Answer for HW8

1 双摆

参见朗道力学 $§ 23$ 习题 2

1.1 (1)

选取系统为 $m_{1}$ 和 $m_{2}$ ，它有两个自由度，选取广义坐标为 $φ_{1}$ 和 $φ_{2}$ 。有质点 $m_{1}$ 和 $m_{2}$ 的坐标分别为 $(l_{1} \sin φ_{1}, l_{1} \cos φ_{1}), (l_{1} \sin φ_{1} + l_{2} \sin φ_{2}, l_{1} \cos φ_{1} + l_{2} \cos φ_{2})$ 。由此可得 $m_{1}$ 的动能为

T_{1} = \frac{1}{2} m_{1} l_{1}^{2} {\dot{φ}}_{1}^{2}

$m_{2}$ 的动能为

\begin{aligned} T_{2} & = \frac{1}{2} m_{2} [l_{1}^{2} {\dot{φ}}_{1}^{2} + l_{2}^{2} {\dot{φ}}_{2}^{2} + 2 l_{1} l_{2} {\dot{φ}}_{1} {\dot{φ}}_{2} \cos (φ_{1} - φ_{2})] \\ \approx \frac{1}{2} m_{2} (l_{1}^{2} {\dot{φ}}_{1}^{2} + l_{2}^{2} {\dot{φ}}_{2}^{2} + 2 l_{1} l_{2} {\dot{φ}}_{1} {\dot{φ}}_{2}) \\ \Rightarrow T & = T_{1} + T_{2} = \frac{1}{2} (m_{1} + m_{2}) l_{1}^{2} {\dot{φ}}_{1}^{2} + \frac{1}{2} m_{2} l_{2}^{2} {\dot{φ}}_{2}^{2} + m_{2} l_{1} l_{2} {\dot{φ}}_{1} {\dot{φ}}_{2} \end{aligned}

于是 $T$ 的惯性系数矩阵 $M$ 为，

M = (\begin{array}{ll} (m_{1} + m_{2}) l_{1}^{2} & m_{2} l_{1} l_{2} \\ m_{2} l_{1} l_{2} & m_{2} l_{2}^{2} \end{array})

系统的势能为

\begin{aligned} U & = - m_{1} g l_{1} \cos φ_{1} - m_{2} g (l_{1} \cos φ_{1} + l_{2} \cos φ_{2}) \\ \approx \frac{1}{2} (m_{1} + m_{2}) g l_{1} φ_{1}^{2} + \frac{1}{2} m_{2} g l_{2} φ_{2}^{2} - (m_{1} g l_{1} + m_{2} g l_{1} + m_{2} g l_{2}) \end{aligned}

于是 $U$ 的刚性系数矩阵 $U$ 为

U = (\begin{array}{ll} (m_{1} + m_{2}) g l_{1} & 0 \\ 0 & m_{2} g l_{2} \end{array})

拉格朗日函数为

L = \frac{1}{2} (m_{1} + m_{2}) l_{1}^{2} {\dot{φ}}_{1}^{2} + \frac{1}{2} m_{2} l_{2}^{2} {\dot{φ}}_{2}^{2} + m_{2} l_{1} l_{2} {\dot{φ}}_{1} {\dot{φ}}_{2} - \frac{1}{2} (m_{1} + m_{2}) g l_{1} φ_{1}^{2} - \frac{1}{2} m_{2} g l_{2} φ_{2}^{2}

其中略去了势能中对运动无影响的常数项．

1.2 (2)

将 $L$ 代入拉格朗日方程得系统的运动微分方程为

\begin{array}{r} (m_{1} + m_{2}) l_{1} {\ddot{φ}}_{1} + m_{2} l_{2} {\ddot{φ}}_{2} + (m_{1} + m_{2}) g φ_{1} & = 0 \\ l_{1} {\ddot{φ}}_{1} + l_{2} {\ddot{φ}}_{2} + g φ_{2} & = 0 \end{array}

设方程的解为 $φ_{1} = A_{1} e^{i ω t}, φ_{2} = A_{2} e^{i ω t}$ ，将它们代入运动微分方程可以得到

\begin{array}{r} A_{1} (m_{1} + m_{2}) (g - l_{1} ω^{2}) - A_{2} m_{2} l_{2} ω^{2} & = 0 \\ - A_{1} l_{1} ω^{2} + A_{2} (g - l_{2} ω^{2}) & = 0 \end{array}

