Question

The Problem The single pendulunm Consider the single pendulum shown below. There is a bob at the end of the pendulum, of mass For small oscillations in the case of a frictionless pivot and a rod with negligible mass, the motion of the pendulum over time can be described by the second-order linear ODE where θ is the angle between the rod and the vertical, g is gravity, t is the length of the rod and θ dag /dt2 Q1 By substituting (2) into the ODE, veriy that the general solation to equation (1) is given by θ c1 cos(wt) + c2 sin(wt) where w Vg/ The double pendulum A double pendulum consists of two pendulums attached end-to-end, as shown below.

Answer 1

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Theta = -g/l theta substituting squareroot g/l = w then theta = w^2 theta rightdoublearrow theta + w^2 theta = 0 (1) This is second order linear ODE then complementary function m^2 + w^2 = 0 m^2 - (i w)^2 = 0 m = plusminus i w therefore solution of eqn (1) theta (t) = C_1 cos (w t) + C_2 sin (w t) since theta_1 = -2w^2 theta_1 + w^2 theta_2 theta_2 = 2w^2 theta_1 - 2w^2 theta_2 In matrix form [theta_1 theta_2] = [- 2w^2 2w^2 w^2 - 2w^2] [theta_1 theta_2] [theta_1 theta_2] = -w^2 [+ 2 - 2 - 1 + 2] [theta_1 theta_2] [theta_1 theta_2] = -w^2 [2 - 2 - 1 2] [theta_1 theta_2] because emptyset = [theta_1 theta_2] emptyset'11 = [theta_1 theta_2] A = [2 - 2 - 1 2] therefore emeptyset'11 = -w^2 A emptyset