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Elliptic Integrals Section 4.4 & Appendix B • Brief math interlude: – Solutions to certain types of nonlinear oscillator problems, while not expressible in closed form in terms of “elementary” functions (trig functions, etc.), they are expressible in terms of Elliptic Integrals – There is nothing mysterious about these! They are just special functions which have been studied completely & thoroughly by mathematicians 150 or more years ago. – All properties are known (derivatives, Taylor’s series, integrals, etc.) & tabulated for common values of the arguments, ….. • Here, because of the application to the Plane Pendulum problem, we are interested in the Elliptic Integral of the 1st Kind: F(k,φ). • From Appendix B: F(k,φ) ∫dθ[1- k2 sin2θ]-½ (limits: 0 < θ < φ), (k2 < 1) Or, with z = sinθ, F(k,x) = ∫dz [(1- z2)(1- k2z2)]-½ (limits: 0 < z < x), (k2 < 1) Plane Pendulum Section 4.4 • Pendulum: A mass m, constrained by a massless, extensionless rod to move in a vertical circle of radius . • The gravitational force acts downward, but the component of this force influencing the motion is to the support rod. • This is a nonlinear oscillator system with a symmetric restoring force. – Only for very small angular displacements is this is linear oscillator! Figure of Plane Pendulum Motion • Component of the gravitational force involved in the motion = F(θ) = - mg sinθ • Equation of motion (rotational version of Newton’s 2nd Law): Torque about the axis = (moment of inertia) (angular acceleration) N = I(d2θ/dt2) = F(θ); I = m2 m2 (d2θ/dt2) = - mg sinθ • Or (d2θ/dt2) + (g/) sinθ = 0 Define: ω02 (g/) So: Or: (d2θ/dt2) + ω02 sinθ = 0 θ + ω02 sinθ = 0 (“natural frequency”) • If & only if the angular displacement θ is small, then sinθ θ & the equation of motion becomes: θ + ω02θ 0 • This is simple harmonic motion for the angular displacement θ. Frequency ω0 (g/)½ Period τ 2π (/g)½ • General equation of motion for the plane pendulum: θ + ω02 sinθ = 0 (1) A VERY nonlinear differential eqtn! A VERY nonlinear oscillator eqtn! • We could try to solve (1) directly. However, instead, follow the text & use energy methods! • The restoring force for the motion: F(θ) = - mg sinθ (upper figure). This is a conservative system A potential energy function U(θ) exists (lower figure). Taking the zero of energy at bottom of the path at θ = 0 & using: F(θ) = - (dU/dθ) U(θ) = mg (1- cosθ) U(θ) = mg (1- cosθ) • Kinetic energy: T = (½)I(dθ/dt)2 = (½)m2(θ)2 • Total energy E = T + U is conserved • Let the highest point of the motion (determined by initial conditions!) be θ θ0. θ0 is the amplitude of the oscillatory motion. – By definition, T(θ0) 0. Also, U(θ0) = E = mg(1- cosθ0) = 2mgsin2[(½)θ0] (trig identity) – Similarly a trig identity gives: U(θ) = 2mgsin2[(½)θ] • Conservation of total energy E = 2mgsin2[(½)θ0) = T + U = (½)m2(θ)2 + 2mgsin2[(½)θ] • So 2mgsin2[(½)θ0] = 2mgsin2[(½)θ]+ (½)m2(θ)2 m cancels out! (2) • Solving (2) for θ = θ(θ) (gives the phase diagram of the pendulum!) & using the frequency for small angles: ω02 (g/) (dθ/dt) = 2ω0{sin2[(½)θ0] - sin2[(½)θ]}½ (3) • We could integrate (3) & get t(θ) rather than θ(t) using the period for small angles: τ0 = (2π/ω0) 2π(/g)½ dt = [τ0/(4π)]{sin2[(½)θ0] - sin2[(½)θ]}-½ dθ (4) • Instead of t(θ), use (4) to get the period τ. Using the fact that the motion is symmetric & also the definition of the period: τ = (τ0/π)∫{sin2[(½)θ0] - sin2[(½)θ]}-½dθ (limits 0 θ θ0) (5) τ = (τ0/π)∫{sin2[(½)θ0] - sin2[(½)θ]}-½dθ (5) (limits 0 θ θ0) • (5) is an Elliptic Integral of the 1st Kind: F(k,x = 1) τ (τ0/π) F(k,1) F(k,1) = ∫dz [(1- z2)(1- k2z2)]-½, (limits: 0 < z < 1), (k2 < 1) where k sin[(½)θ0], z {sin[(½)θ]/sin[(½)θ0]} This is tabulated in various places. • For oscillatory motion, we must have |θ0| < π or -1< k < 1; {k sin[(½)θ0]}; (k2 < 1) • Why? What happens if |θ0| = π? τ (τ0/π) F(k,1) F(k,1) = ∫dz [(1- z2)(1- k2z2)]-½, (limits: 0 < z < 1), (k2 < 1) where k sin[(½)θ0], z {sin[(½)θ]/sin[(½)θ0]} • Consider small displacements from equilibrium (but not necessarily so small that sinθ = θ!) (small kz < 1): – Expand the (1-k2z2)-½ part of the integrand in a Taylor’s series, & integrate term by term: (1-k2z2)-½ 1 + (½)k2z2 + (3/8)k4z4+ ... τ (τ0/π)∫dz(1-z2)-½[1+ (½)k2z2 + (3/8)k4z4 ..] Using tables (& skipping steps) gives: τ τ0[1 + (¼)k2 + (9/64)k4 + ..] • The period is (approximately): τ τ0[1 + (¼)k2 + (9/64)k4 + ..] (6) • We had: k sin[(½)θ0], θ0= amplitude of the oscillations (max angular displacement). In terms of the amplitude, the period is: τ τ0{1 + (¼)sin2[(½)θ0]+(9/64)sin4[(½)θ0] +..