# 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 s t m π = ( )( )( ) max 0.002 2 512 6.43 m m x s s π = = ± ( )( ) ( ) 2 2 4 max 2 2 0.002 2 512 2.07 10 m m x s s π = = × ±± 3.2 0.1sin x t ω = D [ m ] 0.1 cos m x t s ω ω = D D ± When t = 0, x = 0 and 0.5 0.1 m x s ω = = D ± 1 5 s ω = D 2 1.26 T s π ω = = D 3.3 ( ) cos sin x x t x t t ω ω ω = + D D D D ± D and 2 f ω π = D ( ) ( ) [ ] 0.25cos 20 0.00159sin 20 x t t π π = + m 3.4 ( ) cos cos cos sin sin α β α β α = + β ( ) cos cos cos sin sin x A t A t A t ω φ φ ω φ = = + D D ω D cos sin x t t ω ω = Α + Β D D , cos A φ Α = , sin A φ Β = 3.5 2 2 2 2 1 1 2 1 1 1 1 2 2 2 2 mx kx mx kx + = + ± ± 2 ( ) ( ) 2 2 2 2 1 2 2 1 k x x m x x = ± ± 1 2 2 2 2 1 2 2 1 2 k x x m x x ω = = D ± ± 2 2 1 1 1 1 1 2 2 2 kA mx kx = + ± 2 2 2 2 2 2 2 2 1 1 2 1 1 1 2 2 2 1 m x x x x A x x k x x = + = + ± ± ± ± ± 2 1 x 1 2 2 2 2 2 1 2 2 1 2 2 2 1 x x x x A x x = ± ± ± ± 1

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3.6 1 1 2.5 9.8 2 6 l T s g π π = = D s 3.7 For springs tied in parallel: ( ) ( ) 1 2 1 2 s F x k x k x k k x = − = − + ( ) 1 2 1 2 k k 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 k x 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 k ′ = x . Meanwhile, The upward force at P is 1 k x . The downward force at P is ( ) 2 k x x . Therefore, ( ) 1 2 k x k x x = 2 1 2 k x x k k ′ = + And eq k 2 1 1 2 k x x k k k = + ( ) 1 2 1 2 1 2 eq k k k m k k m ω = = + 3.8 For the system ( ) M m + , ( ) kX M m X = + ±± The position and acceleration of m are the same as for ( ) M m + : m m k x x M m = − + ±± cos cos m k k x A t d t M m M δ = + = + + m r The total force on m , m m F mx mg F = = ±± 2
cos r m mk mkd k F mg x mg t M m M 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 = + ( ) g M m d k + = 3.9 ( ) cos t d x e A t γ ω φ = ( ) ( ) sin cos t t d d d dx e A t e A t dt γ γ ω ω φ γ ω φ = − maxima at ( ) ( ) 0 sin cos d d d t t dt dx ω ω φ γ ω φ + = = ( ) tan d d t γ ω φ ω = − thus the condition of relative maximum occurs every time that t increases by 2 d π ω : 1 2 i i d t t π ω + = + For the i th maximum: ( ) cos i t i d x e A t γ i ω φ = ( ) 1 2 1 1 cos i d t i d i i x e A t e π γ γ ω ω φ + + + = x = 2 1 d d T i i x e e x π γ ω γ + = = 3.10 (a) 1 3 2 c s m γ = = 2 2 25 k s m ω = = D 2 2 2 16 d s ω ω γ 2 = = D 2 2 2 7 r d s ω ω γ 2 = = 1 7 r s ω = (b) max 48 0.2 60.4 d F

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• Fall '04
• mokhtari
• Cos, Tan, Trigraph, k2, TI

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