final_1 - Name ID# ME461 Mechanical Vibration 7 Term I Exam...

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Unformatted text preview: Name ID# ME461 Mechanical Vibration 7 Term I Exam (Dr. Zhu) Fall 2010 —-—> Problem 1 (351 kl XI A solid disk, with radius r and mass m rolls without /\ i slipping and is restrained by a spring k]. a damper c. 7 and a massless wire connected to a bar with mass M — and length 2]. The other side of the bar is restrained ab by a spring k2. C X = :2 M9 = Q 95 No slipping W»— k: What is the equation of motion in terms ofr, m, kl, k3. c. l. M and unknowns 6). 9 and é ? kind}; Ema-66 2 _. ‘ Z ‘ 3. ‘ z T . :5- < ,45 (qe’mmb + :‘i m mg) + g (gmrl) 9 - J. M 2‘ ‘ ” :2<‘§)(~20 62 + —_;—(.g_m)rlaz: é(f%ir3 +5gnrl>82 _. .L a .2 n. U ~. K (9 .L x _. 1 a l“ > + 2K2 amfi‘a +_LC4KQV3)82 = éWW‘ , ‘Phob. ’1, New’i‘bn’s (Mome : F€>C>=t41, x ‘4 TCD 21'0un — TY‘ +Fr-Y‘ = J—mr’e RJ‘G 2 cré C?) E Page .. T~ Fb— KLrG—créz mré Fr (WP—>262) mr‘e +<1m9 +r<ma = 2—1— _%mm‘9 Oh m \‘ . ® 79 +CG +t<|9 —= QT/r. #50442. T ><=2r~8 X=Q‘9b v U M u Q 2‘33“ =~T~Q wax-fl: 73-May 8b=€Q'X 9b ' @ T: 495?:‘8 « %2r“8 Q @963 E \ ax +C®+K.© =~4~h18 ~39“ =~>E©M (3: +4g438 +CB+CR+4LQJBQ =0 Problem 3 (352 For the system shown, the damping coefficient is c = 600 (N-S/m); both springs are [(1 2 k2 = k = 3000 (N/m), and the solid disk radius is r =1 (m) with mass M = 40 (kg). The disk rolls without slipping on the surface. The horizontal coordinate x is associated with the center ofthe disk. X 1‘ b8- What is the equation ofmotion in terms of]: m. In. kg. 0, and unknowns xi A" and KI crew Haze ‘ ‘ \ 1 l ‘ 2. --— i.- ‘ 1 T :{MK + gezfirfle - 2(3—3x .. Q 2 U — igl‘KX +5:le ~g—LKfi-lQBKZ—4 fipuqxz KC 2 C i . g _ _ . — —- - ' v “a” a ' i i — ' . _ F50 Neflbnsfleory. X- r6 “ Ix @ZPQ—Cku+l<z)¥—C§(-FP=M¥ KY \\ l @ZT =R~r tbl'Mra‘S = Ell-Mirr‘ <____.<-‘-- <-Rv< @lr 4 Q3 ex 5M§<+Q5<+QKX=O Fr- .3 What is the natural frequency ( a)" ) of the system? m” : l 0 Fad/s Ql< F;— j4x 3000 a —- M :: ~——- = = l m 0‘3" " l 554/? SM 5x40 0 /5 What is the damping ratio (4’ ) ofthe system? J : 212* C C ngn :3 SM g S = -— 600 —. .L /1’ SMwn SXADXIO ~2 Name Name ID# What is the damped frequency (cad )‘? (ad : SB wa=u3nit~ s‘ = to xi“2\¥ = 55 What is the system response ifthe initial condition is .\‘(0) = 0.8011) and x'(0) :1 (m/s)? xc+)= e" 3®“t[c. coswat + cl snoth mm scon=5 wa=5J—S, °.=Xo=0~%,ca=élco%w2 3 Kat): -51: J“. ‘ : I+5xo. _. e «454.5. -yfi 0“ %(+) = A9: " C CostcoA~<123) I A =,iC;’1-c2’ = 0.987 ‘St g 2 hon—K c‘le) 2: +ah4<ég73~ 10,635 =7 xcc) = 0.9379. costsrgt_ 0,525). m . What is the system logarithmic decrement ( in )" ln = :23“ XE . X2 fig x. __ Q“ 2 ‘2“3 —_._ JE— _ 12.3. i072: = SODA-C - Swnwmfiz-Sz. TIT? E — fig" :. be:ka $596? a” 4150 BE use} To FIND E'E-JILUMW S.- (SME' 51W) —.!