test1 - Name ID ME461 Mechanical Vibration 7 Term I Exam(Dr...

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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" :. ...
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test1 - Name ID ME461 Mechanical Vibration 7 Term I Exam(Dr...

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