我无法弄清楚如何编写脚本来绘制KE,PE和TE。我在我的代码部分中包含了多个###########,我觉得问题所在。
def pendulum_runge_kutta(theta0,omega0,g,tfinal,dt):
# initialize arrays
t = np.arange(0.,tfinal+dt,dt) # time array t
npoints = len(t)
theta = np.zeros(npoints) # position array theta
omega = np.zeros(npoints) # position array omega
Ke = np.zeros(npoints)
Pe = np.zeros(npoints)
L=4
g = 9.81
t2=np.linspace(0,tfinal,1000)
theta0=0.01
omega0=0
# exact solution for
thetaExact = theta0*np.cos((g/L)**(1/2)*t2)
# SECOND ORDER RUNGE_KUTTA SOLUTION
theta[0] = theta0
omega[0] = omega0
#Ke[0] = #######################################
#Pe[0] =######################################
m=1.0
for i in range(npoints-1):
# compute midpoint position (not used!) and velocity
thetamid = theta[i] + omega[i]*dt
omegamid = omega[i] - (g/L)*np.sin(theta[i])*dt/2
# use midpoint velocity to advance position
theta[i+1] = theta[i] + omegamid*dt
omega[i+1] = omega[i] -(g/L)*np.sin(thetamid)*dt/2
###########calculate Ke, Pe, Te############
Ke[i+1] = 0.5*m*(omega[i+1]*L)**2
Pe[i+1] = m*g*L*np.sin(theta[i+1])
Te = Ke+Pe
#plot result of Ke, Pe, Te
pl.figure(1)
pl.plot(t,Ke,'c-',label='kinetic energy')
pl.plot(t,Pe,'m-',label='potential energy')
pl.plot(t,Te,'g-',label='total energy')
pl.title('Ke, Pe, and Te')
pl.legend(loc='lower right')
pl.show()
#now plot the results
pl.figure(2)
pl.plot(t,theta,'ro',label='2oRK')
pl.plot(t2,thetaExact,'r',label='Exact')
pl.grid('on')
pl.legend()
pl.xlabel('Time (s)')
pl.ylabel('Theta')
pl.title('Theta vs Time')
pl.show()
#call function
pendulum_runge_kutta(0.01, 0, 9.8, 12, .1)
答案 0 :(得分:0)
中点方法的公式统一适用于一阶系统的所有组件。因此,对于两个中点值,它应为dt/2
,对于全时步长,应为dt
。
潜在能量应包含sin(x)
,C-cos(x)
的积分。例如,
Pe[i+1] = m*g*L*(1-np.cos(theta[i+1]))
时间点i+1
的能量公式也适用于时间点0
。您必须向上移动质量声明,例如在长度L=4
行之后。
最后一点,指出的精确解是用于小角度近似,对于一般物理摆没有有限可表达的精确解。对于theta0=0.01
的给定幅度,小角度近似应该足够好,但要注意更大的摆动。