‘|.. '7. J... __L 11— TL _1 1._ Ema-t 4» am“): +£3.19 _ mex i=f‘é é," Eéir-EL T = firm->1“ avmziu a;er :mw .. ii1[m1+mz*% séétmg .. T O" _ —m'x: (“M 35%“ mrfix ms; = Mur“ flfz + F TL “UT HERE!) W F -- 1 X. =._.. Xirre 6:“ gr 9 (2:, Flt-ht): I '- r-. xlfi £19 x3 = %‘(—FF): ‘2pr 3352(3); tjfifiom 1M T‘EEMS O-F X. z z w t .. _ j“. n xl=£a x'fifi/fihéfi; n-fix x1 = {19‘ ab: 1;: ‘ %L(rr)bifcfifi 31-. sue. @953 [um G.) ‘ H w *Egfi.._mfl-. Fp(m"+ H") + £|(W“-*"‘X“m*%[fi 3 la "2‘ “3 F31) (renews mum} fi" ,0 (fli)+flm¢(%“)+ft{3ix)+J—M “1&5? ff“ mrykafixlr rm: r, r ,x . z FF 2 ? kin -‘ _- fill '. hH-n _ 3'90?) .. [m _ :9 runs In ,9: . .. . - Elxtfqr)-EIX.P+T¢=305 (1) mi‘ + __-——+ _=__.L film I: m if} + I :9 [ P Fe rp*‘zrf*r;*€fi¥[”v‘—$X _ airline/fix} - Fk.><._ fink: Tait?) MUMT '3‘" T we W955; 3171* ftm‘i4r r‘k.x+ kgrikgx:o LC'EE‘ + (twig '* (f‘kfi- Ric-1.711613% :Q-‘l ; ._fls:aumE Emu-19 - pawn-IE .9 rs magma?- EMT Hum 5T. ta. I] J33)— j;____._._-—-—-—-— All in: in? TIGL‘W/ 71,5139 _ T5 a)“ KR kas T5 jean v cg/§J-kx¢{§)= fileé 1 a." 1- 1 TEWE ‘5 “iii—a + K§a=o gram CWSTIL 69.1mm! tnfiish aims +gfb=o _ ci‘ 1 '5qu: T i JIGW'JO-Cé) Flu-D: STEMT mm; 265mg;er 5W (%;M)£“é + = £Es-uwt '5 {(-3 Ma“ (30; 4 Shh-Jflq‘: 53.333 Ira-v3 X15 " {9550 Makes; + .5‘,‘) ' {-512.5— M-m @9450 EESPOLHE {5 5mm gr am: X mm) : __.§:'"—. {J X J't/i‘ll F=E‘%= H.341; (we a) L)”: j—E-fii-‘SIWB mus 1}}1fi3'n‘ E1: = m: Hit-g 55'» ‘_ val = F it) In. F 51 1-53!— : Lil—- F 0.32 mi may u-n (Aufivmfi 056m» A 9) {Ma X = Hi :. 0-31 Pad- 3 ‘ F" 3. film: = "(100599? 965): —8_s?xm‘¥ 5m (1041.12 t) 1. (1e [:15] Auniformbsrw'rfltkmwn msssm=3kgantl [engthLisnttneltedtotpin A spring eftmltnenm stifl‘nemiis attached to the midpoint oftlte her. Dampi'ngis negligible. The mnfimtmien is slteished below. [M static equilibrium, the her is minimal.) The bar is given some initial emditiens, and its fi‘ee respume occurs at a frequency of? r15. Estimate the value et‘lr. Wire— i .1: : 175ml, “- i 9 '- kh thZB 4 ELIE) _ 0 1+ '1 '2 It. gin-edema larva “Ir (“i-.16 Z O Uni : E :1 gram m1 4. 'L 1*- k : F. (3191.31 At) a 324 try/1 3 S 3. (ll) p15} Ameehanieal system is sketched below. The difl'eremial equation ufmotion is. given. The mass is m - Ill leg. and the undamped natural fi'equerntguI is 111.520 n‘s. The dsmpingfanorisdmmihbd. inaseparuflestbe£=QL Themsisflleufinmd sinusoidslly. It is faund than, four all Frequencies afexeitatiun, the maximum possible steady—5m: rcspome consists ofthe mass uscillsling with an amplimde uf-l mm. Estimate the magnitude ofthe applied force. F,. (You may use the assumption that: i; is small). this Ci 4- kES‘I = Flt} File} : Ft :3» Owl’- M6" lfl mum “lander - s ital-1°. PM pu’iffl. Gn-pli'rl-ud e :1, I F _ :3 -,cl:'.l IJ 5" _—_.-_._.—-—~'—" ——_' I _' —_' : ‘l “A (EFF/l P n in L J [i--r1‘3“+t_a.rr; |'_ 1‘ l‘ei‘é 'ka" r. 1 k‘LL.‘ Iii—(“1 - nae-J h f \) km 11' fur-vi W” ' ; hi I 13—1 - n - \0,| I'flflnj «fr Then 'li’am ll!) eJr relannflcz. F0: 1‘; Lm‘lr : 71515;: (0904”) 5 J) [r40 23.31;?) 