Question

Objectives To have a better understanding of direction fields, phase portraits, and graph of solutions functions of the equations describing a nonlinear pendulum. Introduction A pendulum consists of a small ball attached to the end of a rigid rod, the latter can swing from its other end, as shown in the picture below. We assume that the ball at the end of the pendulum has mass m > 0 and the pendulum rod has length l > 0. Also assume that the there is a friction force acting on the ball, with damping coefficient d > 0. Recall that a friction force opposes the movement and is proportional to the speed of the object. The mass at the end of the pendulum will move on a circle of radius l centered at the oscillation point. Our main variable is 0(t), the angle between the rod and the downward vertical, positive in the counter clockwise direction. The displacement of the pendulum mass along the the circle where it moves is s(t) = le(t). Newton's equation of motion is m (l0)" = -mg sin(0) – d (10)'. The negative sign in the first term of the right-hand side above is because the component of the weight tangent to the circle is in the direction opposite to the direction of increasing 8. The negative in the second term above is because the friction force opposes the movement. We rewrite the equation as 0" = - sin(0) – 46. m In this Lab we will focus on the case g/l = 1 and m = 1, that is 0" = - sin(0) - de'.

Answer 1

pendulum equation with friction is given as 02_q sind door We tane q-m=1 heel sindado D Now we reduce this to two first order es in by taking 2,= 0 = 0 = x2 = 24 Xa=0 0" x = - sino-do -- Sin -day Tag =- sinx-daal clb' Now to find ерш l bn um bon- We write x,=0, q=0 8,30 – 22= a=0 ³-dinay-dazzo 7 sinay=0 1 x,=nit (12) lineari be matrix Lor1/4) ( 9 )=(cos4 -d)| 20/ Where ! is the Jacobs-matrix (-cosse-a) ei 077 au (1 d.) Now to find the eigen value We adil J-allo 1-1 + |- cos, -d-a | a(df) 4c0122-D reda + costo a=-de Id - 4 cose - 2 "even - 2 Now for 24 = 2K17 I never att de and for x, =(2K+1) IT _ ( 37 % : - / hence lao-de Jåte ode cd e.) for no friction . we put d=0 then de vente and Hodd = id) So for n= (akt) there will be two critical pt corresponding to a=fl but I n=2k there is hо съtc. 1. Last modified: 7:06 pm