As depicted in Fig. 2, a double pendulum consists of a pendulum attached to another pendulum. We indicate the upper and lower pendulums by subscripts 1 and 2, respectively, and we place the origin at the pivot point of the upper pendulum with y increasing upward. We further assume that the system oscillates in a vertical plane subject to gravity, that the pendulum rods are massless and rigid, and the pendulum masses are considered to be point masses. Under these assumptions, force balances can be used to derive the following equations of motion Ö₁ -g (2m1m2) sin 01 - - m29 sin (01 – 202) — 2 sin (01 - 02) m2 [032L2+0² L₁ cos (01 − 02) L1 [2m1m2m2 cos (201-02)] 2 sin (01-02) L1 (m₁ + m2) +9 (m₁ + m2) cos 01 +02½ L2m2 cos (01 -02) L2 [2m1m2m2 cos (201 - 02)] (2a) སྐྱེ་སྐྱེ (2b) where the superscript dot (-) represents the time derivative, 0 is the angle (radians), t represents time [s], g the gravitational acceleration (9.81 ms¹), m is the mass kg, and L is the length [m]. Note that the x and y coordinates of the masses are functions of the angles as in x₁ = L₁ sin 01, Y₁ = -L1 cos 01 x2 = x1 + L2 sin 02, y2 = y1 - L2 cos 02 m 0₂ ma Figure 2: A double pendulum. Use ode45 to solve for the angles and angular velocities of the masses as a function of time from 0s to 40 s. Create a plot with a time series of the angles and a state space plot of 02 versus 01. Test your code for: (a) Small displacement: L₁ 02 = 0₁ = 02 = 0. = L₂ = 1m, m₁ = m2 = 0.25 kg, with initial conditions, 01 = 0.5 m and (b) Large displacement: L₁ = L2 = 1m, m₁ = 0.5 kg, m2 = 0.25 kg, with initial conditions, 0₁ = 1m and 02 = 0₁ = 02 = 0.