用Python画双摆

双摆问题

所谓双摆，就是两个连在一起的摆。

设小球$m_1$和$m_2$的坐标分别为$(x_1,y_2),(x_2,y_2)$，则

$$\begin{gathered} x_1=l_1\sin {\theta }_1 \\ {y}_1=-l_1\cos {\theta }_1 \\ {x}_2=x_1+l_2\sin {\theta }_2 \\ {y}_2=y_1-l_2\cos {\theta }_2 \end{gathered}$$

如图所示，设$O$为坐标原点，如果这两个连在一起的摆在同一个平面内，则$m_1$和$m_2$点的坐标为

$$\begin{array}{rlr}& x_1=l_1\sin {\theta }_1& x_2=x_1+l_2\sin {\theta }_2\\ & y_1=-l_1\cos {\theta }_1& y_2=y_1-l_2\cos {\theta }_2\end{array}$$

接下来本来是要推公式的，考虑考虑到大家可能会有公式恐惧症，同时又喜欢看图，所以把公式挪到后面。

所以，只需知道角速度的微分方程，就可写出对应的代码，其方程如下

$$\begin{gathered} {\dot{\omega }}_1=\frac{(-m\frac{g}{l_1}\sin {\theta }_1+\mu \lambda {\omega }_2^2\sin \delta )-\mu \cos \delta (-\frac{g}{l_1}\sin {\theta }_2-{\omega }_1^2\sin \delta )}{m-\mu {\cos }^2\delta } \\ {\dot{\omega }}_2=\frac{-\cos \delta (-m\frac{g}{l_1}\sin {\theta }_1+\mu \lambda {\omega }_2^2\sin \delta )+m(-\frac{g}{l_1}\sin {\theta }_2-{\omega }_1^2\sin \delta )}{m\lambda -\mu \lambda {\cos }^2\delta } \end{gathered}$$

从而转为代码得到

# 其中，lam,mu,G_L1,M为全局变量
def derivs(state, t):
    dydx = np.zeros_like(state)
    th1,om1,th2,om2 = state
    dydx[0] = state[1]
    delta = state[2] - state[0]
    cDelta, sDelta = np.cos(delta), np.sin(delta)
    sTh1,_,sTh2,_ = np.sin(state)
    den1 = M - mu*cDelta**2
    dydx[1] = (mu * om1**2 * sDelta * cDelta
                + mu * G_L1 * sTh2 * cDelta
                + mu * lam * om2**2 * sDelta
                - M * G_L1 * sTh1)/ den1
    dydx[2] = state[3]
    den2 = lam * den1
    dydx[3] = (- mu * lam * om2**2 * sDelta * cDelta
                + M * G_L1 * sTh1 * cDelta
                - M * om1**2 * sDelta
                - M * G_L1 * sTh2)/ den2
    return dydx

接下来根据微分方程的解，便可进行绘图。

# 这段代码用于设置初值，并调用integrate求解微分方程组
import numpy as np
import scipy.integrate as integrate

G = 9.8
L1,L2 = 1.0, 1.0
G_L1 = G/L1
lam = L2/L1   #杆长度比L2/L1
mu = 1.0      #质量比M2/M1
M = 1+mu

# 生成时间
dt = 0.01
t = np.arange(0, 20, dt)

th1,th2 = 120.0, -10.0  #初始角度
om1,om2 = 0.0, 0.00       #初始角速度
state = np.radians([th1, om1, th2, om2])

# 微分方程组数值解
y = integrate.odeint(derivs, state, t)

# 真实坐标
x1 = L1*sin(y[:, 0])
y1 = -L1*cos(y[:, 0])
x2 = L2*sin(y[:, 2]) + x1
y2 = -L2*cos(y[:, 2]) + y1

至此，就得到了所有位置处的坐标，从而可以观察到双摆的轨迹如图所示

绘图代码为

import matplotlib.pyplot as plt
plt.scatter(x1,y1,marker='.')
plt.scatter(x2,y2,marker='.')
plt.show()

若将时间设置得长一点，然后在画图的时候更改一下颜色，就会看到双摆的运动区间，可见自然界还是挺有情怀的

其绘图代码为

plt.plot(x1,y1,marker='.',alpha=0.2, linewidth=0.2)
plt.plot(x2,y2,marker='.',alpha=0.2, linewidth=2, c='r')
plt.axis('off')
plt.show()

当然，也可以将其运动轨迹以一种三维的形式绘制出来

ax = plt.gca(projection='3d')
ax.plot3D(t,x1,y1,linewidth=1)
plt.show()

额……好吧，看来并没有什么情怀。

但是，如果把这两个小球分别当作两个星球，而我们又在一颗星球上，那么所观测到的另一颗星球的运动大致如下，不出意外是个圆，毕竟圆形二者之间的距离是恒定的。

绘图代码为

ax = plt.gca(projection='3d')
ax.plot3D(t,x2-x1,y2-y1,linewidth=0.5)
plt.show()

如果更改一下初值，则图形将有如下变化

初值设为

th1,th2 = 0, 0  #初始角度
om1,om2 = 120.0, 108.00       #初始角速度

运动过程

最后，还是传统技能，绘制一下双摆的运动过程如下

代码为

import matplotlib.animation as animation
# 下面为绘图过程
fig = plt.figure(figsize=(12,12))
ax = fig.add_subplot(111, autoscale_on=False, xlim=(-2, 2), ylim=(-2, 2))
ax.set_aspect('equal')
ax.grid()

line, = ax.plot([], [], 'o-', lw=2)
time_template = 'time = %.1fs'
time_text = ax.text(0.05, 0.9, '', transform=ax.transAxes)

