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Exam1_solutions

# Exam1_solutions - ME 391 FALL 2007 EXAM 1 WEDNESDAY NAME...

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Unformatted text preview: ME 391: FALL 2007 EXAM 1: WEDNESDAY OCT 10, 2007 NAME: 50LUTlONF-C PID Note: i. There are 5 questions with equal weight (Total: 100 points) ii. A table of Laplace transforms has been attached at the end for use in questions 4 and 5. iii. Use of calculators is not allowed. iv. Only one page (8.5 ” x 11 ”, both sides) of notes is allowed. v. Please show your work in clear and logical steps in order to get partial credit. 1. Solve the following linear, first—order differential equation by finding the integrating factor. 2. The angle of twist (9(2‘) of a circular shaft subject to a torque T is given by the following linear, constant coefficient, homogeneous, second order differential equation 2 d 26+ci€+k0=0 dt dt where I is the moment of inertia, c and k are the constants of proportionality for the clamping and elastic torques, respectively. Given 1:20, c=4 and k=1, solve the above differential equation to find an expression for the angle of twist 0(2‘) . Your answer should have two arbitrary constants. I lot/£0 it Come + [email protected]:o —— J ILTZ—OJ (LE—Le) k1, 0&1 &£ 200W. u0g+ 9.0 alt? 0M: ZOM’L-l—Ltm—klzo m: ~l+i HQ- Mac/1m _L,+ oat. 2X20 — lea UL 3. Find a particular solution of the following differential equation by the method of undetermined coefficients y'+3y'+ 2y = e” Ci‘ w1+3m+220 x yczC,Q +C‘2Q__x ABYW 7’2 : AGEX lhoauwa v‘» m 6a W oily/Qty? Subxiw‘l-Mmy 5n W 09;; _x - ‘ FLAG. +AXQ“ +36AQ~XAAX2~Xj‘€ZAXQ'—x ~ Q’X -X — A2 +MK'WX’fZﬁ/JX :2“ a) All 4. Determine fm=r1{ 3s+10 } s2 +12s+100 W W! fzﬂzj‘ Hoo : S? {42; +26+61r ~ (§+6)Z+g? Fm LaP£OCQ Tmféw Tag/23 (f {Qafcﬁktjébj Q:*6/ kt—A) K 1 § em‘ﬂw H] 2 Qr—a)‘+ k1 211:, ; A 0w + 9 5’ 51" \ZS*1°° @6‘)2+ 8’2 €+6)24?l ’?;+\o : ACMKJ +83 s A =3 5° [0 m+8§ :3 E? 5—2A 2,; L; 25+“) : 2‘ 5+6 _ 3 52—;‘15‘41470 é+g)2+a>2 é+gji+gz ferf'lé 25445 : g f" 3+6 j‘k 51 52mm“? 66”)“31f ﬂ géwmf 5. Solve the following initial value problem by taking the Laplace Transform of the equation, solving for Y(s) and taking inverse Laplace transform to find y(t). y’+9y = t+cos3t+cos2t,y(0) =1,y'(0) = 3 TalH—r7 Lapﬁoae TWAW 0% We \$2an 5327"} *qofé7jiafi‘chri’écoS3’ff whorze) Fm Tﬁmﬁn W) {52 y(5) #5 ’yr(o):) + q 7. L + S + 5 52 ;Z_(q SZ+LP \$178J-—S~—3 +<3VQ7 2 4“? r + 5? S‘Z‘t‘i J‘Z—t-L, 2 Q *qJVCV‘) 7. §+ 3 .9 I + I + I 52 S‘z—f? 52A,, jffﬁr) : 5‘ +, 43 \ 52+? \$z+q 4b 1 2 4' S + I ,_ 5‘ («MW @149)" €2+L)G?+4/ Fm W, I4é2£.}:,Qakt q IJg,5 j: cmgt 52”” 5%?" 1"? 9L j: 5M.Bt g i4 3 ,\Lm\3t \$24kz 574-31}— f k3 : k‘t—L'hkt ZEUHG) ,t l 3 Q f E _—L-’}: J—-¥'é 3 i I 3% —& 2t 6014‘?) 33 \$201+?) “ 9:7 ( m J i"; Mr }: emu h1)1 5'1} .3334 5 Z L {(29); : L,’c9vn3t €7.44?! )2 6*?1)1 6‘ ,1 1 of f M T cowt , cabt Ghawuw qfé152j: ( Q ’HPXS “31/ 31,27— 62+21/(;?437')j : LCC/OSZ’C’CdSEt) 5 jec): fféYGJJ 1&5 3+1" 2 of" _L “' ' 2+4 34%; + EIZOme‘Li { j f+<3€J A“, 51+3lj(y%8/j €2+qjl : C033{ AQMxEt _L ~ 417 (N Jawhrj +%£s.xn3(: +§Lfcm2t~1w 36) 13 k3 kt—sin kt I— 14 2k3 sin kt—ktcos kt l— 15 2ks tsin kt I— 16 ZkSZ sin kt + ktcos kt I— 17 (b2_a2)s 2 2 cosat—cosbt Some Important Theorems 73m» = s2F<s> — sf(0) — f’(0) Z{e"’f(t)}=F(s-a) I{f(t—a)u(t—a)}=e"“F(s) ...
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