Solutions_HW11

Solutions_HW11 - December 09, 2010 Theoretical Dynamics...

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Unformatted text preview: December 09, 2010 Theoretical Dynamics Homework 11 Instructor: Dr. Thomas Cohen Submitted by: Vivek Saxena Problem 1 ˙ ˙ ˙ Let φ, θ and ψ denote the Euler angles, and ω φ = φ, ω θ = θ and ω ψ = ψ denote their angular velocities. First, we express the components of these angular velocities along the body axes (x , y , z ), i.e. we find (ωx , ωy , ωz ). The full transformation can be depicted as ˙ ωφ = φ ωx T1 ˙ ω −→ ω =θ θ y ˙ ωz ωψ = ψ body system (1) body system The transformation, T1 , is a transformation from “body spherical coordinates” to “body Cartesian coordinates”. Referring to diagram (c) below, Figure defining the Euler angles. Figure courtesy Goldstein, Poole and Safko, Classical Mechanics, 3rd Edn. 11 - 1 and using the fact that angular velocities for infinitesimal rotations are directed along the normal to the plane of rotation (consistent with the right hand screw rule), we can write ˙ (2) (ω φ )x = φ sin θ sin ψ ˙ (3) (ω φ )y = φ sin θ cos ψ ˙ (ω φ )z = φ cos θ ˙ (ω θ )x = θ cos ψ (4) (5) ˙ (ω θ )y = −θ sin θ (6) (ω θ )z = 0 (7) (ω ψ )x = 0 (8) (ω ψ )y = 0 ˙ (ω ψ )z = ψ (9) (10) Adding the corresponding components, we get body ˙ ˙ ωx = φ sin θ sin ψ + θ cos ψ ˙ ˙ ω body = φ sin θ cos ψ − θ sin ψ y body ωz ˙ ˙ = φ cos θ + ψ (11) (12) (13) Problem 2 The principal moments of inertia are I1 = I2 = I0 and I3 = I0 /2. (a) ˙ ˙ In this case, θ = 0 and φ = 2ω0 . The angular momentum along the body-fixed z-axis is given to be constant and equal to I0 ω0 . This implies I0 I3 ω3 = ω3 = I0 ω0 =⇒ ω3 = 2ω0 (14) 2 So, from Eqn. (13) above, ˙ ˙ φ cos θ + ψ = 2ω0 (15) ˙ =⇒ 2ω0 cos θ + ψ = 2ω0 (16) The condition for regular precession is ˙ ˙ mgl = φ(I3 ω3 − I1 φ cos θ) =⇒ mgl = 2ω0 (17) I0 (2ω0 ) − I0 (2ω0 ) cos θ 2 2 =⇒ mgl = 2I0 ω0 (1 − 2 cos θ) =⇒ 1 = 2(1 − 2 cos θ) 1 =⇒ cos θ = 4 So, θ = cos−1 1 . 