特征方程为上述方程组的系数行列式等于零，或者 $\det (U - ω^{2} M) = 0$ ，即

| \begin{array}{cc} (m_{1} + m_{2}) (g - l_{1} ω^{2}) & - m_{2} l_{2} ω^{2} \\ - m_{2} l_{1} ω^{2} & m_{2} (g - l_{2} ω^{2}) \end{array} | = 0

由此可解得本征频率为

\begin{aligned} ω_{1, 2}^{2} = & \frac{g}{2 m_{1} l_{1} l_{2}} \times {(m_{1} + m_{2}) (l_{1} + l_{2}) \pm \sqrt{(m_{1} + m_{2}) [(m_{1} + m_{2}) {(l_{1} + l_{2})}^{2} - 4 m_{1} l_{1} l_{2}]}} \end{aligned}

将上式再代入运动微分方程可以得到 $A_{2} = α_{1, 2} A_{1}$ ， $α$ 大概率是一坨丑陋的常数式子(这里偷个懒)，它们简单写是：

α_{1} = \frac{l_{1} ω_{1}^{2}}{g - l_{2} ω_{1}^{2}}, α_{2} = \frac{l_{1} ω_{2}^{2}}{g - l_{2} ω_{2}^{2}}

于是

[\begin{matrix} φ_{1} \\ φ_{2} \end{matrix}] = [\begin{matrix} A_{1} |_{ω_{1}} & A_{1} |_{ω_{2}} \\ A_{2} |_{ω_{1}} & A_{2} |_{ω_{2}} \end{matrix}] {[\begin{matrix} e^{i ω_{1} t} \\ e^{i ω_{2} t} \end{matrix}]}_{r e a l} = [\begin{matrix} A_{1} |_{ω_{1}} & A_{1} |_{ω_{2}} \\ A_{2} |_{ω_{1}} & A_{2} |_{ω_{2}} \end{matrix}] [\begin{matrix} \cos (ω_{1} t + ϕ_{1}) \\ \cos (ω_{2} t + ϕ_{2}) \end{matrix}]

记 $C_{1} = A_{1} |_{ω_{1}}, C_{2} = A_{1} |_{ω_{2}}$ ，上式变为

[\begin{matrix} φ_{1} \\ φ_{2} \end{matrix}] = [\begin{matrix} C_{1} & C_{2} \\ α_{1} C_{1} & α_{2} C_{2} \end{matrix}] [\begin{matrix} \cos (ω_{1} t + ϕ_{1}) \\ \cos (ω_{2} t + ϕ_{2}) \end{matrix}]

1.3 (3)

反解上式得到简振坐标

[\begin{matrix} Q_{1} \\ Q_{2} \end{matrix}] = [\begin{matrix} C_{1} \cos (ω_{1} t + ϕ_{1}) \\ C_{2} \cos (ω_{2} t + ϕ_{2}) \end{matrix}] = {[\begin{matrix} 1 & 1 \\ α_{1} & α_{2} \end{matrix}]}^{- 1} [\begin{matrix} φ_{1} \\ φ_{2} \end{matrix}] = \frac{1}{α_{2} - α_{1}} [\begin{matrix} α_{2} & - 1 \\ - α_{1} & 1 \end{matrix}] [\begin{matrix} φ_{1} \\ φ_{2} \end{matrix}]

对于 $l_{1} = l_{2} = l, m_{1} = m_{2} = m$ 的情况会很简单，变成

[\begin{matrix} Q_{1} \\ Q_{2} \end{matrix}] = \frac{\sqrt{2}}{4} [\begin{matrix} \sqrt{2} & - 1 \\ \sqrt{2} & 1 \end{matrix}] [\begin{matrix} φ_{1} \\ φ_{2} \end{matrix}]

