Famous Mathematica Probems and Their Stories Simpe Penduum (Lecture 3) Department of Appied Mathematics Nationa Chiao Tung University Hsin-Chu 30010, TAIWAN 23rd September 2009
History penduus: (hanging, Latin) (1) Gaieo Gaiei (1602) first studied the properties of penduum. (2) Christiaan Huygens (1656, Dutch) buit the first penduum cock.
Equation of motion F = mg sin θ: force, : ength of penduum x = θ: arc ength, v = ẋ = θ: veocity a = v = ẍ = θ: acceeration Newton s 2nd aw m θ = mg sin θ θ + g sin θ = 0 noninear D.E. The equation is independent of mass m.
Euer-Lagrange Equation h = (1 cos θ) : T = 1 2 m(ẋ)2 = 1 2 ( θ) 2 V = mgh = mg(1 cos θ) height (kinetic energy) (potentia energy) L = T V = 1 2 ( θ) 2 mg(1 cos θ) (agrangian) Euer-Lagrange equation ( ) d L L dt θ θ = d dt (m2 θ) 2 = mg sin θ = 0 θ + g sin θ = 0
Dimensiona anaysis Θ = 2π g period of simpe penduum period of penduum Θ [Θ] = T ength of penduum [] = L mass of penduum m [m] = M acceeration of gravity g [g] = LT 2 anguar ampitude θ [θ] = 1 (dimensioness) 5 parameters, 3 independent units, so we expect 5 3 = 2 dimensioness parameters. It is easy to show that [Θ] = [] 1/2 [g] 1/2. Thus we have two dimensioness parameters Π 1 = θ, Π 2 = Θ 1/2 g 1/2
Dimensiona anaysis(continue) Buckingham Pi Theorem Π 2 = p(π 1 ) = Θ = p(θ) g p(θ) = 2π, Θ = 2π g C. Huygens (1629-1695) d 2 θ dt 2 + g sin θ = 0 or d 2 θ dt 2 + g θ = 0, θ 1. [θ] T 2 = g [sin θ] = g 1 [θ] = T 2 = g = Θ g
Soving penduum: approximation θ + g sin θ = 0 noninear (difficut) sin θ θ, θ 1 θ + g sin θ = 0 θ + g θ = 0 inear ẍ + g x = 0 (x = θ) Try: x = A cos(ωt + δ) (harmonic wave,natura in physics) A: ampitude, ω: frequency ([ω] = 1 T ), δ: phase shift ω 2 = g ([ω] 2 = [g]/[] = T 2 Two parameters {A, δ} (2nd-order D.E.) ẍ + ω 2 x = 0 = x = c 1 cos ωt + c 2 sin ωt
Soving penduum: reduction of order v = ẋ = dx dt, dv dt = dv dx dv v = dt = d 2 x dt 2 = g x dx dt = v dv dt d 2 x dt 2 + g dv x = 0 = v dx + g x = 0 separation of variabe (tota differentia) vdv + g dx = d ( 1 2 v 2 + 1 2 ) g = 0 (1st-order D.E.) 1 2 v 2 + 1 g 2 = C 1 g 2 A2 Hamiton-Jacob equation
Soving penduum: reduction of order (continue) v = dx dt = Integration technique sin 1 x A = g A 2 x 2, A x A dx A 2 x 2 = g dt = g t + C ( ) g x = A sin t + C d 2 x dt 2 + g dx x = 0 dt = v, dv dt = g x
Conservation of energy Theorem d 2 x dt 2 + g x = d 2 x dt 2 + δ δx T (t) = 1 2 ( dx dt ( ) 1 g 2 x 2 = 0 ) 2 (kinetic energy) V (t) = 1 g 2 x 2 (potentia energy) E(t) = T (t) + V (t) = 1 2 ( dx dt ) 2 + 1 g 2 x 2 = constant t-transation invariant = energy conservation aw
Euer-Lagrange equation and Hamitonian system L = 1 2 1 2ẋ 2 g x 2 (Lagrangian) Euer-Lagrange equation ( ) d L L dθ ẋ x = d ( dt (ẋ) g ) x = ẍ + g x = 0 Hamitonian system (p=momentum, q=position) H = 1 2 + 1 g 2ẋ 2 x 2 = 1 2 p2 + 1 g 2 q2 dq dt = H p = p, {p, q}: conjugate variabes dp dt = H q = g q [pq] = [ ] = [action]
Back to origina equation Theorem E(θ, θ) = 1 2 m( θ) 2 + mg(1 cos θ) = C = constant θ = 0, θ = θ 0 = C = mg(1 cos θ 0 ) Hamiton-Jacobi equation θ 2 = 2g (cos θ cos θ 0) = 4g (sin2 θ 0 2 sin2 θ 2 ) Soving 1st-order D.E. t = 1 2 θ g 0 dθ k 2 sin 2 θ 2, k = sin θ 0 2
Period of penduum: Eiptic function T = 1 2 θ0 g 0 dθ k 2 sin 2 θ 2, (period, eiptic function) Asymptotic expansion ( T = 2π 1 + 1 θ g 2 2 sin2 0 2 + 32 θ ) 2 2 4 2 sin4 0 2 +
Eiptic function vis trigonometric function dx 1 + x 2 tan 1 x dx sin 1 x, cos 1 x 1 x 2 dx eiptic function 1 x 4 Differentia equation point of view dx ax = 0 dt x = eat d 2 x dt 2 a2 x = 0 x = sinh ax, cosh ax d 2 x dt 2 + a2 x = 0 x = sin ax, cos ax
Eiptic function: Hamiton-Jacobi equation 2nd-order D. E. 1st-order D. E. (conservation aw) ( ) dx 2 = dt N a n x n, n=1 Hamiton-Jacobi equation (1) N 2: x= eementary function (2) N = 3, 4: x is not eementary function! (Legendre)
Arc-ength of an eipse: eiptic integra s = ( ) x 2 + a dx 2 + dy 2 = η η = a = a 0 η 0 ( ) y 2 = 1, a, b > 0 b 1 a2 b 2 a 2 sin 2 φdφ 1 k 2 sin 2 φdφ 0 a 2 cos 2 φ + b 2 sin 2 φdφ a k = 2 + b 2 (eccentricity) a η E(k, η) 1 k 2 sin 2 φdφ, (eiptic integra) 0