VERTICALLY AND HORIZONTALLY DRIVEN PENDULUMS

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1 VERTICALLY AND HORIZONTALLY DRIVEN PENDULUMS RAVITEJ UPPU Abstract. In this I will be doing two cases majorly where the pendulum s support moves horizontally and vertically with a frequency ν. Date: 6 March, Chennai Mathematical Institute. 1

2 2 RAVITEJ UPPU 1. Pendulum with a vibtrating support Definition Of A Fixed Point : Here we see that the systems are also dependent on time. Thus our fixed ppoints here are those whose ψ and ψ are null at a time t when the particle reaches that point in phase space given a set of initial conditions Vertically Driven Pendulum. In all the cases our pendulum has a light stiff rod to which a mass m is connected. Let the distance of support from origin O be given by the function D(t). Then our x and y will be the distance functions of the pendulum x = lsinψ ẋ = (l ψcosψ) y = lcosψ D(t) ẏ = (l ψsinψ Ḋ(t)) Hence v 2 = (ẋ 2 + ẏ 2 ) v 2 = (l 2 ψ 2 cos 2 ψ + (l ψsinψ Ḋ(t))2 ) We have Potential Energy V = mg(lcosψ + D(t)) Kinetic Energy T = 1 2 m(l2 ψ 2 cos 2 ψ + (l ψsinψ Ḋ(t))2 ) T = 1 2 m(l2 ψ 2 ) + Ḋ2 2l ψsinψḋ) Let s write the Lagrangian L = T V L = 1 2 m((l2 ψ 2 ) + Ḋ2 2l ψsinψḋ) + mg(lcosψ + D(t)) But we know that any two Lagrangians which differ by a total differential of time give same equations of motion. Hence we can remove the terms Ḋ2 and mgd(t). Hence my Lagrangian becomes L = 1 2 m(l2 ψ 2 2l ψsinψḋ) + mglcosψ Let s do a small trick here. Let s replace ψḋsinψ by ( Dcosψ d dt (Ḋcosψ)). We get L = 1 2 m(l2 ψ 2 2l(+ Dcosψ d (Ḋcosψ)) + mglcosψ) dt

3 VERTICALLY AND HORIZONTALLY DRIVEN PENDULUMS 3 Therefore L = 1 2 ml2 ψ2 + ml(g D)cosψ Behold this is like a time-variant gravitational field. And we already know the first part of the equation which is simple pendulum (i.e when D = 0). Now let s try to get the equation of motion from Hence on calculating we have d dt ( L L ) = ψ ψ ψ = ψ l sinψ( D g) Here we have that ψ and ψ are the variables. Let s try to analyze this motion putting ψ = R and ψ = S. From this we have that Ṙ = S Ṡ = 1 l ( D g)sinr At the fixed points Ṙ = Ṡ = 0 from this we get that S = 0 and R = nπ Therefore we can compute the A-matrix to be ( ) 0 1 A = 1 l ( D g)cosr 0 This gives us At the point (π, 0) λ 2 = 1 l ( D g)cosr λ 2 = 1 (g D) l So unless g > D this not negative hence we get that at the point (π, 0) if the particle reaches at a time t π we get the point to be an elliptic point and hence stable Reference: This can be found as an introduction of the paper written by M.V.Bartuccelli,G.Gentile, K.V.Georgiou with topic On the dynamics of Vertically Driven Damped Planar Pendulum submitted in August 2000.A reference can be found at

4 4 RAVITEJ UPPU 1.2. Horizontally Driven Pendulum. Let s look at what happens when we have a horizontal driving force now.now our x and y will be x = lsinψ D(t) ẋ = l ψcosψ Ḋ y = lcosψ D(t) ẏ = l ψsinψ from tjis we get T to be T = m 2 (l2 + Ḋ2 2l ψḋcosψ) And we have Vas This gives us the Lagrangian as V = mglcosψ L = ml2 ψ ml( Dsinψ + gcosψ) Hence from this by using the Euler-Lagrangian equation of motion we have that ψ = ψ l ( Dcosψ gsinψ) which is just a term afar from the simple pendulum case. Now let s try to do the same analysis as above.for fixed points Ṙ = Ṡ = 0 hence we have S = 0 and tanr = D D g which says our R = tan 1 g. If we compare with simple pendulum, i.e it has an unstable point at (π, 0)which says that R = π hence we get that D(t π ) = 0. Now this is a fixed point of my system. Now if we try to write the A-matrix at time t π and at point (π, 0) we have it as ( A ( π, 0) = l ( D(t π )sinπ + gcosπ) 0 ) Therefore we get that λ 2 = g l. This is always negative hence my point has become stable under the taken conditions.

5 VERTICALLY AND HORIZONTALLY DRIVEN PENDULUMS 5 2. The support moving on a circle in x-y plane Let t take the distance function on x and y to be D x (t) and D y (t) where for a circle D x (t) = Asinωt and D y (t) = Acosωt where ω is the angular velocity. Now our x and y are Our Kinetic Energy (T) is and Potential Energy (V) is x = lsinψ D x ẋ = l ψcosψ Ḋx y = lcosψ D y ẏ = l ψsinψ Ḋy T = m 2 (l2 ψ 2 + D x 2 + D y 2 2l ψ( D x cosψ + D y sinψ) V = mg(lcosψ D) Now we write the Lagrangian and use a similar trick whick we used in 1.1 and we get L L = m 2 l2 ψ 2 + ml( D x sinψ + (g D y )cosψ) Hence we get the Euler-Lagrangian Equation of motion as ψ = 1 l ( D x cosψ + ( D y g)sinψ) Here the variables are ψ and ψ. Let s call them as R and S respectively. Then we get the equations as Ṙ = S Ṡ = 1 l ( D x cosr + ( D y g)sinr) D x (g D y) hence the fixed points are S = 0 and tanr =. Here if we want (π, 0) to be a fixed point then D x (t π ) = 0 at time t π. Hence we get the A-matrix to be ( ) 0 1 A π = 1 l ( D x (t π )sinπ + (g D y (t π )cosπ 0 Hence we get λ 2 = 1 l (g D y (t π )) which says that we have this point to be an elliptic fixed point when we reach this point at time t π and only if g > D y (t π ). Hence under these conditions our point is

6 6 RAVITEJ UPPU a stable fixed point. Now as the motion of the support we have considered to be on the circle hence at time t π we have Aω 2 sinωt π = 0 t π = 2π ω and Aω2 < g These are the conditions for our point to be a stable fixed point.