Analytical Mech Homework Solutions 25

Analytical Mech Homework Solutions 25 - 3.11 17 2 mx = 0 2...

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3.11 (a) 2 17 30 2 mx mx mx ββ ++ = ±± ± 3 2 γ β = and 22 17 2 ω = D 2 24 r 2 ωγ =− = D 2 r = (b) max 2 d F m A γω = D 222 25 4 d 2 =−= D 5 2 d = 2 2 15 A = 3.12 1 2 d T e = 1 ln 2 ln 2 d d f T == (a) 1 2 () d =− D So, 1 2 d =+ D 11 2 ln 2 1 dd ff f ππ     = +         D 100.6 f Hz = D (b) 1 2 rd 2 ln 2 1 d f = − 99.4 r f Hz = 3.13 Since the amplitude diminishes by e d T in each complete period, 1 1 d n T ee e 1 d Tn = 1 2 d d n π Now 1 2 2 2 d D So 1 1 2 2 1 1 4 n =+ = + D ωω 4
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