s11f1026

s11f1026 - = A A ψ = √ 2-A A ˙ φ = ω B ω-B ˙ ψ =...

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ADVANCED DYNAMICS — PHY 4241/5227 HOME AND CLASS WORK – SET 6 Solution for assignment 26: Double pendulum solution and plot (continuation of 25). Let us take minors with respect to the first row of the determinant. For the ω + frequency the ratio of the two minors is 4 1+ 4 2+ = - 1 - 2 2 + 2 = ( - 1 - 2) (2 + 2) (1 - 2) (1 - 2) = 1 - 2 and for ω - it is 4 1 - 4 2 - = - 1 + 2 2 - 2 = ( - 1 + 2) (2 - 2) (1 + 2) (1 + 2) = 1 2 . Therefore, the solutions (real part) can be written φ + ( t ) = A + cos( ω + t ) + B + sin( ω + t ) , φ - ( t ) = A - cos( ω + t ) + B - sin( ω + t ) , φ ( t ) = φ + ( t ) + φ - ( t ) , ψ ( t ) = - 2 φ + ( t ) + 2 φ - ( t ) . The four constants are determined by the four initial value, e.g., φ 0 , ˙ φ 0 , ψ 0 , ˙ ψ 0 at time t = 0: φ
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Unformatted text preview: = A + + A-, ψ = √ 2 (-A + + A-) , ˙ φ = ω + B + + ω-B-, ˙ ψ = √ 2 (-ω + B + + ω-B-) , which gives A + = φ 2-ψ 2 √ 2 , A-= φ 2 + ψ 2 √ 2 , B + = ˙ φ 2 ω +-˙ ψ 2 √ 2 ω + , B-= ˙ φ 2 ω-+ ˙ ψ 2 √ 2 ω-. For the initial conditions φ = 0 , ˙ φ = 1 , ψ = 0 , ˙ ψ =-1 at time t = 0 the figure gives a plot up to t = 50 q l/g (next page).-1-0.8-0.6-0.4-0.2 0.2 0.4 0.6 0.8 1 10 20 30 40 50 Angles [rad] t [(l/g) 1/2 ] φ ψ...
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s11f1026 - = A A ψ = √ 2-A A ˙ φ = ω B ω-B ˙ ψ =...

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