solving these equations. anaily‘ . '011 sure first sin a”! immmpfilude oflhc resulting forced oscillations as a functiOn 0mm 4; andmshaw the graph oflhe Function am. 5-293 = 2. -2mA+(2-9J2)3 = 0 " i8!) 40) r ' B = I in 4+9) ‘QI-Jh-m‘ beginsm'th C(0)=l and steadily decreases as m incrfiase is mpmctical Insomnce frequency. 5' l6. (5 -aJ') A+ “JR- lols—m'! _ ' 25+5w1+m“ cm) = 101J25+6m3+m‘ C A 17_ (45rmszJr6mB = 50, —6wA+(4S—a)2)3 = 0 50(45—(61) B 2025— 54502 + m“ _ mus—54a;2 +w‘ ('l ml : 50/\}2025~ 54m2 + (9‘ so, to find its maximum value,w‘ ‘ derivative ' —100m(—27+m‘) C' = _—__.———. . (m) (2025434af+m‘)m » Hence the practical resonance frequency (where the derivative v r m = ‘5‘} = 3J3, The graph of Clear) is shown at the top of 18. leO—m2)A+10mB = 100, —10mA+(650-m’)3 =11: 100(650—afi) B ‘ 100m, A _ 422500—1200m1m“ _ 422500'-1200m‘- . C(m) = 100/J422500—1200m’m‘ so,tu fins! I . ‘ | . I _. 200 («600+w’) Differentiation, substitution of a a, 5-,“ a, and aim . . _ w i . F. -_-_—_._——l———_——_. 51") ' (422500— imam? + w‘)“ 9'” W + ail-)9 = 0 t can = JMm+gIL 22, Let x denote the displacement of the mass from its equt . veloglty, and a) = via the angular velocity of the pulley. yicl s ‘ 50 tul- mmm frequency (where the derivative vanishes) is mv1/2+Ia12l2+lcx1!2 —mgx = c. When we differentiate both sides with respect to t and simpliff.‘r differential equation (m+Ila2)Jf'+kx = mg. Hence (a = kl(m+llal). 23. (a) In ft—lb-sec units we have m = 1000 and k = 100005.333 = :- 050 Hz. " “ w (h) We are given that a; = 21:12.25 :3 2.79 radisec, " ' mx"+kx = F(:‘) Simplifies to sown, so the critical frequency is we : m x'+10x = (unwasinax. mil" as 3.12 Hz. When we substitute x0) = A sin ax we find '- - Then the force F _ .5 = h "'3 = 9-891 th ' I ' - cm 11200 meters, so Hookersflaimacmne s walghu (3) Therefore, if e 211$ 50 C(m) steadily deereasesaa m‘iu I s Acosta: + Bsin wt in the differential usual yields the equations e{Nation L' .014, 23_ (a) The given differential equation corresponds to Equafib It therefore follows from Equation (21) that the amplimde'of: ' vibrations at frequency w is F0 Coo) = = :M - mtol)2 + (an)2 mAa)z (k — mm: )2 + (on) h = I A, sa = k7 3 ‘. It- I [fiande 6:70 J— 90 ( “’7‘” ) VA- (b) Nowwecalculate fl: and pzl/‘Kk-mmlhwmf. Then ‘ w] I C(60) = 1 1 m 1 :,(t) = pfinooqm—ahpfisinmr—a) ' [(k "m ) +07“) ] = pGn[g£cos(wt —a)+ gsjnm; _ m] and u. :3 see that the numerator vanishes when 0 o =paelcosflcostwr—ausmfismm,m] ; m : f 272 E q 2m ]> Ji— 2 we. _ '_ 1,0) = pGocoqwrhg_fi) Emk —c1 m ka—c2 m - ‘ ‘ 29. We need only substitute E0 = new and F0 2 air in the result of Problem 30. When we substitute the values a) 2 21nd L, m = 800, k = 7x 10‘, c I, z 10, u = 0.05 in the formula of Problem 29, simplify, and squares 25(sm'v2 +122500) 16(167r‘v‘ — 643751r2v’ + 76562500)2 C'SCJIV) = c (in mm) as a function of [he Velocity b . _. “1" 1;. 45000311:2 - 535937500) ' . . . l ‘ +2 . Iwgr'v 64315.12?! 46562509): . IF " (1 _ - - 2 . . mmmmmmnsaquadrancm v. it IS easy to when the maximum amplitude occ um; We fin 436 m/sec a 32.i2 mJJhr. The Cunespond @0436) :5 0.1364 :11 : 13.64 cm 30", he iflg d “Inca-ii CHAPTER 6 ElgENVALUES AND EIGENVECT., . .h SECTION 6.1 mrnonuc‘rton T0 EIGENVALUES ' r. In each 0f Problems 1—32 we first list the characteristic polynomial matrix A, and then the roots of pa.) —— which are theeiflenvalnes‘fl. that appear in Problems 1—26 are integers, st) each chm1msficmlyn caCh eigenvalue A, of the matrix A, we determine the associate eiBEl'l ' for the solution space of the linear system (A — 111% = n_ we m form in terms nfthe components of v = [a b Inmos’tcasesan, then apparent. it‘ A is a 2 x 2 matrix, for instance, then our two scalareqtmtie- onc ofthe other. so we can substitute a convenient numerical value lb: and then solve either equation for the second component b (or vice versa). 1_ Characteristic polynomial: pH.) = )3 —S)l.+6 = (A —?.)(Ji. -3) Eigenvalues: 11 = 2, 11 = 3 _ 20??!) = 0 l w th 2; z 1 3“ aub = o } V' H w u A 3 6H2!) = 0 2 _‘ '1 t : : v = 4" a—2b = o ’ l - 2. Characteristic polynomial: p0.) = 11171—2 = (At-DUL-Z) : Eigenvalues: A, = —l, ii? = 2 w,l l_ éaefib = 0 v = l m ’1‘ ‘ " ' 39—31: = o i 1 ‘ 2_ 3a—6b = o = [I] W“ ’2? = ' 39—61. = 0 a ...
View Full Document

Page1 / 9

chapter 5c - 4/4 - 3]) in the text take [h - g 129, WWW“...

This preview shows document pages 1 - 9. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online