Test 3 with solutions

Test 3 with solutions - MA 252~NH Test 3 (July 31 (Thu),...

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Unformatted text preview: MA 252~NH Test 3 (July 31 (Thu), 2008) This test has 8 pages with 8 problems (139% each for Profilems 1 ~ 6 and 2013135 each for Problems 7 and 8). Show All Your Work! Name Probiem 1. Find £(f) when f(t) : t6 + 008(5t). £%%Cm9®w:£fifl%£flgfigi m 5f s g m 311$: ~— m an... .7204 g 3? S3a§ Problem 2. Find £(f) When f(t) = 8"”15 cos(3t).‘ (Hint: COmbiue items 23 and 52 Tabie of Laplace Transforms.) GEL?) 2:» Answer: L1 (mi: r? "" .mwplF-D-HN-Mu-m- Wm Wan-WAWI-vamm—J-m‘n-KWWu ..,,., ,, 52 r.‘ [:4 +7 “ax. "L— (f =31 (~sz (35% “JR (S ’5‘?) 2:: WW, " b and 3&5}. {1 i Answer: LI” F m I + - Problem 4. Find £”1(F) when F(3) = :S;8:. 3 § § mm 6 29 W? 3 Mi Slag-3% “52+? 2" W}; EmsmflméfE/MMH Answer: L“1 F m 103 + 3 52(5 + 3)‘ Problem 6. Do the partial fraction decomposition of F(8) = . Q ‘ “a ' a; A 4. .12 + C MM) alimmggf M 316+» '" 5 31 5+3, Ems) 10“ 33 ‘3 Mg"? wag-$.35: S w. I: _ 3 Jr % Eb JVHH’JJ S S1 S+ 3 Mi“ 1“ , ’ g C 1 "an; ( :2 {WWMW—w , , 422‘ ‘ 3;, ‘ £63 8215.“) (/03 Problem 7. Use Laplace transform to solve the initial value problem y’ —— 3y 2 138in(2t), y(0) 2 ——1. t\:i<m:w-:x m m jig (§"3W(5} 1‘1 "=5" {WOEWENCW}: {9) 2L W. :3 AW 4%, a fifififlfigfig Cad—3mm 3 5 3"Ptér—4i (83mm) & "2. _ 1*“ Wfifi’é‘SWWfiCsw egg—ac} .2 7 ,1». G, E; v ,, mummy? +833 L 3 )3} +434 35} A+B =-~= 0 =39 A == *8 935:0 £2; czar?) mlrzfistsmeajmwgsm 4A ~39 22:16 d £78142 U A: 43:2, 1P):ng C: 3g, 3.362;)3'wg [ *5? m m ggéjrggg) {MSW} mama) {$43)} { I: “éaiis’lwg AS 232;? gmfigaflg 3 Y;§~%%§”fi 5 L 3' 1% (Continuation of Problem 7.) S :— .. #1....“ .1315 w ZS T“ b {H 8—3 "‘ 3—3 32% m a _~ 2,2 9 m NEW €13 $34» 2?“ 13%;)? *2 OK“ ( $33.. )m B if! L .. 3 eat.— 2W€2€jmg MPfi') i” am: {LAW—.7251 m 3» Q a" 3&2: '5? ML i- w) m 36% 4 (12% 3 m 6 CW2” WW mm m 383“ 4% WW ®th13a- » 7 Problem 8. (i) A lfikiiogram mass is attached to a spring whose constant is 16 N/m, and the mass is released from the rest 2 meters below the equilibrium position. Find the equation of the motion. wfiféw/gx @9ng “EM {(3. e "mg/#7 e “114:5 me) =2 q C293 (46) + C2 W W") CREE; { (Lg Cam E'E’ieh {/3 KM) = 2 awed 96(0); 0 65 we “Er :2, CE .232! 701%}- == $4.6 me) + 452,090 {4-15) I’M = weer- o + 4C2» I 0 £2 462:0‘ - "@920 ML (ii) A 1-kiiogram mess is attached to a spring whose constant is 16 N/m, and the entire system is then submerged in a liquid that imparts a. damping force numerically equal to 10 times the instantaneous velocity. Determine the equation of motion if the mass is initially released from a point 2 meter above the equilibrium position With the downward velocity of 12 m/s. (mark/0 144+ M m C) “we; {swefliirleg} neg-O ' 7% M”: "2) m We moot Watts-mm: WW2) ec’f‘fOHQ M Wum.m“Mmsm-mfljfiiww”‘”"W'mew:mwmw m“ *“ “Wm—WM m. C 2 7LCH==4Cte ewczefif 5‘ 5 is 4w '— I 3" 3—”.de xiii) e (me + 163m)“ €2t(4ci“10‘1“65t) ‘ > C2 M13“ - *éfltCQgécfh 3Q£E"§.,L{é m £§+~Qm I 3 I we; i ewe Answer 5375 = w *7C[O):wémm%mM-%w “fie ~Www “Mg 3W «- _r 2 x 0) l(-§)n-f§-(fl§£) :53 F32“ng i ...
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Test 3 with solutions - MA 252~NH Test 3 (July 31 (Thu),...

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