} (7) If k is large, we need many terms for an accurate result. For small k, this rapidly converges. k = sin( θ0) is determined by the initial conditions!!! • PHYSICS: Unlike the simple pendulum (where sinθ θ), the period for a real pendulum depends STRONGLY on the amplitude! – For the simple pendulum, the period τ0 = 2π(/g)½ is “isochronous” (independent of amplitude) • The period is (approximately; τ0 = 2π(/g)½) τ τ0{1+ (¼)sin2[(½)θ0] + (9/64)sin4[(½)θ0] +..} (8) • For small k = sin[(½)θ0] we can also make the small θ0 approximation & expand sin[(½)θ0] for small θ0: sin[(½)θ0] (½)θ0 - (1/48)(θ0)3 Put this into (8) & keep terms through 4th order in θ0 τ τ0[ 1 + (1/16)(θ0)2 + (11/3072)(θ0)4 + .. ] Finally the period as a function of amplitude θ0 for small θ0: τ τ0[ 1 + (0.0625)(θ0)2 + (0.00358)(θ0)4 + .. ] Phase Diagram for the Plane Pendulum • From conservation of energy, we had: E = 2mgsin2[(½)θ0] = T + U = (½)m2(θ)2 + 2mgsin2[(½)θ] So 2mgsin2[(½)θ0] = 2mgsin2[(½)θ] + (½)m2(θ)2 (2) • Solving (2) for θ = θ(θ) (gives the phase diagram of the pendulum!) – Using the frequency for small angles: ω02 (g/) (dθ/dt) = 2ω0{sin2[(½)θ0] - sin2[(½)θ]}½ (3) (dθ/dt) = 2ω0{sin2[(½)θ0] - sin2[(½)θ]}½ ; ω02 (g/) defining E0 2mg Phase Diagram for the Plane Pendulum • Qualitative Discussion Energy Eqtn: 2mgsin2[(½)θ0] = 2mgsin2[(½)θ] + (½)m2(θ)2 (2) ω02 (g/) (dθ/dt) = 2ω0{sin2[(½)θ0] - sin2[(½)θ]}½ (3) • For θ & θ0 small, eqtn (2) becomes: (θ/ω0)2 + θ2 (θ0)2 Using coordinates (θ/ω0) = (/g)½θ & θ, phase paths are ellipses (this is a SHO!). • For general –π < θ < π, we have E < 2mg E0. m is bound in a well: U(θ) = mg(1 - cosθ) Phase paths are closed curves given by (3) • U(θ) is periodic in θ, so we only need to plot – π < θ < π. From (dU/dθ) = 0 & looking at (d2U/dθ2), points θ = 2nπ, 0 are positions of stable equilibrium. Also, when damping exists (as in a real pendulum) these points become attractors (for long times, the phase paths will spiral towards these points). • If E > 2mg E0, the motion no longer oscillatory, but it is still periodic! This corresponds ------------------------- E to the pendulum making complete (circular) revolutions about the support axis. We still have U(θ) = mg(1 - cosθ) but the particle has enough energy to move from one periodic valley to the next (“over the hill”; see figure). • We still have conservation of energy E = T + U = (½)m2(θ)2 + 2mgsin2[(½)θ] But, now, θ0 is not defined! Instead, E is just some constant determined by initial conditions. The phase paths are open curves, still given by: (dθ/dt) = 2ω0{sin2[(½)θ0) - sin2[(½)θ)]}½ (3) • If E = 2mg E0 θ0 = π (mass initially vertical!) The phase path eqtn (dθ/dt) = 2ω0{sin2[(½)θ0] - sin2[(½)θ]}½ Becomes: (dθ/dt) = 2ω0cos[(½)θ] The phase paths in this case are 2 simple cosine functions (the heavy curves in the figure) • The phase paths with E = E0 don’t represent actual continuous motions of the pendulum! These are paths of unstable equilibrium. If the pendulum were at rest with θ0 = π, any small disturbance would cause it to move on some path E = E0 + δ (δ very small). • If the motion were on the phase path E = E0, the pendulum would reach θ = nπ with 0 velocity {(dθ/dt) = 2ω0cos[(½)nπ] = 0} but only after an infinite time! Proof: We had the period (limits 0 θ θ0) τ = (τ0/π)∫{sin2[(½)θ0] - sin2[(½)θ]}-½dθ Set θ0 = π & get τ • A phase path separating locally bounded motion from locally unbounded motion (like E = E0 for the pendulum) is called a “SEPARATRIX” – A separatrix always passes through a point of unstable equilibrium. Motion in the vicinity of a separatrix is extremely sensitive to the initial conditions. Points on either side of the separatrix have very different trajectories (like pendulum case just described!)