The Chaotic Pendulum I Continuous Nonlinear Dynamics Rubin H Landau - PowerPoint PPT Presentation

ODE Free Pend Phase Space Implementation-Assessment The Chaotic Pendulum I Continuous Nonlinear Dynamics Rubin H Landau Sally Haerer, Producer-Director Based on A Survey of Computational Physics by Landau, Páez, & Bordeianu with Support from the National Science Foundation Course: Computational Physics I 1 / 58

l1 m1 m2 θ 2 l2 θ 1 m I θ α α f ODE Free Pend Phase Space Implementation-Assessment Problem: Realistic Single or Double Pendulum Simulate Nonlinear, Chaotic System loading TwoPend Driven single pendulum Large oscillations, even over-the-top Free, double pendulum 2 / 58

I m f α α θ ODE Free Pend Phase Space Implementation-Assessment Chaotic Pendulum ODE � τ = I d 2 θ Newton’s Laws for Rotational Motion dt 2 Gravitation τ : − mgl sin θ − β ˙ Friction τ : θ External τ : τ 0 cos ω t I d 2 θ dt 2 = − mgl sin θ − β d θ dt + τ 0 cos ω t (1) d 2 θ 0 sin θ − α d θ dt 2 = − ω 2 dt + f cos ω t (2) ω 0 = mgl α = β f = τ 0 , I , I I 3 / 58

I m f α α θ ODE Free Pend Phase Space Implementation-Assessment Chaotic Pendulum ODE � τ = I d 2 θ Newton’s Laws for Rotational Motion dt 2 Gravitation τ : − mgl sin θ − β ˙ Friction τ : θ External τ : τ 0 cos ω t I d 2 θ dt 2 = − mgl sin θ − β d θ dt + τ 0 cos ω t (1) d 2 θ 0 sin θ − α d θ dt 2 = − ω 2 dt + f cos ω t (2) ω 0 = mgl α = β f = τ 0 , I , I I 4 / 58

I m f α α θ ODE Free Pend Phase Space Implementation-Assessment Chaotic Pendulum ODE � τ = I d 2 θ Newton’s Laws for Rotational Motion dt 2 Gravitation τ : − mgl sin θ − β ˙ Friction τ : θ External τ : τ 0 cos ω t I d 2 θ dt 2 = − mgl sin θ − β d θ dt + τ 0 cos ω t (1) d 2 θ 0 sin θ − α d θ dt 2 = − ω 2 dt + f cos ω t (2) ω 0 = mgl α = β f = τ 0 , I , I I 5 / 58

I m f α α θ ODE Free Pend Phase Space Implementation-Assessment Chaotic Pendulum ODE � τ = I d 2 θ Newton’s Laws for Rotational Motion dt 2 Gravitation τ : − mgl sin θ − β ˙ Friction τ : θ External τ : τ 0 cos ω t I d 2 θ dt 2 = − mgl sin θ − β d θ dt + τ 0 cos ω t (1) d 2 θ 0 sin θ − α d θ dt 2 = − ω 2 dt + f cos ω t (2) ω 0 = mgl α = β f = τ 0 , I , I I 6 / 58

I m f α α θ ODE Free Pend Phase Space Implementation-Assessment Chaotic Pendulum ODE � τ = I d 2 θ Newton’s Laws for Rotational Motion dt 2 Gravitation τ : − mgl sin θ − β ˙ Friction τ : θ External τ : τ 0 cos ω t I d 2 θ dt 2 = − mgl sin θ − β d θ dt + τ 0 cos ω t (1) d 2 θ 0 sin θ − α d θ dt 2 = − ω 2 dt + f cos ω t (2) ω 0 = mgl α = β f = τ 0 , I , I I 7 / 58

α f α m I θ ODE Free Pend Phase Space Implementation-Assessment Chaotic Pendulum ODE ˙ y = � � f ( � y , t ) Standard ODE Form (rk4): d 2 θ 0 sin θ − α d θ dt 2 = − ω 2 dt + f cos ω t (1) 2 nd O t-dependent nonlinear ODE sin θ ≃ θ − θ 3 / 3 ! · · · Nonlinearity: y ( 0 ) = θ ( t ) , y ( 1 ) = d θ ( t ) dt dy ( 0 ) = y ( 1 ) (2) dt dy ( 1 ) 0 sin y ( 0 ) − α y ( 1 ) + f cos ω t = − ω 2 (3) dt 8 / 58

