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315 HW_2_Solution

# 315 HW_2_Solution - k A1E1 —(o.05)2(30(105 1 6’1 30(12...

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Unformatted text preview: k _ A1E1 — %(o.05)2 (30 (105)) 1 _ _...___ __ ______.___. 6’1 30 (12) = 163.6250 lb /in k = A2 E2 = 163.625 (g5) =13”? lb/m _ ‘ \ \ 2 (2 363\$. ‘ a \ \ \ 176.3: 393.17! w _ 9‘ keq = k1 + k2 = 163.6250 + W = W lb/in -- ------- C3. Let x be measured from the unstretched length of the springs. The equation of motion is: mSE=—(k1 +k2) (x+5,t) +Wsin0 where (k1 + kg) 6, = W sin 6 i.e., mii +(k1 +k2)x =0 Thus the natural frequency of vibration of the cart is given by 23 . ”'5’ {.374 k1 + k2 W (386.4) “’1 “ V m " T ‘é’i’f-‘md/m (3.) Velocity of anvil = v = 50 ft/sec- = 600 in/sec. x = 0 at static equilibrium~ Q position. _ _ _ BEE. = _ ‘.___m 3 no — x(t=0) keq 4 1:. Conservation of momentum: (M +m)io =mv or in =i(t=o) .—. Mm+vm Natural frequency: w __ 4 k n M + m Complete solution: x(t) =Ao sin (wn t + «m where 2 % 2 2 2 2 '1' __ 2 x0 = m g m v 2 A°_x°+—_ {1318 +(M+m)4k} and 58 Since v = 600, m = 12/386.4, M = 100/386.4, k = 100, we ﬁnd 1 2 2 2 Ao=[ 12(386-4) ] [12000)] 32m =1.7308in 4 (100) (386.4) 386.4 112 (400) ¢o = tan"1 -- 386A 112 = tan'1 (— 0.01734) = — 0.9934 deg V3864 (600) V400 (b) x = 0 at static equilibrium position: x0 = x(t=0) = 0. Conservation of momentum gives: Miro =mv or 5:0 =i(t=0) = 911.41 ‘ Complete solution: ' x(t) = A0 sin (w, t + (1)0) where 1 . 2 ‘2' i = 2 x0 __ m2 V” (M) 2 m v 12 (600) V 386,4 can M 4 1: K74 k M 386.4 4 (100) (100) Xown 4’0 =tan-l T =tan'1(0).-=o 77 Equation of motion: Massm: mg—T=mii PulleyJo: Jo 6=Tr—k4r(0+90)4r where 00 = angular deﬂection of the pulley under the weight, mg, given by: mg mgr=k(4r90)4r or 90= 16rk Substituting Eqs. (1) and (3) into (2), we obtain "= _ u _ 2 mg J00 (mg mx)r k16r(0+16rk) Using x = r 0 and ii = r 3, Eq. (4) becomes (Jo +m1'z)b.-‘l-(ltlr2 k)9==0 .m wn= [5? (L) ".1254, kg} sin 9+ m3? sine =0 3, 77:12; + («al+myf) 9:0 w" = «at-9mg! ml; (c) mil 5 + «a? sine - "213‘." 9‘0 i m1: 5 +(kaf— "3!) 9:0 by“ = "ka?- 1:2! m1 Iconfzn‘ﬁurai'ion (It) has the In‘gkesl’ natural freguency. Eguai'u‘on of motion as =-wae_z«(_§.9)§ _. 2i 2.! 2" 3 9)'—"“t9 Where For given claim , w - 9(‘°)("") 5%) +'0 (2000) (s)2 +9(Iooo) ' n- ' =45-1547 . < 197(5)a ...
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315 HW_2_Solution - k A1E1 —(o.05)2(30(105 1 6’1 30(12...

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