8 without damping 0 forcing frequency mx kx

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Unformatted text preview: ency Forced vibrations (3.8) • Without damping ( γ = 0 ). forcing frequency mx + kx = F0 cos(ω t) ￿￿ mx + kx = 0 ￿￿ xh (t) = C1 cos(ω0 t) + C2 sin(ω0 t) Forced vibrations (3.8) • Without damping ( γ = 0 ). forcing frequency mx + kx = F0 cos(ω t) ￿￿ mx + kx = 0 ￿￿ xh (t) = C1 cos(ω0 t) + C2 sin(ω0 t) ω0 = ? Forced vibrations (3.8) • Without damping ( γ = 0 ). forcing frequency mx + kx = F0 cos(ω t) ￿￿ mx + kx = 0 ￿￿ xh (t) = C1 cos(ω0 t) + C2 sin(ω0 t) ω0 = ￿ k m Forced vibrations (3.8) • Without damping ( γ = 0 ). forcing frequency mx + kx = F0 cos(ω t) ￿￿ mx + kx = 0 ￿￿ xh (t) = C1 cos(ω0 t) + C2 sin(ω0 t) ω0 = ￿ k m natural frequency Forced vibrations (3.8) • Without damping ( γ = 0 ). forcing frequency mx + kx = F0 cos(ω t) ￿￿ mx + kx = 0 ￿￿ xh (t) = C1 cos(ω0 t) + C2 sin(ω0 t) • Case 1: ω ￿= ω0 ω0 = ￿ k m natural frequency Forced vibrations (3.8) • Without damping ( γ = 0 ). forcing frequency mx + kx = F0 cos(ω t) ￿￿ mx + kx = 0 ￿￿ xh (t) = C1 cos(ω0 t) + C2 sin(ω0 t) • Case 1: ω ￿= ω0 xp (t) = A cos(ω t) + B sin(ω t) ω0 = ￿ k m natural frequency Forced vibrations (3.8) • Without damping ( γ = 0 ). forcing frequency mx + kx = F0 cos(ω t) ￿￿ mx + kx = 0 ￿￿ xh (t) = C1 cos(ω0 t) + C2 sin(ω0 t) • Case 1: ω ￿= ω0 xp (t) = A cos(ω t) + B sin(ω t) ￿￿ xp (t) = −ω A cos(ω t) − ω B sin(ω t) 2 2 ω0 = ￿ k m natural frequency Forced vibrations (3.8) • Without damping ( γ = 0 ). forcing frequency mx + kx = F0 cos(ω t) ￿￿ mx + kx = 0 ￿￿ xh (t) = C1 cos(ω0 t) + C2 sin(ω0 t) • Case 1: ω ￿= ω0 xp (t) = A cos(ω t) + B sin(ω t) ￿￿ 2 xp (t) = −ω A cos(ω t) ￿...
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This note was uploaded on 02/12/2014 for the course MATH 256 taught by Professor Ericcytrynbaum during the Spring '13 term at The University of British Columbia.

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