α f α m I θ ODE Free Pend Phase Space Implementation-Assessment Chaotic Pendulum ODE ˙ y = � � f ( � y , t ) Standard ODE Form (rk4): d 2 θ 0 sin θ − α d θ dt 2 = − ω 2 dt + f cos ω t (1) 2 nd O t-dependent nonlinear ODE sin θ ≃ θ − θ 3 / 3 ! · · · Nonlinearity: y ( 0 ) = θ ( t ) , y ( 1 ) = d θ ( t ) dt dy ( 0 ) = y ( 1 ) (2) dt dy ( 1 ) 0 sin y ( 0 ) − α y ( 1 ) + f cos ω t = − ω 2 (3) dt 9 / 58

α f α m I θ ODE Free Pend Phase Space Implementation-Assessment Chaotic Pendulum ODE ˙ y = � � f ( � y , t ) Standard ODE Form (rk4): d 2 θ 0 sin θ − α d θ dt 2 = − ω 2 dt + f cos ω t (1) 2 nd O t-dependent nonlinear ODE sin θ ≃ θ − θ 3 / 3 ! · · · Nonlinearity: y ( 0 ) = θ ( t ) , y ( 1 ) = d θ ( t ) dt dy ( 0 ) = y ( 1 ) (2) dt dy ( 1 ) 0 sin y ( 0 ) − α y ( 1 ) + f cos ω t = − ω 2 (3) dt 10 / 58

α f α m I θ ODE Free Pend Phase Space Implementation-Assessment Chaotic Pendulum ODE ˙ y = � � f ( � y , t ) Standard ODE Form (rk4): d 2 θ 0 sin θ − α d θ dt 2 = − ω 2 dt + f cos ω t (1) 2 nd O t-dependent nonlinear ODE sin θ ≃ θ − θ 3 / 3 ! · · · Nonlinearity: y ( 0 ) = θ ( t ) , y ( 1 ) = d θ ( t ) dt dy ( 0 ) = y ( 1 ) (2) dt dy ( 1 ) 0 sin y ( 0 ) − α y ( 1 ) + f cos ω t = − ω 2 (3) dt 11 / 58

α f α m I θ ODE Free Pend Phase Space Implementation-Assessment Chaotic Pendulum ODE ˙ y = � � f ( � y , t ) Standard ODE Form (rk4): d 2 θ 0 sin θ − α d θ dt 2 = − ω 2 dt + f cos ω t (1) 2 nd O t-dependent nonlinear ODE sin θ ≃ θ − θ 3 / 3 ! · · · Nonlinearity: y ( 0 ) = θ ( t ) , y ( 1 ) = d θ ( t ) dt dy ( 0 ) = y ( 1 ) (2) dt dy ( 1 ) 0 sin y ( 0 ) − α y ( 1 ) + f cos ω t = − ω 2 (3) dt 12 / 58

m I θ ODE Free Pend Phase Space Implementation-Assessment Start Simply: Free Oscillations (Test Algorithm & Physics) Ignore Friction & External Torques ( f = α = 0) ¨ θ = − ω 2 0 sin θ (1) ¨ θ ≃ − ω 2 0 θ (linear, θ ≃ 0) ⇒ θ ( t ) = θ 0 sin ( ω 0 t + φ ) (2) (1): ”Analytic solution”; sort of: � θ m d θ T ∝ (3) � 1 / 2 sin 2 ( θ m / 2 ) − sin 2 ( θ/ 2 ) � 0 13 / 58

m I θ ODE Free Pend Phase Space Implementation-Assessment Start Simply: Free Oscillations (Test Algorithm & Physics) Ignore Friction & External Torques ( f = α = 0) ¨ θ = − ω 2 0 sin θ (1) ¨ θ ≃ − ω 2 0 θ (linear, θ ≃ 0) ⇒ θ ( t ) = θ 0 sin ( ω 0 t + φ ) (2) (1): ”Analytic solution”; sort of: � θ m d θ T ∝ (3) � 1 / 2 sin 2 ( θ m / 2 ) − sin 2 ( θ/ 2 ) � 0 14 / 58

