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Unformatted text preview: ency Forced vibrations (3.8)
â€¢ Without damping ( Î³ = 0 ). forcing frequency mx + kx = F0 cos(Ï‰ t)
ï¿¿ï¿¿ mx + kx = 0
ï¿¿ï¿¿ xh (t) = C1 cos(Ï‰0 t) + C2 sin(Ï‰0 t) Forced vibrations (3.8)
â€¢ Without damping ( Î³ = 0 ). forcing frequency mx + kx = F0 cos(Ï‰ t)
ï¿¿ï¿¿ mx + kx = 0
ï¿¿ï¿¿ xh (t) = C1 cos(Ï‰0 t) + C2 sin(Ï‰0 t) Ï‰0 = ? Forced vibrations (3.8)
â€¢ Without damping ( Î³ = 0 ). forcing frequency mx + kx = F0 cos(Ï‰ t)
ï¿¿ï¿¿ mx + kx = 0
ï¿¿ï¿¿ xh (t) = C1 cos(Ï‰0 t) + C2 sin(Ï‰0 t) Ï‰0 = ï¿¿ k
m Forced vibrations (3.8)
â€¢ Without damping ( Î³ = 0 ). forcing frequency mx + kx = F0 cos(Ï‰ t)
ï¿¿ï¿¿ mx + kx = 0
ï¿¿ï¿¿ xh (t) = C1 cos(Ï‰0 t) + C2 sin(Ï‰0 t) Ï‰0 = ï¿¿ k
m natural frequency Forced vibrations (3.8)
â€¢ Without damping ( Î³ = 0 ). forcing frequency mx + kx = F0 cos(Ï‰ t)
ï¿¿ï¿¿ mx + kx = 0
ï¿¿ï¿¿ xh (t) = C1 cos(Ï‰0 t) + C2 sin(Ï‰0 t)
â€¢ Case 1: Ï‰ ï¿¿= Ï‰0 Ï‰0 = ï¿¿ k
m natural frequency Forced vibrations (3.8)
â€¢ Without damping ( Î³ = 0 ). forcing frequency mx + kx = F0 cos(Ï‰ t)
ï¿¿ï¿¿ mx + kx = 0
ï¿¿ï¿¿ xh (t) = C1 cos(Ï‰0 t) + C2 sin(Ï‰0 t)
â€¢ Case 1: Ï‰ ï¿¿= Ï‰0 xp (t) = A cos(Ï‰ t) + B sin(Ï‰ t) Ï‰0 = ï¿¿ k
m natural frequency Forced vibrations (3.8)
â€¢ Without damping ( Î³ = 0 ). forcing frequency mx + kx = F0 cos(Ï‰ t)
ï¿¿ï¿¿ mx + kx = 0
ï¿¿ï¿¿ xh (t) = C1 cos(Ï‰0 t) + C2 sin(Ï‰0 t)
â€¢ Case 1: Ï‰ ï¿¿= Ï‰0 xp (t) = A cos(Ï‰ t) + B sin(Ï‰ t)
ï¿¿ï¿¿
xp (t) = âˆ’Ï‰ A cos(Ï‰ t) âˆ’ Ï‰ B sin(Ï‰ t)
2 2 Ï‰0 = ï¿¿ k
m natural frequency Forced vibrations (3.8)
â€¢ Without damping ( Î³ = 0 ). forcing frequency mx + kx = F0 cos(Ï‰ t)
ï¿¿ï¿¿ mx + kx = 0
ï¿¿ï¿¿ xh (t) = C1 cos(Ï‰0 t) + C2 sin(Ï‰0 t)
â€¢ Case 1: Ï‰ ï¿¿= Ï‰0 xp (t) = A cos(Ï‰ t) + B sin(Ï‰ t)
ï¿¿ï¿¿
2
xp (t) = âˆ’Ï‰ A cos(Ï‰ t)
ï¿¿ï...
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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.
 Spring '13
 EricCytrynbaum
 Differential Equations, Equations

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