2. 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, …..

3. Here, because of the application to the Plane Pendulum problem, we are interested in the Elliptic Integral of the 1 st Kind: F(k,φ). From Appendix B: F(k,φ)  ∫ dθ[1- k 2 sin 2 θ] -½ (limits: 0 < θ < φ), (k 2 < 1) Or, with z = sinθ, F(k,x) = ∫ dz [(1- z 2 )(1- k 2 z 2 )] -½ (limits: 0 < z < x), (k 2 < 1)

4. 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!

5. Component of the gravitational force involved in the motion = F(θ) = - mg sinθ Equation of motion (rotational version of Newton’s 2 nd Law): Torque about the axis = (moment of inertia)  (angular acceleration) N = I(d 2 θ/dt 2 ) = F(θ); I = m 2  m 2 (d 2 θ/dt 2 ) = - mg sinθ Figure of Plane Pendulum Motion

6. Or (d 2 θ/dt 2 ) + (g/ ) sinθ = 0 Define: ω 0 2  (g/ ) (“natural frequency”) So: (d 2 θ/dt 2 ) + ω 0 2 sinθ = 0 Or: θ + ω 0 2 sinθ = 0 If & only if the angular displacement θ is small, then sinθ  θ & the equation of motion becomes: θ + ω 0 2 θ  0 This is simple harmonic motion for the angular displacement θ. Frequency ω 0  (g/ ) ½ Period τ  2π ( /g) ½

7. General equation of motion for the plane pendulum: θ + ω 0 2 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)

8. U(θ) = mg (1- cosθ) Kinetic energy: T = (½)I(dθ/dt) 2 = (½)m 2 (θ) 2 Total energy E = T + U is conserved Let the highest point of the motion (determined by initial conditions!) be θ  θ 0. θ 0 is the amplitude of the oscillatory motion. –By definition, T(θ 0 )  0. Also, U(θ 0 ) = E = mg (1- cosθ 0 ) = 2mg sin 2 [( ½) θ 0 ] (trig identity) –Similarly a trig identity gives: U(θ) = 2mg sin 2 [(½)θ] Conservation of total energy  E = 2mg sin 2 [(½)θ 0 ) = T + U = (½)m 2 (θ) 2 + 2mg sin 2 [(½)θ] So 2mg sin 2 [(½)θ 0 ] = 2mg sin 2 [(½)θ]+ (½)m 2 (θ) 2 (2) m cancels out!

9. Solving (2) for θ = θ(θ) (gives the phase diagram of the pendulum!) & using the frequency for small angles: ω 0 2  (g/ ) (dθ/dt) =  2ω 0 {sin 2 [(½)θ 0 ] - sin 2 [(½)θ]} ½ (3) We could integrate (3) & get t(θ) rather than θ(t) using the period for small angles: τ 0 = (2π/ω 0 )  2π( /g) ½ dt = [τ 0 /(4π)]{sin 2 [(½)θ 0 ] - sin 2 [(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 = (τ 0 /π) ∫ {sin 2 [(½)θ 0 ] - sin 2 [(½)θ]} -½ dθ (5) (limits 0  θ  θ 0 )

10. τ = (τ 0 /π) ∫ {sin 2 [(½)θ 0 ] - sin 2 [(½)θ]} -½ dθ (5) (limits 0  θ  θ 0 ) (5) is an Elliptic Integral of the 1 st Kind: F(k,x = 1) τ  (τ 0 /π) F(k,1) F(k,1) = ∫ dz [(1- z 2 )(1- k 2 z 2 )] -½, (limits: 0 < z < 1), (k 2 < 1) where k  sin[(½)θ 0 ], z  {sin[(½)θ]/sin[(½)θ 0 ]} This is tabulated in various places. For oscillatory motion, we must have |θ 0 | < π or -1< k < 1; {k  sin[(½)θ 0 ]}; (k 2 < 1) Why? What happens if |θ 0 | = π?

11. τ  (τ 0 /π) F(k,1) F(k,1) = ∫ dz [(1- z 2 )(1- k 2 z 2 )] -½, (limits: 0 < z < 1), (k 2 < 1) where k  sin[(½)θ 0 ], z  {sin[(½)θ]/sin[(½)θ 0 ]} Consider small displacements from equilibrium (but not necessarily so small that sinθ = θ!) (small kz < 1): –Expand the (1-k 2 z 2 ) -½ part of the integrand in a Taylor’s series, & integrate term by term: (1-k 2 z 2 ) -½  1 + (½)k 2 z 2 + ( 3/8 )k 4 z 4 +...  τ  (τ 0 /π) ∫ dz(1-z 2 ) -½ [1+ (½)k 2 z 2 + ( 3/8 )k 4 z 4..] Using tables (& skipping steps) gives: τ  τ 0 [1 + (¼)k 2 + ( 9/64 )k 4 +..]

