Forced Pendulum Numerical approach

Transcript

1 Numerical approach UiO April 8, 2014

2 Physical problem and equation We have a pendulum of length l, with mass m. The pendulum is subject to gravitation as well as both a forcing and linear resistance force. We denote the acceleration of gravity as g, and the angle of the pendulum as θ. The magnitude of the forcing is F and has an angular frequency ω, while α is a measure of the resistance force. This gives us the following equation of motion: ml d 2 θ dθ + αl + mg sin(θ) = F cos(ωt) dt2 dt

3 Scaling d 2 θ dt 2 + α dθ m dt + g sin(θ) = F l ml cos(ωt) We assume that the forcing is near the resonance frequency: g/l = ω + ω g l = ω 2 0 = 1 + 2ω ω + ω 2 We also assume that the system has an periodic response with small amplitude. When we rescale the equation we assume that the leading non-linear term, the forcing, the damping, and the term associated with ω are of the same order of magnitude.

4 Scaling First we change variable to τ = ωt, and scale the angle θ = Bz. We also taylor expand the sine term. z + α mω z + ω2 0 B2 (z ω2 6 z3 ) = F Bmlω 2 cos(τ) We want the non-linear term to be of the same order as the forcing, and we obtain this be choosing: B 2 = F Bmlω 2 We use our assumptions and introduce: ɛ = B 2, βɛ = α mω, κ = 1 6, γɛ = 2ω ω

5 Scaling This gives the equation on the form: + O(ɛ 2 ) where the term βz represents linear damping, (1 + γɛ)z represents linear detuning, κɛz 3 represents nonlinear detuning and ɛ cos(τ) represents the forcing.

6 Asymptotic solution We use: z = z 0 + ɛz 1 + O(ɛ 2 ) This gives the leading order terms: ɛ 0 : z 0 + z 0 = 0 z 0 = ae iτ + c.c To decide a we need to use the terms linear in ɛ: ɛ 1 : z 1 + z 1 = γz 0 βz 0 + κz3 0 + cos(τ) = e iτ ( γa iβa κa a 2 ) + κa 3 e 3iτ + c.c Since e iτ and its complex conjugated solves the homogeneous equation for z 1 we need to exclude them from the RHS in order to avoid secular terms in the particular solution.

7 Asymptotic solution We write a = a e iδ γa iβa κa a 2 = 0 γ a iβ a + 3κ a 3 = e iδ 2 γ a iβ a + 3κ a 3 2 = e iδ 2 2 (3κ a 2 γ) 2 a 2 + β 2 a 2 = 1 4 This is a third order equation for a 2 = ρ 9κ 2 ρ 3 6κγρ 2 + (γ 2 + β 2 )ρ 1 4 = 0

8 Asymptotic solution When we have found a 2 numerically we find δ by: sin(δ) = 2β a With δ and a we can write the leading order solution as: z 0 (τ) = 2 a cos(τ + δ)

9 Numerical solution In order to solve the equation with a numerical ODE solver we rewrite it as two coupled equations. We do this by introducing x = dz dτ : z = x x = ɛβx (1 + ɛγ)z + ɛβz 3 + ɛ cos(τ)

10 Numerical solution, case 1 Here we have γ = 0, β = 0 and κ = 1. This gives the asymptotic solution: z 0 = 2 ( ) cos(τ)

11 Numerical solution, case 1 3 with γ=0, β =0 and κ=1 2 1 z Start as z 0 Start at rest z τ

12 Numerical solution, case 1 Comments: - No damping, and the system will therefore not stabilize with a fixed amplitude. - Not a periodic solution, but almost

13 Phase plots case 1 γ = 0, β = 0 and κ = 1

14 Phase plots case 1 γ = 0, β = 0 and κ = 1

15 Numerical solution, case 2 Here we have γ = 1, β = 0 and κ = 0. This gives the asymptotic solution: z 0 = cos(τ) We also have an exact solution: z = 1 γ cos(τ) + A cos( (1 + γɛ)τ) where A = 1 γ = 1 if we start at rest, and A = 1 γ + 2 a = 0 if we start like z 0

16 Numerical solution, case with γ=1, β =0 and κ=0 Start as z 0 Start at rest z z τ

17 Numerical solution, case 2 Comments: - Solution that starts as z 0 stays as z 0, the forces balance so that the amplitude is conserved - The solution that starts at rest consist of two solution with slightly different angular frequency, hence convolution. Max amplitude is two.

18 Phase plots case 2 γ = 1, β = 0 and κ = 0

19 Numerical solution, case 3 Here we have γ = 0, β = 1 and κ = 1. This gives the asymptotic solution: z 0 = cos(τ )

20 Numerical solution, case with γ=0, β =1 and κ=1 Start as z 0 Start at rest z z τ

21 Numerical solution, case 3 Comments: - Because of the damping term the solution for all initial conditions stabilizes at a fixed amplitude. - The angular frequency is slightly different from the asymptotic solution.

22 Phase plots case 3 γ = 0, β = 1 and κ = 1

23 Phase plots case 3 γ = 0, β = 1 and κ = 1

24 Numerical solution, case 4 Here we have γ = 0, β = 0.1 and κ = 1. This gives the asymptotic solution: z 0 = cos(τ )

25 Numerical solution, case with γ=0, β =0.1 and κ=1 Start as z 0 Start at rest z z τ

26 Numerical solution, case 4 Comments: - Because of the damping term the solution for all initial conditions stabilizes at a fixed amplitude, but it takes a long time. - Compared to case3 it takes a lot longer for the amplitude to stabilize at a fixed value

27 Phase plots case 4 γ = 0, β = 0.1 and κ = 1

28 Maximum excursion For the case γ = 0 and κ = β = 1 we want to find the value for ɛ that gives an maximum excursion close to 10 and 30. θ 1,max = 10 z 1,max = π 18 ɛ θ 2,max = 30 z 2,max = π 6 ɛ Iterating over ɛ (0, 1] while testing for z numeric,max z i,max < χ where χ 1 and i = 1, 2 gives this values for ɛ θ 1,max ɛ = 0.04 θ 2,max ɛ = 0.33

29 Plots of Maximum excursin γ = 0, β = 1 and κ = ǫ = z ForwardEuler RungeKutta2 RungeKutta4 Asymptotisk τ

30 Plots of Maximum excursin γ = 0, β = 1 and κ = ǫ = z ForwardEuler RungeKutta2 RungeKutta4 Asymptotisk τ

31 Case 1 Maximum excursin γ = 0, β = 0 and κ = 1 2 Case 1 with epsilon = z 4 6 ForwardEuler RungeKutta2 RungeKutta4 Asymptotisk τ

32 Case 1 Maximum excursin Case 1 with ɛ = 0.33 θ 2,max = 30 Comments: Here we can see that the damping term is the key factor that makes the solution stable.