# 初始化图形
def init():
    line.set_data([], [])
    time_text.set_text('')
    return line, time_text

def animate(i):
    thisx = [0, x1[i], x2[i]]
    thisy = [0, y1[i], y2[i]]
    line.set_data(thisx, thisy)
    time_text.set_text(time_template % (i*dt))
    return line, time_text

ani = animation.FuncAnimation(fig, animate, range(1, len(y)),   
        interval=dt*1000, blit=True, init_func=init)
ani.save("dua_1.gif",writer='imagemagick')
plt.show()

附 公式推导过程

双摆的动能和势能分别为

T = ∑ i ∈ { 1 , 2 } m i 2 ( x ˙ i 2 + y ˙ i 2 ) V = − ∑ i ∈ { 1 , 2 } m i g y i \begin{aligned} T=&\sum_{i\in\{1,2\}}\frac{m_i}{2}(\dot x^2_i+\dot y^2_i)\\ V=&-\sum_{i\in\{1,2\}}m_igy_i \end{aligned} T=V=​i∈{ 1,2}∑​2mi​​(x˙i2​+y˙​i2​)−i∈{ 1,2}∑​mi​gyi​​

系统的拉格朗日量为$L=T-V$，带入角度坐标可得

$$\begin{gathered} L=\frac{m_1+m_2}{2}{l}_1^2{\dot{\theta }}_1^2+\frac{m_2}{2}{l}_2^2{\dot{\theta }}_2^2+m_2{l}_1{l}_2{\dot{\theta }}_1{\dot{\theta }}_2\cos ({\theta }_1-{\theta }_2) \\ +(m_1+m_2)l_{1}g\cos {\theta }_1+m_2{l}_{2}g\cos {\theta }_2 \end{gathered}$$

根据拉格朗日方程

$$\frac{\text{d}}{\text{d}t}\frac{\partial L}{\partial {\dot{\theta }}_i}-\frac{\partial L}{\partial {\theta }_i}=0$$

则有

$$\begin{gathered} \frac{\text{d}}{\text{d}t}((m_1+m_2)l_1^2{\dot{\theta }}_1+m_2{l}_1{l}_2{\dot{\theta }}_2\cos ({\theta }_1-{\theta }_2))=-m_2{l}_1{l}_2{\dot{\theta }}_1{\dot{\theta }}_2\sin ({\theta }_1-{\theta }_2)-(m_1+m_2)l_{1}g\sin {\theta }_1 \\ \frac{\text{d}}{\text{d}t}(m_2{l}_2^2{\dot{\theta }}_2-m_2{l}_1{l}_2{\dot{\theta }}_1\cos ({\theta }_1-{\theta }_2))=m_2{l}_1{l}_2{\dot{\theta }}_1{\dot{\theta }}_2\sin ({\theta }_1-{\theta }_2)-m_2{l}_{2}g\sin {\theta }_2 \end{gathered}$$

记$\mu =\frac{m_2}{m_1}，\lambda =\frac{l_2}{l_1},\delta ={\theta }_1-{\theta }_2$，可简化为

$$\begin{gathered} (1+\mu ){\ddot{\theta }}_1+\mu \lambda [{\ddot{\theta }}_2\cos \delta -{\dot{\theta }}_2^2\sin \delta ]=-(1+\mu )\frac{g}{l_1}\sin {\theta }_1 \\ \lambda {\ddot{\theta }}_2+{\ddot{\theta }}_1\cos \delta +{\dot{\theta }}_1^2\sin \delta =-\frac{g}{l_1}\sin {\theta }_2 \end{gathered}$$

令${\omega }_1={\dot{\theta }}_1,{\omega }_2={\dot{\theta }}_2$，记$m=1+\mu$，上式可写为矩阵形式

$$\left[\begin{array}{cc}m& \mu \lambda \cos \delta \\ \cos \delta & \lambda \end{array}\right]\left[\begin{array}{c}{\dot{\omega }}_1\\ {\dot{\omega }}_2\end{array}\right]=\left[\begin{array}{c}-m\frac{g}{l_1}\sin {\theta }_1+\mu \lambda {\omega }_2^2\sin \delta \\ -\frac{g}{l_1}\sin {\theta }_2-{\omega }_1^2\sin \delta \end{array}\right]$$

记$\omega =[{\omega }_1,{\omega }_2{]}^T$，$M=\left[\begin{array}{cc}m& \mu \lambda \cos \delta \\ \cos \delta & \lambda \end{array}\right]$，$F=\left[\begin{array}{c}-m\frac{g}{l_1}\sin {\theta }_1+\mu \lambda {\omega }_2^2\sin \delta \\ -\frac{g}{l_1}\sin {\theta }_2-{\omega }_1^2\sin \delta \end{array}\right]$，则

$$M\dot{\omega }=F\to \dot{\omega }=M^{-1}F$$

其中，

$$M^{-1}=\frac{1}{m\lambda -\mu \lambda {\cos }^2\delta }\left[\begin{array}{cc}\lambda & -\mu \lambda \cos \delta \\ -\cos \delta & m\end{array}\right]$$

展开可得则

$$\begin{gathered} {\dot{\omega }}_1=\frac{(-m\frac{g}{l_1}\sin {\theta }_1+\mu \lambda {\omega }_2^2\sin \delta )-\mu \cos \delta (-\frac{g}{l_1}\sin {\theta }_2-{\omega }_1^2\sin \delta )}{m-\mu {\cos }^2\delta } \\ {\dot{\omega }}_2=\frac{-\cos \delta (-m\frac{g}{l_1}\sin {\theta }_1+\mu \lambda {\omega }_2^2\sin \delta )+m(-\frac{g}{l_1}\sin {\theta }_2-{\omega }_1^2\sin \delta )}{m\lambda -\mu \lambda {\cos }^2\delta } \end{gathered}$$