Fowles03

# Fowles03 - CHAPTER 3 OSCILLATIONS 3.1 x = 0.002sin 2 ( 512...

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CHAPTER 3 OSCILLATIONS 3.1 () [ ] 1 0.002sin 2 512 x stm π  =  ( ) ( ) max 0.002 2 512 6.43 mm x ss  ==   ± ( ) ( ) 22 4 max 0.002 2 512 2.07 10 x × ±± 3.2 0.1sin x t ω = D [ m ] 0.1 cos m xt s ωω = DD ± When t = 0, x = 0 and 0.5 0.1 m x s  D ± 1 5 s = D 2 1.26 Ts D 3.3 cos sin x x tx t t =+ D D ± D and 2 f = D ( )[ ] 0.25cos 20 0.00159sin 20 x tt ππ m 3.4 cos cos cos sin sin α βα β −= + cos cos cos sin sin x AtA t A t φφ φ =− = + D cos sin x , cos A Α = , sin A Β = 3.5 11 2 1111 2222 mx kx mx kx += + 2 12 21 kx x mx x 1 2 x mxx    D 111 222 kA mx kx ± 2 2 x x x Ax x x = + ± 2 1 x 1 2 xx A = 1

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3.6 11 2.5 9.8 2 6 l Ts g ππ == D s 3.7 For springs tied in parallel: () ( ) 12 1 2 s Fx k xkx k kx =− + () 1 2 kk m ω +  =   For springs tied in series: The upward force is m eq k x . Therefore, the downward force on spring is 2 k eq k x . The upward force on the spring is 2 k 1 kx where x is the displacement of P, the point at which the springs are tied. Since the spring is in equilibrium, 2 k 1 eq k ′ = x . Meanwhile, The upward force at P is 1 . The downward force at P is ( ) 2 kxx . Therefore, k x x ′′ 2 x ′ = + And eq k 2 1 xk  =  +  1 2 eq k mk k m + 3.8 For the system M m + , ( ) kX M m X −= + ±± The position and acceleration of m are the same as for ( ) M m + : mm k x x M m + cos cos m x At d t M mM δ =+ = ++ m r The total force on m , Fm xm gF = 2
cos rm mk mkd k Fm g x m g t M mM m M =+ ++ m + For the block to just begin to leave the bottom of the box at the top of the vertical oscillations, 0 r F = at m x d = − : 0 mkd mg M m =− + () gM m d k + = 3.9 cos t d xeA t γ ω φ () () sin cos tt dd d dx eA t t dt γγ ωφ −− maxima at ()() 0s i n c o s d dt dx − + == tan d d t −= thus the condition of relative maximum occurs every time that t increases by 2 d π : 1 2 ii d + For the i th maximum: ( ) cos i t id t i = 1 2 11 cos t i i x t e γω + x = 2 1 T i i x ee x + 3.10 (a) 1 3 2 c s m 22 25 k s m D 222 16 d s ωω 2 =−= D 2 7 rd s 2 1 7 r s = (b) max 48 0.2 60.4 d F C = D Am m (c) 2 n 23 rr r r = D 7 = ta 41.4 D 3

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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
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## Fowles03 - CHAPTER 3 OSCILLATIONS 3.1 x = 0.002sin 2 ( 512...

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