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HW2_Solution

# HW2_Solution - Problem 2.5[1 With E measured from the...

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Unformatted text preview: Problem 2.5 [1 With E measured from the static equiiibrium position of m1 and k, the equation of motion after impact is {ml + 1711);? + [at = mag {a} The general solution is “(1‘) a Acosta”: + Bsin a)“: + ":g (b) w “ k (C) " m + mg The initial conditions are “(0) m 0 {1(0) :iL-ngh (d) m; +1112 The initiat veiocity in Eq.(d) was determined by conservation of momenmm during impact: min x (ml + mﬁtKO) where at; = 1High Impose initial conditions to determine A and B: mag [£(0)T»0:>A=—k (3) 2n] mo): (0an 32\$ “1 {f} ml + mg (on Substituting Eqs. (e) and (f) in Eq. (b) gives J2 I “0) = mag (1 — cos Cunt) + g] \$sin cunt a)” In] -i- m2 Problem 2.15 1.Dererm£ne { and (on. 4' ML: 1E] .51. = 1 In J». : o.o123=1.23% 231'} u- , 27r(20) 0.2 1‘.- Therefore the assumption of smaii damping implicit in the above equaﬁon is valid. TD : ~5— = 0.155%; I; 2 TD == Olisec; a)“ 2 £31, 2 41.89 fads/sec 0.13 2. Determine stiﬁ’izess coeﬂicienr. k = wﬁm 2 (4139ij = 173.51bs/En. 3. Deremzfne damping coeﬁcienr. C" a 21116:)” = 2am {41.89} = 8.3771b-—sec/in. c =9” = 0.0128(8.377) = 0.1071b—sec/in. Problem 2.17 Reading values directly from Fig. i.§.4b: Peak Time, F631;, ifj (g) 1,.(sec) 1 ‘ 0.80 I 0.78 M 31 7.84 0.50 To aw = 0235 sec E7 ...
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