12. The period is (approximately): τ  τ 0 [1 + (¼)k 2 + ( 9/64 )k 4 +..] (6) We had: k  sin[(½)θ 0 ], θ 0 = amplitude of the oscillations (max angular displacement). In terms of the amplitude, the period is:  τ  τ 0 {1 + (¼)sin 2 [(½)θ 0 ]+( 9/64 )sin 4 [(½)θ 0 ] +..} (7) If k is large, we need many terms for an accurate result. For small k, this rapidly converges. k = sin(θ 0 ) 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 τ 0 = 2π( /g) ½ is “isochronous” (independent of amplitude)

13. The period is (approximately; τ 0 = 2π( /g) ½ ) τ  τ 0 {1+ (¼)sin 2 [(½)θ 0 ] + ( 9/64 )sin 4 [(½)θ 0 ] +..} (8) For small k = sin[(½)θ 0 ] we can also make the small θ 0 approximation & expand sin[(½)θ 0 ] for small θ 0 : sin[(½)θ 0 ]  (½)θ 0 - ( 1/48 )(θ 0 ) 3 Put this into (8) & keep terms through 4 th order in θ 0 τ  τ 0 [ 1 + ( 1/16 )(θ 0 ) 2 + ( 11/3072 )(θ 0 ) 4 +.. ] Finally the period as a function of amplitude θ 0 for small θ 0 : τ  τ 0 [ 1 + (0.0625)(θ 0 ) 2 + (0.00358)(θ 0 ) 4 +.. ]

14. From conservation of energy, we had: E = 2mg sin 2 [(½)θ 0 ] = T + U = (½)m 2 (θ) 2 + 2mg sin 2 [(½)θ] So 2mg sin 2 [(½)θ 0 ] = 2mg sin 2 [(½)θ] + (½)m 2 (θ) 2 (2) Solving (2) for θ = θ(θ) (gives the phase diagram of the pendulum!) –Using the frequency for small angles: ω 0 2  (g/ ) (dθ/dt) =  2ω 0 {sin 2 [(½)θ 0 ] - sin 2 [(½)θ]} ½ (3) Phase Diagram for the Plane Pendulum

15. (dθ/dt) =  2ω 0 {sin 2 [(½)θ 0 ] - sin 2 [(½)θ]} ½ ; ω 0 2  (g/ ) defining E 0  2mg

16. Energy Eqtn: 2mg sin 2 [(½)θ 0 ] = 2mg sin 2 [(½)θ] + (½)m 2 (θ) 2 (2) ω 0 2  (g/ ) (dθ/dt) =  2ω 0 {sin 2 [(½)θ 0 ] - sin 2 [(½)θ]} ½ (3) For θ & θ 0 small, eqtn (2) becomes: (θ/ω 0 ) 2 + θ 2  (θ 0 ) 2 Using coordinates (θ/ω 0 ) = ( /g) ½ θ & θ, phase paths are ellipses (this is a SHO!). For general –π < θ < π, we have E < 2mg  E 0. 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 (d 2 U/dθ 2 ), 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). Phase Diagram for the Plane Pendulum Qualitative Discussion

17. If E > 2mg  E 0, the motion no longer oscillatory, but it is still periodic! This corresponds 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)m 2 (θ) 2 + 2mg sin 2 [(½)θ] But, now, θ 0 is not defined! Instead, E is just some constant determined by initial conditions. The phase paths are open curves, still given by: (dθ/dt) =  2ω 0 {sin 2 [(½)θ 0 ) - sin 2 [(½)θ)]} ½ (3) ------------------------- E

18. If E = 2mg  E 0  θ 0 =  π (mass initially vertical!) The phase path eqtn (dθ/dt) =  2ω 0 {sin 2 [(½)θ 0 ] - sin 2 [(½)θ]} ½ Becomes: (dθ/dt) =  2ω 0 cos[(½)θ] The phase paths in this case are 2 simple cosine functions (the heavy curves in the figure) The phase paths with E = E 0 don’t represent actual continuous motions of the pendulum! These are paths of unstable equilibrium.  If the pendulum were at rest with θ 0 =  π, any small disturbance would cause it to move on some path E = E 0 + δ (δ very small).

19. If the motion were on the phase path E = E 0, the pendulum would reach θ = nπ with 0 velocity {(dθ/dt) =  2ω 0 cos[(½)nπ] = 0} but only after an infinite time! Proof: We had the period (limits 0  θ  θ 0 ) τ = (τ 0 /π) ∫ {sin 2 [(½)θ 0 ] - sin 2 [(½)θ]} -½ dθ Set θ 0 = π & get τ   A phase path separating locally bounded motion from locally unbounded motion (like E = E 0 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!)