Large angle pendulum plot did not show as expected

Question

I am trying to plot the relationship between period and amplitude for an undamped and undriven pendulum for when small angle approximation breaks down, however, my code did not do what I expected...

I think I should be expecting a strictly increasing graph as shown in this video: https://www.youtube.com/watch?v=34zcw_nNFGU

Here is my code, I used zero crossing method to calculate period:

import numpy as np
import matplotlib.pyplot as plt
from scipy.integrate import solve_ivp
from itertools import chain

# Second order differential equation to be solved:
# d^2 theta/dt^2 = - (g/l)*sin(theta) - q* (d theta/dt) + F*sin(omega*t)
# set g = l and omega = 2/3 rad per second
# Let y[0] = theta, y[1] = d(theta)/dt

def derivatives(t,y,q,F):
    return [y[1], -np.sin(y[0])-q*y[1]+F*np.sin((2/3)*t)]

t = np.linspace(0.0, 100, 10000)

#initial conditions:theta0, omega0
theta0 = np.linspace(0.0,np.pi,100)
q = 0.0       #alpha / (mass*g), resistive term
F = 0.0       #G*np.sin(2*t/3)

value = []
amp = []
period = []

for i in range (len(theta0)):
    sol = solve_ivp(derivatives, (0.0,100.0), (theta0[i], 0.0), method = 'RK45', t_eval = t,args = (q,F))
    velocity = sol.y[1]
    time = sol.t

    zero_cross = 0

    for k in range (len(velocity)-1):
        if (velocity[k+1]*velocity[k]) < 0:
            zero_cross += 1
            value.append(k)

        else:
            zero_cross += 0

    if zero_cross != 0:
        amp.append(theta0[i])
      # period calculated using the time evolved between the first and last zero-crossing detected
        period.append((2*(time[value[zero_cross - 1]] - time[value[0]])) / (zero_cross -1))


plt.plot(amp,period)
plt.title('Period of oscillation of an undamped, undriven pendulum \nwith varying initial angular displacemnet')
plt.xlabel('Initial Displacement')
plt.ylabel('Period/s')
plt.show()

enter image description here

Answer 1

You can use the event mechanism of solve_ivp for such tasks, it is designed for such "simple" situations

def halfperiod(t,y): return y[1]
halfperiod.terminal=True  #  stop when root found
halfperiod.direction=1    #  find sign changes from negative to positive

for i in range (1,len(theta0)): # amp==0 gives no useful result
    sol = solve_ivp(derivatives, (0.0,100.0), (theta0[i], 0.0), method = 'RK45', events =(halfperiod,) )
    if sol.success and len(sol.t_events[-1])>0:
        period.append(2*sol.t_events[-1][0]) # the full period is twice the event time
        amp.append(theta0[i])

This results in the plot