004"!) Ti 0: 0,032 new»? 1. (10 [:45] A unilme rigid T-bar is made oftwo rigidly cont-tamed bars oflength L, esehwidt masher. Thus, thetatsl msssisl‘Jn, snddtemassmmntofinertlsls J¢=SmL1tll TheT-haris anathema spinatpeimfl. Aspeingofstifihesekand a dashpot of damping coefficient c are attached at each end ofthe bar as shown. The configuration is matched below. Find the value old for critical damping. Write c in term of the other parameters. The Free-body diagram is partially given(smsl18). [(r'ul'lcsl ale—winch ____'___ 3. (101315] at msssispringa'deshpotisslee‘tehedbelow. The maism=10 kg. {1335: under its own weight by 9.81 mm. (In other words. its Static deflection in file field of gravity is 9.31m]. The damping factor isdeatet'tninled. in temperate test. to he C - 0.05. The mass is then [breed simiddly. it is [bum that, [be all frequencies afeewitstion, the andJran possible steady-sum response consists on}: mass oscillating about its equilibrium with an amplitude of ICI mm. Estimate the magnitude ofthe applied firms, Fa. (You may usethe mumplim that t: is small]. aghast gull-{Hunt “"5. 5 in: 341;”; k: :2 :Omfifith/fl) k It. ma 1 ii a A ?.BI ado—3F"- - 4 - TKA k - l0 lE-j/{L pill figfilhenmfls I mpg, sl—ewiy-SJW-lt 颣-'is'cl"l-’n Olmp't'tlrwde is lat-hm .1; ml orient: R93'Jnafiw! 1"; l y: E: _ X ti—mflfizrrl‘ _' an M’ 7‘5! 3 i : Etk‘fim, : 2(0.0‘3)(lo+*5/21)(|=)Mu_5m) ii: ill 1. (10 pts] Aimitbrrn bar with length Li's attached inthe upriyn pus'niun to spin at fixed point 0. A spring ofsliffnessk is attached to the bar at the midpoint, and a dashpo‘t is attached at the end. You need to include why in this problem. fi—I us Lie—6‘- k 6' 6 ICE-3; a o "'3 o a?“ (a) Sketch the free-body diagram on the figure given at the fight. Include reaction forces and air mornts, and. label all other forces in terms ofthe parameters given, consistently withthe angle wardinueindiu‘eed. Assume small 9. (13] Find an expression Ebr the undsmped natural frequency. Assume small 9. The mass moment od‘inertis aboutthe center ofmass is l ID=—mL3_ 1: tr"? Ema : Int? To = I( l’ Viv-.(E ‘2- : £5511 .1 L S 3.- u Mal—19ml} _ ELB Lot-L7 -- Elli-Y'e cnsfi :— i—MJB Small. 6? lit-nits E 5} 635$ E 1. L 1" " kit L : imLQt—CL9+(1_"§E)9 o 2. (10 p15) The equation of motion from equilibrium for the unforced 2-1:; ha: 511mm. is given below. A flat tesp-onse is sham belw. It is found that the ratio ofsuccessive peaks is 2.?1323 = 2. Determine the damping mgficient 5'. You can assume small (name) I: in your calmlatiom l 1 lmL=§+_cL‘é+—wa=n_ fl=e=231szs 3 4 4 x, I t '1' I. I. 1: gm (3‘: I!“ 351' : Ewe 5 .1 )‘h 3”? E 5 =‘) E: £— fir smut 0k ?_ ‘l'Tr L a —r sec 2 _. ..- rt“; u 5.3 Md Qua-Efi;-=.wfl (En-.4“ Y) . i 1- 2m? 91‘- .__»4” =32 rug-g; .‘ShL‘ 4 h'u - 8 - 3 l '_ C— - (an r l’ C' 3 9fie-:1 ...
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