4 11 - 2 (18) (b) The equation of motion is 2 ˙ ˙ ¨ 1˙ θ + (ψ + φ cos θ)φ sin θ − ω0 sin θ = 0 2 (19) Assuming that ∆θ << θ, we can write sin(θ + ∆θ) = sin θ cos ∆θ + cos θ sin ∆θ ≈ sin θ + cos θ∆θ √ 15 ∆θ = + 4 4 cos(θ + ∆θ) = cos θ cos ∆θ − sin θ sin ∆θ ≈ cos θ − sin θ∆θ √ 15 1 ∆θ ≈− 4 4 1 Linearizing the equation of motion about θ0 = cos−1 4 , we get √ 2 ¨ + ω0 ∆θ = −ω 2 15 ∆θ 0 4 4 from which it follows that the nutation frequency is (20) (21) (22) ω0 2. Problem 3 The equation of motion is 2 x = f ( x) = − λ2 ω0 x ¨ (23) x = x0 + λx1 + λ2 x2 + λ3 x3 + λ4 x4 + λ5 x5 + . . . (24) Let Substituting this into Eqn. (1) we get 2 2 2 2 x 0 + λ x 1 + λ 2 x 2 + λ3 x 3 + λ 4 x 4 + λ 5 x 5 = − λ2 ω 0 x 0 − λ 3 ω 0 x 1 − λ4 ω 0 x 2 − λ 5 ω 0 x 3 ¨ ¨ ¨ ¨ ¨ ¨ (25) to order λ5 . Comparing coefficients of various powers of λ on both sides, we get x0 = 0 ¨ (26) x1 = 0 ¨ (27) x2 = ¨ x3 = ¨ x4 = ¨ x5 = ¨ 2 − ω0 x 0 2 − ω0 x 1 2 − ω0 x 2 2 − ω0 x 3 11 - 3 (28) (29) (30) (31) Using the boundary conditions given in the problem, the solutions to each of these second order differential equations can be determined successively. The results are: x0 (t) = x(0) (32) x1 (t) = x(0)t ˙ (33) x 2 ( t) = − 2 ω0 x(0)t2 2 2˙ ω0 x(0)t3 x 3 ( t) = − 6 4 ω0 x(0)t4 x 4 ( t) = 24 4˙ ω0 x(0)t5 x 5 ( t) = 120 (34) (35) (36) (37) (a) The solution to order t5 is: x(t) = x(0) + x(0)t − ˙ 2˙ 4 4˙ 2 ω0 x(0)t2 ω0 x(0)t3 ω0 x(0)t4 ω0 x(0)t5 − + + 2 6 24 120 (38) (b) The exact solution is: x(0) ˙ sin ω0 t ω0 ω 3 t3 ω 5 t5 x(0) ˙ ω 2 t2 ω 4 t4 ω 6 t6 ω0 t − 0 + 0 + . . . = x(0) 1 − 0 + 0 − 0 + . . . + 2 24 720 ω0 6 120 2 x(0)t2 2 x(0)t3 4 x(0)t4 4 x(0)t5 ω ω˙ ω ω˙ = x(0) + x(0)t − 0 ˙ −0 +0 +0 + O ( t6 ) 2 6 24 120 xex (t) = x(0) cos ω0 t + (39) (40) (41) So to order t5 , the exact solution agrees with the perturbative solution. Problem 4 The equations of motion are λ2 ∂V m ∂x λ2 ∂V y=− ¨ m ∂y x=− ¨ (42) (43) Define ∂V ∂x ∂V g (x, y ) = − ∂y f (x, y ) = − 11 - 4 (44) (45) Further, let x = x0 + λx1 + λ2 x2 + λ3 x3 + λ4 x4 + . . . (46) y = y0 + λy1 + λ2 y2 + λ3 y3 + λ4 y4 + . . . (47) The functions f (x, y ) and g (x, y ) can be Taylor expanded about the point (x0 , y0 ) (denoted by the subscript 0). For f (x, y ) the expansion is of the form f (x, y ) = f (x0 , y0 ) + ∂2f + ∂x∂y 0 ∂f ∂x + ∂x∂y 0 ∂f ∂y 0 ∂2f (x − x0 )(y − y0 ) + 2 ∂y∂x ∂f = f (x0 , y0 ) + ∂x ∂2f 0 ( x − x0 ) + ∂f ( x − x0 ) + ∂y 0 0 ∂2f ∂x2 ( y − y0 ) + 0 0 ∂2f ( x − x0 ) 2 + 2 ∂y 2 0 ( y − y0 ) 2 2 0 ( y − y0 ) 2 2 (x − x0 )(y − y0 ) + H.O.T. 2 ∂2f ∂x2 ( y − y0 ) + 0 ( x − x0 ) 2 ∂2f + 2 ∂y 2 (x − x0 )(y − y0 ) + H.O.T. (48) Plugging this into