2 耦合谐振子

$L = \frac{1}{2} m_{1} {\dot{x}}^{2} - \frac{1}{2} k_{1} x^{2} + \frac{1}{2} m_{2} {\dot{y}}^{2} - \frac{1}{2} k_{2} y^{2} + α x y$ 代入拉格朗日方程得到

\begin{aligned} m_{1} \ddot{x} + k_{1} x - α y & = 0 \\ m_{2} \ddot{y} + k_{2} y - α x & = 0 \end{aligned}

代入 $x, y = a, b \times \cos (ω t)$ 得到系数方程

\begin{aligned} (- m_{1} ω^{2} + k_{1}) a - α b & = 0 \\ - α a + (- m_{2} ω^{2} + k_{2}) b & = 0 \end{aligned}

得到久期方程

| \begin{array}{cc} - m_{1} ω^{2} + k_{1} & - α \\ - α & - m_{2} ω^{2} + k_{2} \end{array} | = 0

解得本征频率

\begin{aligned} ω_{1, 2}^{2} & = \frac{m_{1} k_{2} + m_{2} k_{1} \pm \sqrt{(m_{1} k_{2} + m_{2} k_{1})^{2} - 4 m_{1} m_{2} (k_{1} k_{2} - α^{2})}}{2 m_{1} m_{2}} \\ = \frac{m_{1} k_{2} + m_{2} k_{1} \pm \sqrt{(m_{1} k_{2} - m_{2} k_{1})^{2} + 4 α^{2} m_{1} m_{2}}}{2 m_{1} m_{2}} \end{aligned}

将上式代回系数方程得到， $b = β_{1, 2} \cdot a$ ，其中

β_{1, 2} = \frac{- m_{1} ω_{1, 2}^{2} + k_{1}}{α} = - \frac{m_{1} k_{2} - m_{2} k_{1} \pm \sqrt{(m_{1} k_{2} - m_{2} k_{1})^{2} + 4 m_{1} m_{2} α^{2}}}{2 m_{2} α}

于是

[\begin{matrix} x \\ y \end{matrix}] = [\begin{matrix} a |_{ω_{1}} & a |_{ω_{2}} \\ b |_{ω_{1}} & b |_{ω_{2}} \end{matrix}] [\begin{matrix} \cos (ω_{1} t + ϕ_{1}) \\ \cos (ω_{2} t + ϕ_{2}) \end{matrix}]

记 $C_{1} = a |_{ω_{1}}, C_{2} = a |_{ω_{2}}$ ，上式变为

[\begin{matrix} x \\ y \end{matrix}] = [\begin{matrix} C_{1} & C_{2} \\ β_{1} C_{1} & β_{2} C_{2} \end{matrix}] [\begin{matrix} \cos (ω_{1} t + ϕ_{1}) \\ \cos (ω_{2} t + ϕ_{2}) \end{matrix}]

反解上式得到简振坐标

[\begin{matrix} Q_{1} \\ Q_{2} \end{matrix}] = [\begin{matrix} C_{1} \cos (ω_{1} t + ϕ_{1}) \\ C_{2} \cos (ω_{2} t + ϕ_{2}) \end{matrix}] = {[\begin{matrix} 1 & 1 \\ β_{1} & β_{2} \end{matrix}]}^{- 1} [\begin{matrix} x \\ y \end{matrix}] = \frac{1}{β_{2} - β_{1}} [\begin{matrix} β_{2} & - 1 \\ - β_{1} & 1 \end{matrix}] [\begin{matrix} x \\ y \end{matrix}] \propto [\begin{matrix} \frac{- m_{1} ω_{2}^{2} + k_{1}}{α} & - 1 \\ \frac{m_{1} ω_{1}^{2} - k_{1}}{α} & 1 \end{matrix}] [\begin{matrix} x \\ y \end{matrix}]

对于 $k_{1} = k_{2} = k, m_{1} = m_{2} = m$ 的情况会很简单，变成

[\begin{matrix} Q_{1} \\ Q_{2} \end{matrix}] = \frac{\sqrt{2}}{2} [\begin{matrix} 1 & 1 \\ 1 & - 1 \end{matrix}] [\begin{matrix} x \\ y \end{matrix}]

3 平面谐振子

参见朗道力学 $§ 23$ 习题3
或 HW7-1 T3 (a)
或令上一题中的 $α = m_{2} = 0$ .

朗道：