315 HW_3_Solution

# 315 HW_3_Solution - For pendulum can 3 in vacuum 0.5 Hz =...

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Unformatted text preview: For pendulum, can: 3/! in vacuum .. 0.5 Hz =. 11‘ rut/sec 1: \$4,: = 9vsl/n'z = 0.9940 m (44 = w“ 1.. T:- ;n viscous medium = 0,4,5 Hz = 0.9“. mJ/Sa: 2.. 1-01 1—130?! “"9 T-__°’nzd =77 , >=l.3m 0" ‘5': %o.4\$§7 1 u o .- Ezua‘bfan of moéu‘on: m9 9 + Ct 9 + M72 9 -0 Cd: = 2(m11) ("In ’- ‘ 0.4}f7 2.706 J S'mce T = f_?___ .3 , Ce W N-m—Aec/ra’ . (a) Viscous d3mping, (b) Coulomb damping. (iii) (a) rd = 0.2 sec, fd = 5 Hz, wd = 31.416 rad/sec. (b) Tn = 0.2 sec, I‘n = 5 Hz, w, = 31.410 rad/sec. xi (a) (a) = e‘ “n ’* xi-e-l xi 2 1r g 1n = In 2 = 0.0931 = [Km 1 ._ {3 or 39.9590 (1 = 0.4804 or g = 0.1090 Since wd =wn V 1 — ,weﬁnd «0 31.410 wn = = = 31.0005 rad/sec 1 _ (-2 V0.9s798 _ 2 _ _5_9_9_ 2 __ 4 k —m w _ 9 81 (31.0005) —5.0910 (10 )N/m 0 s‘ =- —- s: cc 2 m wn Hence 0 = 2 m can 5 = 2 (93:11) (31.0005) (0.1090) = 353.1104 N—s/m (b) From Eq. (2.116): _5_0(_)_ 9.81 Using N = W = 500 N, I”: 0.002 I: a (0.002) (5.0304 (104)) 4 w 4 (500) k = m w: = (31.410)2 = 5.0304 (10‘) N/m = 0.0503 Assume that the bicycle and the be]; {all as a rigid body by 5 cm at point A. Thus the mass will be subjected to an initial downward displacement of 5 cm (t = 0 assumed at point A): xo=0.05m,5:o =0 a)n == VE= = 24.7614 rad /sec me.l 800 100 4038.5566 » wd = 0),, V 1 — g” = 24.7014 1 — 0.24702 = 23.9905 rad/sec Response of the system: x(t) =Xe—W‘t sin (cud t +¢) 2 l , 2 xo + 3’ OJ. xo “:1 where X = x3 + l 2 '2' = (0.05), + 0.2476) (24.7614) 0.05)] =0.051607m 23.9905 ‘ . x0 we! 0.05 (23.9905) d = ‘1 —— = t —1 —_————-—-—- = . o g 3” 4’ tan *0 + g wn x0 . an [0.2470 (24.7614) (0.05) 75 6645 359 _' _ Thus the displacement of the boy (positive downward) in vertical direction is given by x(t) = 0.051007 e" “3" t sin (23.0905 t + 75.0045°) m Consider a small angular displacement of the bar 0 about its static equilibri, position. Newton’s second law gives: 3 -- 3! 3! -6’ 6’ t’ t’ \$M=J°9=‘k[”T TJ‘°[9:HI]‘“‘ 9:][7] I . ~ cl- 3 - I 1 1e, JoQ+—ls 0+:kt’20 0 ‘ V. I 104 7 where Jo = 2; m (’2. The undamped natural frequency of torsional vibration is given by: ___ 3U” = 36k w“ 410 V 7111 Let x = displacement of mass and P = tension in rope on the left of mass. Equations of motion: 2F=mi€=~k1§.-P or P=.-mSE—kx (1) 2M=J09=Pr2—c(9r1) (2) Using Eq. (1) in (2), we obtain -(mii+kx)r2-cl9r1=Jo.0' (3) With x = 9 r2, Eq. (3) can be written as: _ (Jo+mr;)'é+crfb+k:;9=o (4) For given data, Eq. (4) becomes .. 2 . [5 + 10 (0.25)’) 9 + c (0.1) 9 + k (0.25)2 9 = 0 or 5.025 2} + 0.12c ('9 + 0.0625 1: 9 = 0 (5) Since amplitude is reduced by 80% in 10 cycl, X! 1.0 cm c w. r.l —=—=5= 1“ 0.2 1‘1 lnT=ln5=L6094=10§wan (0) 11 Since the natural frequency (assumed to be undamped torsional vibration frequency) is 5 Hz, wn = 2 7r (5) = 31.416 rad/sec. Also 1 2 1r 2 1r 0.2 - (7) “‘4 aTaanl—e VI-e Eq. (6) gives 10‘ 0.2 32.832 g‘ 1.6094 = 10 g (31.410) = _ i.e., 'V 1 — g2 =-- 62‘s” g = 39.0400 g 1.6094 ' Le. 0.02561 c ( r' 9) Thus we obtain: ’ g C 0‘ I 0.0625 C c c g=0.02561 = —— =___=_____ cc 2 0),I 2 (5.625) (31.410) Friction force = p N: 0.2 (5) = 1 N. k = 62756- = 250 N/m. Reduction in amplitude in each cycle = 4 u N = gill = 0.016 m. Number of half-cycles k 250 executed before the motion ceases (r): _. _”‘ N r) X" k = 0.1—0.004 a 12 — 2 u N 0.008 k Thus after 6 cycles, the mass stops at a distance of 0.1 - 6 (0.016) = 0.004 m from the unstressed position of the spring. can =’\/_—l£.-= \ l =- 22.1472 rad/sec In T W “ wn = 2— = 0.2837 sec Thus total time of vibration = 8 1', : 1.7022 sec. ...
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