12 notes- Dynamical Systems Equation.pptx

Current position previous position kprevious position

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current position = previous position + (k*previous position + c)c0 k = -1/2 c = 0
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Dynamical Systems Equation Δx/Δt = kx + c Δx/Δt = -1/2x + 0 initial position = 100 t = x = 0 100 1 50 2 25 3 12.5 4 6.25 5 3.13 6 1.56 7 .78 0 2 4 6 8 10 12 14 0 20 40 60 80 100 120 k = -1/2 c = 0
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Dynamical Systems Equation Δx/Δt = kx + c Δx/Δt = -1/2x + 0 initial position = 200 t = x = 0 1 2 3 4 5 6 7 Speed changes at each position. current position = previous position + (k*previous position + c) k = -1/2 c = 0
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Dynamical Systems Equation Δx/Δt = kx + c Δx/Δt = -1/2x + 0 initial position = 200 t = x = 0 1 2 3 4 5 6 7 0 2 4 6 8 10 12 14 0 50 100 150 200 250 k = -1/2 c = 0
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Dynamical Systems Equation Δx/Δt = kx + c Δx/Δt = -1/2x + 0 0 2 4 6 8 10 12 14 0 50 100 150 200 250 initial position: 100 200 Different initial positions, Same k:
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Dynamical Systems Equation Δx/Δt = kx + c Δx/Δt = -1/2x + 0 0 2 4 6 8 10 12 14 0 50 100 150 200 250 initial position: 100 200 /æt/ vs. /it/
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Dynamical Systems Equation Δx/Δt = kx + c Δx/Δt = -1/4x + 0 initial position = 100 t = x = 0 1 2 3 4 5 6 7 Speed changes at each position. current position = previous position + (k*previous position + c) k = -1/4 c = 0
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Dynamical Systems Equation Δx/Δt = kx + c Δx/Δt = -1/4x + 0 initial position = 100 t = x = 0 100 1 75 2 56.25 3 42.19 4 31.64 5 23.73 6 17.19 7 13.35 0 2 4 6 8 10 12 14 16 18 20 0 20 40 60 80 100 120 k = -1/4 c = 0
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Dynamical Systems Equation Δx/Δt = kx + c Δx/Δt = kx + 0 initial position = 100 0 2 4 6 8 10 12 14 16 18 20 0 20 40 60 80 100 120 k Bigger k =
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Dynamical Systems Equation Speed of a spring: Δx Δt = kx +c x= a position of the system k= stiffness acceleration of the system
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  • Spring '14
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