m I θ ODE Free Pend Phase Space Implementation-Assessment Start Simply: Free Oscillations (Test Algorithm & Physics) Ignore Friction & External Torques ( f = α = 0) ¨ θ = − ω 2 0 sin θ (1) ¨ θ ≃ − ω 2 0 θ (linear, θ ≃ 0) ⇒ θ ( t ) = θ 0 sin ( ω 0 t + φ ) (2) (1): ”Analytic solution”; sort of: � θ m d θ T ∝ (3) � 1 / 2 sin 2 ( θ m / 2 ) − sin 2 ( θ/ 2 ) � 0 15 / 58

m I θ ODE Free Pend Phase Space Implementation-Assessment Start Simply: Free Oscillations (Test Algorithm & Physics) Ignore Friction & External Torques ( f = α = 0) ¨ θ = − ω 2 0 sin θ (1) ¨ θ ≃ − ω 2 0 θ (linear, θ ≃ 0) ⇒ θ ( t ) = θ 0 sin ( ω 0 t + φ ) (2) (1): ”Analytic solution”; sort of: � θ m d θ T ∝ (3) � 1 / 2 sin 2 ( θ m / 2 ) − sin 2 ( θ/ 2 ) � 0 16 / 58

ODE Free Pend Phase Space Implementation-Assessment ¨ θ = − ω 2 Free Pendulum Implementation 0 sin θ Solve ODE with rk4 Initial conditions: { θ = 0, ˙ θ ( 0 ) � = 0}; increase ˙ θ ( 0 ) 1 Verify SHM ¨ θ = − ω 2 0 θ ⇒ ω = ω 0 = 2 π/ T = constant 2 Devise algorithm to determine period T ( 3 × θ = 0) 3 Determine T ( θ ) for realistic pendulum, compare 4 Verify as KE ( 0 ) ≤ 2 mgl : non harmonic oscillations 5 Verify ⇒ separatrix ( KE ( 0 ) → 2 mgl ), T → ∞ 6 Listen harmonic & anharmonic motion (Hear now) 7 Hear Data applet 8 17 / 58

ODE Free Pend Phase Space Implementation-Assessment ¨ θ = − ω 2 Free Pendulum Implementation 0 sin θ Solve ODE with rk4 Initial conditions: { θ = 0, ˙ θ ( 0 ) � = 0}; increase ˙ θ ( 0 ) 1 Verify SHM ¨ θ = − ω 2 0 θ ⇒ ω = ω 0 = 2 π/ T = constant 2 Devise algorithm to determine period T ( 3 × θ = 0) 3 Determine T ( θ ) for realistic pendulum, compare 4 Verify as KE ( 0 ) ≤ 2 mgl : non harmonic oscillations 5 Verify ⇒ separatrix ( KE ( 0 ) → 2 mgl ), T → ∞ 6 Listen harmonic & anharmonic motion (Hear now) 7 Hear Data applet 8 18 / 58

ODE Free Pend Phase Space Implementation-Assessment ¨ θ = − ω 2 Free Pendulum Implementation 0 sin θ Solve ODE with rk4 Initial conditions: { θ = 0, ˙ θ ( 0 ) � = 0}; increase ˙ θ ( 0 ) 1 Verify SHM ¨ θ = − ω 2 0 θ ⇒ ω = ω 0 = 2 π/ T = constant 2 Devise algorithm to determine period T ( 3 × θ = 0) 3 Determine T ( θ ) for realistic pendulum, compare 4 Verify as KE ( 0 ) ≤ 2 mgl : non harmonic oscillations 5 Verify ⇒ separatrix ( KE ( 0 ) → 2 mgl ), T → ∞ 6 Listen harmonic & anharmonic motion (Hear now) 7 Hear Data applet 8 19 / 58

ODE Free Pend Phase Space Implementation-Assessment ¨ θ = − ω 2 Free Pendulum Implementation 0 sin θ Solve ODE with rk4 Initial conditions: { θ = 0, ˙ θ ( 0 ) � = 0}; increase ˙ θ ( 0 ) 1 Verify SHM ¨ θ = − ω 2 0 θ ⇒ ω = ω 0 = 2 π/ T = constant 2 Devise algorithm to determine period T ( 3 × θ = 0) 3 Determine T ( θ ) for realistic pendulum, compare 4 Verify as KE ( 0 ) ≤ 2 mgl : non harmonic oscillations 5 Verify ⇒ separatrix ( KE ( 0 ) → 2 mgl ), T → ∞ 6 Listen harmonic & anharmonic motion (Hear now) 7 Hear Data applet 8 20 / 58