the differential equation for x, we get (retaining terms up to second order in the partial derivatives) x 0 + λ x 1 + λ 2 x 2 + λ3 x 3 + λ 4 x 4 = − ¨ ¨ ¨ ¨ ¨ + ∂f λ2 f (x0 , y0 ) + m ∂x ∂2f ∂x2 0 0 ( x − x0 ) 2 ∂2f + 2 ∂y 2 ( x − x0 ) + 0 ∂f ∂y 0 ( y − y0 ) ( y − y0 ) 2 ∂2f + 2 ∂x∂y 0 (x − x0 )(y − y0 ) (49) =− + λ2 ∂f f (x0 , y0 ) + m ∂x ∂f ∂y (λx1 + λ2 x2 + λ3 x3 + λ4 x4 ) 0 (λy1 + λ2 y2 + λ3 y3 + λ4 y4 ) + 0 1 ∂2f + 2 ∂y 2 ∂2f + ∂x∂y 1 ∂2f 2 ∂x2 (λx1 + λ2 x2 + λ3 x3 + λ4 x4 )2 0 (λy1 + λ2 y2 + λ3 y3 + λ4 y4 )2 0 (λx1 + λ2 x2 + λ3 x3 + λ4 x4 )(λy1 + λ2 y2 + λ3 y3 + λ4 y4 ) 0 (50) Equating the coefficients of various powers of λ on both sides, we get x0 = 0 ¨ (51) x1 = 0 ¨ (52) 1 x2 = − f (x0 , y0 ) ¨ m 1 ∂f ∂f x3 = − ¨ x1 + m ∂x 0 ∂y x4 = − ¨ 1 m ∂f ∂x x2 + 0 ∂f ∂y (53) (54) y1 0 y2 + 0 11 - 5 1 ∂2f 2 ∂x2 x2 + 1 0 1 ∂2f 2 ∂y 2 2 y1 0 (55) Following a similar procedure for the differential equation for y , we get y0 = 0 ¨ (56) y1 = 0 ¨ (57) 1 y2 = − g (x0 , y0 ) ¨ m 1 ∂g ∂g y3 = − ¨ x1 + m ∂x 0 ∂y y4 = − ¨ 1 m ∂g ∂x x2 + 0 ∂g ∂y (58) (59) y1 0 1 ∂2g 2 ∂x2 y2 + 0 x2 + 1 0 1 ∂2g 2 ∂y 2 2 y1 (60) 0 The solutions are x 0 ( t) = c 1 t + c 2 , y0 (t) = d1 t + d2 (61) x 1 ( t) = c 3 t + c 4 , y1 (t) = d3 t + d4 1 x 2 ( t) = − f (x0 , y0 )t2 + c5 t + c6 2m 1 y 2 ( t) = − g (x0 , y0 )t2 + d5 t + d6 2m 1 ∂f 1 ∂f ∂f x 3 ( t) = − + d4 + d3 t2 − c4 c3 2m ∂x 0 ∂y 0 6m ∂x 0 ∂g 1 ∂g ∂g 1 + d4 t2 − + d3 c4 c3 y 3 ( t) = − 2m ∂x 0 ∂y 0 6m ∂x 0 (62) x 4 ( t) = 1 ∂f 2m2 ∂x f (x0 , y0 ) + 0 1 ∂f 2m2 ∂y 0 (63) (64) ∂f ∂y ∂g ∂y t3 + c 7 t + c 8 t3 + d 7 t + d 8 1 ∂f +− m ∂x 1 ∂f c5 − m ∂y 0 1 ∂2f d5 − m ∂x2 0 1 ∂2f c3 c4 − m ∂y 2 0 1 ∂f +− m ∂x 1 ∂f c6 − m ∂x 0 1 ∂2f d6 − 2m ∂x2 0 1 ∂2f − 2m ∂y 2 0 1 ∂g 2m2 ∂x g (x0 , y0 ) + y 4 ( t) = 0 1 ∂g 2m2 ∂y 0 f (x0 , y0 ) − 1 ∂g +− m ∂x 1 ∂g c5 − m ∂y 0 1 ∂2g d5 − m ∂x2 0 1 ∂g +− m ∂x 1 ∂g c6 − m ∂x 0 1 ∂2g d6 − 2m ∂x2 0 0 c2 − 3 1 ∂2f 2m ∂y 2 ∂2f d3 d4 − ∂x∂y 0 1 ∂2f d4 − m ∂x∂y 0 1 ∂2g 2m ∂x2 1 ∂2g c3 c4 − m ∂y 2 0 1 ∂2g − 2m ∂y 2 0 (66) 0 1 ∂2f 2m ∂x2 g (x0 , y0 ) − (65) 0 0 c2 − 3 0 0 ∂2g d3 d4 − ∂x∂y 0 1 ∂2f m ∂x∂y ( c 3 d4 + c 4 d3 ) c 4 d4 1 ∂2g 2m ∂y 2 1 ∂2g d4 − m ∂x∂y 0 0 d2 − 3 0 d2 − 3 c 4 d4 0 0 t3 6 1 ∂2g m ∂x∂y c 3 d3 0 t3 6 t2 + d9 t + d10 2 (68) So, the general solution for x(t) is x(t) = x1 (t) + x2 (t) + x3 (t) + x4 (t), that is, x ( t) = α 0 + α 1 t + α 2 t2 + α 3 t 3 + α 4 t 4 11 - 6 t4 12 t2 + c9 t + c10 2 (67) ( c 3 d4 + c 4 d3 ) 0 c 3 d3 (69) t4 12 where α0 = c2 + c4 + c6 + c8 + c10 = x(0) (70) α1 = c1 + c3 + c5 + c7 + c9 = x(0) ˙ (71) α2 = − 1 1 ∂f f (x0 , y0 ) − c4 2m 2m ∂x 1 + 2 α3 = α4 = 1 ∂f − m ∂x 1 6 − 1 12 1 ∂f m ∂x 0 1 ∂f 2m2 ∂x 0 ∂f ∂y 0 1 ∂2f d6 − 2m ∂x2 0 1 ∂f c6 − m ∂x 0 c5 − + d4 1 ∂f m ∂y d5 − 0 f (x0 , y0 ) + 0 1 ∂2f m ∂x2 1 ∂f 2m2 ∂y 0 − 0 0 1 ∂2f 2m ∂y 2 c3 c4 − g (x0 , y0 ) − 0 d4 − 1 ∂2f m ∂y 2 0 1 ∂2f 2m ∂x2 1 ∂2f m ∂x∂y d3 d4 − 0 c2 − 3 (72) c 4 d4 0 ∂2f ∂x∂y ( c 3 d4 + c 4 d3 ) (73) 0 1 ∂2f 2m ∂y 2 0 d2 − 3 1 ∂2f m ∂x∂y c 3 d3 0 (74) Similarly, the general solution for y (t) is y ( t ) = β 0 + β 1 t + β 2 t 2 + β 3 t3 + β 4 t 4 (75) where β0 = d2 + d4 + d6 + d8 + d10 = y (0) (76) β1 = d1 + d3 + d5 + d7 + d9 = y (0) ˙ (77) β2 = − 1 1 ∂g g (x0 , y0 ) − c4 2m 2m ∂x 1 + 2 β3 = β4 = 1 6 1 12 1 ∂g − m ∂x − 1 ∂g m ∂x 1 ∂g c6 − m ∂x 0 0 1 ∂g 2m2 ∂x c5 − ∂g ∂y + d4 0 0 1 ∂2g d6 − 2m ∂x2 0 1 ∂g m ∂y 0 g (x0 , y0 ) + 0 d5 − 1 ∂2g m ∂x2 1 ∂g 2m2 ∂y 0 0 0 − 1 ∂2g 2m ∂y 2 c3 c4 − f (x0 , y0 ) − 0 d4 − 1 ∂2g m ∂y 2 0 1 ∂2g 2m ∂x2 1 ∂2g m ∂x∂y d3 d4 − 0 c2 − 3 (78) c 4 d4 0 ∂2g ∂x∂y ( c 3 d4 + c 4 d3 ) (79) 0 1 ∂2g 2m ∂y 2 0 d2 − 3 1 ∂2g m ∂x∂y c 3 d3 0 (80) From the differential equations, we have f (x0 , y0 ) f (x0 , y0 ) =⇒ α2 = − m 2m g (x0 , y0 ) g (x0 , y0 ) =⇒ β2 = − y (t = 0) = − ¨ m 2m x(t = 0) = − ¨ (81) (82) and d3 x dt3 d3 y dt3 t=0 =− 1 m ∂f ∂x t=0 =− 1 m ∂g ∂x x(0) + ˙ 0 x(0) + ˙ 0 11 - 7 ∂f ∂y 0 ∂g ∂y 0 y (0) = 6α3 ˙ (83) y (0) = 6β3 ˙ (84) d4 x dt4 t=0 =− 1∂ m ∂t ∂f ∂f x+ ˙ y ˙ ∂x ∂y + t=0 ∂ ∂x ∂f ∂f x+ ˙ y ˙ ∂x ∂y x(0) + ˙ t=0 ∂ ∂y ∂f ∂f x+ ˙ y ˙ ∂x ∂y y (0) ˙ t=0 = 24α4 d4 y dt4 t=0 =− (85) 1∂ m ∂t ∂g ∂g x+ ˙ y ˙ ∂x ∂y + t=0 ∂ ∂x ∂g ∂g x+ ˙ y ˙ ∂x ∂y x(0) + ˙ t=0 ∂ ∂y ∂g ∂g x+ ˙ y ˙ ∂x ∂y = 24β4 y (0) ˙ t=0 (86) In this manner, we can determine the αi ’s and βi ’s from the boundary conditions, and hence determine x(t) and y (t) to order t4 . Problem 5 The equation of motion is ¨ λV0 sin θ = 0 θ+ I (87) θ(t) = ω0 t + λθ1 + λ2 θ2 + . . . (88) Now, As we are interested only in a solution to O(λ2 ), sin(θ) ≈ sin(ω0 + λθ1 ) = sin(ω0 t) cos(λθ1 ) + sin(λθ1 ) cos(ω0 t) ≈ sin(ω0 t) + λθ1 cos(ω0 t) (89) Substituting ¨ ¨ λθ 1 + λ2 θ 2 + λV0 (sin(ω0 t) + λθ1 cos(ω0 t)) = 0 I (90) Equating coefficients of like powers of λ on both sides, we get V0 ¨ θ1 = − sin(ω0 t) I V0 ¨ θ2 = − θ1 (t) cos(ω0 t) I (91) (92) ˙ Using the given boundary conditions, θi (0) = 0 and θi (0) = 0, these equations can be solved analytically, to yield the solutions V0 2 [sin(ω0 t) − ω0 t] Iω0 5ω 0 t V 2 sin(2ω0 t) − ω0 t cos(ω0 t) + 2 sin(ω0 t) − θ 2 ( t) = 2 0 2 8 4 I ω0 θ 1 ( t) = 11 - 8 (93) (94) Problem 6 (a) The solution θ(t) for an anharmonic oscillator must satisfy the differential equation V0 ¨ θ = − sin θ I (95) The solution θ(t) must be periodic with some period T such that if θ(t = 0) = 0 then θ(t = T ) = 2π . Since the energy, 1˙ E = I θ2 − V0 cos θ 2 (96) ˙ ˙ is conserved, we must have θ(t = 0) = θ(t = 2π ). That is, the angular speed after one period will be identical to that at t = 0. If V0 is negligibly small, then the solution is θ0 = ω0 t where ω0 = 2π , with T the understanding that θ(t + T ) = θ(t) (that is, the angles are measured modulo 2π ). This ensures periodicity. If V0 is not negligibly small, then this solution still predicts the correct angular position at times t = nT where n is an integer, because the potential energy itself is a periodic function. Deviations from this solution will occur between two successive periods, i.e. nT < t < (n + 1)T , due to the variation of the potential energy term, and will vanish at t = nT . So, the deviations can be expanded in a Fourier sine series of the form n cn sin(nωt), where nωT = 2πm where n and m are integers, so that the argument of the sine term ensures that the deviations vanish at integral multiples of the period T . That is, the solution can be written as ∞ θ(t) = ωt + (97) cn sin(nωt) n=1 In general, ω = ω0 . This is because of the fact that in the anharmonic regime, we cannot approximate sin θ as θ and so the effect of the gravitational potential V0 is to slow the pendulum down so that it takes longer to complete one cycle than it would if the oscillation frequency were ω0 . Therefore, ω < ω0 . Note that the existence of higher Fourier modes emphasizes the anharmonicity of the solution. (b) We have ω = ω0 + λω1 + λ2 ω2 + . . . cn = λ n (c(0) n + λc(1) n + λ2 c(2) n (98) + . . .) (99) So, up to second order in λ, we can write (0) (1) (0) θ(t) = (ω0 + λω1 + λ2 ω2 )t + λ(c1 + λc1 ) sin[(ω0 + λω1 + λ2 ω2 )t] + λ2 c2 sin[2(ω0 + λω1 + λ2 ω2 )t] (100) 11 - 9 Since the sine terms are premultiplied by λ, they can be approximated as sin[(ω0 + λω1 + λ2 ω2 )t] ≈ sin[(ω0 + λω1 )t] = sin(ω0 t) cos(λω1 t) + cos(ω0 t) sin(λω1 t) ≈ sin(ω0 t) + λω1 t cos(ω0 t) 2 sin[2(ω0 + λω1 + λ ω2 )t] ≈ sin[2(ω0 + λω1 )t] ≈ sin(2ω0 t) (101) (102) as we are only interested in terms to order λ2 . Substituting these into the expression for θ(t) and rearranging, we get (0) (0) (1) (0) θ(t) = ω0 t + λ[ω1 t + c1 sin(ω0 t)] + λ2 [ω2 t + c1 ω1 t cos(ω0 t) + c1 sin(ω0 t) + c2 sin(2ω0 t)] (103) Comparing Eqn. (103) with Eqns. (88), (93) and (94) we get V0 V2 5 ω0 , ω2 = − 2 0 4 × ω 0 2 4 Iω0 I ω0 2 V V0 (1) c1 = 2 0 4 × 2 = 2, Iω0 I ω0 2 1 V = 20 4 × 8 I ω0 ω1 = − (104) (0) (105) c1 (0) c2 Let α = V0 2. Iω0 (106) Then, 5 ω = ω0 1 − α − α 2 4 (107) 1 θ(t) = ωt + (α + 2α2 ) sin(ωt) + α2 sin(2ωt) 8 (108) Eqn. (107) confirms tha ω < ω0 as reasoned in part (a). Problem 7 The energy is 1˙ E = I θ2 − V0 cos θ 2 (109) 12 Iω − V0 = 10V0 20 (110) ˙ At t = 0, θ(t) = ω0 and θ = 0, so implies that ω0 = 22V0 I (111) So, α= V0 1 2 = 22 Iω0 We assume that V0 = 1 and I = 1 in suitable units, so that ω0 = for problems 5 and 6 are: 11 - 10 (112) √ 22 rad/s and α = 1/22. The plots Figure for problem 5 Figure for problem 6 In the plot for problem 6, the large magnitudes of the numerical and series solution and zero-th order terms mask out the first and second order oscillatory terms, which are shown separately below: For detailed calculations, and plots of all the terms, please refer to the attached Mathematica notebook, problem-7.nb. From the plots, it follows that the solutions are very close for large values of t. 11 - 11 Clear "Global` " ; Problem 5 (Numerical Solution): The differential equation is Θ’’ + ifun First y . NDSolve y '' t 1 InterpolatingFunction Plot ifun t , Sin y t 0., 300. V0 I 0, y 0 sin Θ = 0. Let k = 0, y ' 0 V0 I . We solve for the case k = 1. Sqrt 22 1 , t, 0, 300 1200 1000 800 600 400 200 50 100 150 200 250 300 Series Solution obtained in Problem 5. V0 f1 Ω_ , t_ Sin Ω t Ωt ; Ω2 J V0 Sin 2 Ω t 5Ωt f2 Ω_ , t_ Ω t Cos Ω t J J Ω2 ; 2 Sin Ω t 8 4 V0 s Ω_ , t_ Ω t Sin Ω t Ωt J Ω2 Sin 2 Ω t V0 5Ωt Ω t Cos Ω t J J f1s t_ Ω2 8 f1 Sqrt 22 , t ; 2 Sin Ω t 4 . V0 1, J 1 . V0 1, J 1 1 22 t Sin 22 t 22 f2s t_ 1 5 f2 Sqrt 22 , t 11 1 t 22 2 2 22 t Cos 22 t 2 Sin 22 t Sin 2 8 22 t , y , t, 0, 300 2 p robl em -7.nb sol t_ s Sqrt 22 , t . V0 1, J 1 FullSimplify 1 2 22 t 79 4 Cos 22 t 24 Sin 22 t Sin 2 22 t 176 PlotLegends` Plot ifun t , sol t , Sqrt 22 t, f1s t , f2s t , t, 0, 20 , AxesLabel "t", "Θ" , PlotLegend "Numerical", "Series", "Zero Order : Ω0 t", "First Order : Θ1 t ", "Second Order : Θ2 t " , LegendShadow None, PlotStyle Red , Blue, Black, Green, Purple Θ Numerical Series 80 Zero Order : Ω 0 t First Order : Θ 1 t Second Order : Θ 2 t 60 40 20 t 5 10 15 Series Solution obtained in Problem 6. 5 Α2 Ω6 Ω0 1 ; Α 4 Ω6 . Α 1 22, Ω0 Sqrt 22 FullSimplify 1843 88 22 term1 t_ 6 Α 1843 t Sin 121 88 22 2 Α 2 Sin Ω6 t .Α 1 22, Ω0 Sqrt 22 20 p roblem -7.nb 1 Α 2 Sin 2 Ω6 term2 t_ t .Α 1 22, Ω0 Sqrt 22 8 1843 t Sin 44 22 3872 1843 t term1 t sol6 t_ 88 22 Sin 1843 t 6 1843 t 121 22 88 1843 t 44 Sin 88 term2 t 22 3872 22 1843 t, term1 t , term2 t , 88 22 t, 0, 20 , AxesLabel "t", "Θ" , PlotLegend "Numerical", "Series", "Zero Order : Ω0 t", "First Order : Θ1 t ", "Second Order : Θ2 t " , Plot ifun t , sol6 t , LegendShadow None, PlotStyle Red , Blue, Black, Green, Purple Θ Numerical 80 Series Zero Order : Ω 0 t First Order : Θ 1 t 60 Second Order : Θ 2 t 40 20 t 5 10 15 Plot sol6 t , t, 0, 20 , AxesLabel "t", "Θ" , PlotLegend "Series" , LegendShadow None, PlotStyle 20 Blue 3 4 p robl em -7.nb Θ 80 Series 60 40 20 t 5 10 15 20 1843 Plot t, t, 0, 30 , AxesLabel "t", "Θ" , 22 88 PlotLegend "Zero Order : Ωt" , LegendShadow None, PlotStyle Black Θ 120 Zero Order : Ωt 100 80 60 40 20 t 5 10 15 20 Plot term1 t , t, 0, 30 , AxesLabel "t", "Θ" , PlotLegend "First Order : Θ1 t " , LegendShadow 25 30 None, PlotStyle Green p roblem -7.nb Θ 0.04 0.02 t 5 10 15 20 25 30 0.02 0.04 First Order : Θ 1 t Plot term2 t , t, 0, 30 , AxesLabel "t", "Θ" , PlotLegend "Second Order : Θ2 t " , LegendShadow None, PlotStyle Θ 0.0002 0.0001 t 5 10 15 0.0001 0.0002 Second Order : Θ 2 t 20 25 30 Purple 5 ...
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This note was uploaded on 12/29/2011 for the course PHYSICS 601 taught by Professor Hassam during the Fall '11 term at Maryland.

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