分析力学 第7次作业

Chasse_neige

4.8 单摆振动方程为 $\ddot{\theta }+\frac{g}{l}\sin \theta =0$, 若振动角度不太大, 利用微扰近似方法给出二阶精度的近似解, 包括周期的二阶精度修正。

展开

$$\sin \theta =\theta -\frac{1}{6}{\theta }^3+o({\theta }^5)$$

假设

$$\theta ={\theta }_0+ϵ{\theta }_1+{ϵ}^2{\theta }_2$$

令 ${\omega }_0=\sqrt{\frac{g}{l}}$

$$\omega ={\omega }_0+ϵ{\omega }_1+{ϵ}^2{\omega }_2$$$$\begin{array}{c}\ddot{\theta }=-{\omega }_0^2\theta +{ϵ}^2\frac{{\omega }_0^2}{6}{\theta }^3\\ \ddot{\theta }+{\omega }^2\theta =({\omega }^2-{\omega }_0^2)\theta +{ϵ}^2\frac{{\omega }_0^2}{6}{\theta }^3\end{array}$$

作代换 $\tau =\omega t+\varphi$，得到

$${\theta }^{″}+\theta =(1-\frac{{\omega }_0^2}{{\omega }^2})\theta +{ϵ}^2\frac{{\omega }_0^2}{6{\omega }^2}{\theta }^3$$

分离得到

$$\begin{array}{c}{\theta }_0^{″}+{\theta }_0=0\\ {\theta }_1^{″}+{\theta }_1=2\frac{{\omega }_1}{{\omega }_0}{\theta }_0\\ {\theta }_2^{″}+{\theta }_2=(2\frac{{\omega }_2}{{\omega }_0}-3{\left(\frac{{\omega }_1}{{\omega }_0}\right)}^2){\theta }_0+2\frac{{\omega }_1}{{\omega }_0}{\theta }_1+\frac{1}{6}{\theta }_0^3\end{array}$$

解得0阶解

$${\theta }_0(t)=A\cos (\tau )$$

所以1阶方程为

$${\theta }_1^{″}+{\theta }_1=2A\frac{{\omega }_1}{{\omega }_0}\cos \tau$$

采用拉普拉斯变换，舍弃初值相关项

$$p^{2}L[{\theta }_1]+L[{\theta }_1]=L[2A\frac{{\omega }_1}{{\omega }_0}\cos \tau ]$$

所以

$$\begin{array}{c}{\theta }_1(t)=L^{-1}\left[\frac{1}{p^2+1}L[2A\frac{{\omega }_1}{{\omega }_0}\cos \tau ]\right]\\ ={\int }_0^{\tau }2A\frac{{\omega }_1}{{\omega }_0}\cos (\tau -{\tau }^{\prime })\sin {\tau }^{\prime }\mathrm{d}{\tau }^{\prime }\\ =A\frac{{\omega }_1}{{\omega }_0}\tau \sin \tau \end{array}$$

为了消除非物理项，得到

$${\omega }_1=0$$

所以

$${\theta }_1(t)=0$$

不存在1阶项。对于2阶

$${\theta }_2^{″}+{\theta }_2=2A\frac{{\omega }_2}{{\omega }_0}\cos \tau +\frac{A^3}{6}{\cos }^3\tau$$

所以

$$\begin{array}{c}{\theta }_2(t)=L^{-1}\left[\frac{1}{p^2+1}L[2A\frac{{\omega }_2}{{\omega }_0}\cos \tau +\frac{A^3}{6}{\cos }^3\tau ]\right]\\ ={\int }_0^{\tau }2A\frac{{\omega }_2}{{\omega }_0}\cos (\tau -{\tau }^{\prime })\sin {\tau }^{\prime }+\frac{A^3}{6}{\cos }^3(\tau -{\tau }^{\prime })\sin {\tau }^{\prime }\mathrm{d}{\tau }^{\prime }\\ =A\frac{{\omega }_2}{{\omega }_0}\tau \sin \tau +\frac{A^3}{192}(12\tau \sin \tau +\cos \tau -\cos 3\tau )\end{array}$$

消去非物理项，得到

$${\omega }_2=-\frac{A^2}{16}{\omega }_0$$

得到结论

$$\theta (t)=A\cos \tau +\frac{A^3}{192}(\cos \tau -\cos 3\tau )$$$$\omega ={\omega }_0(1-\frac{A^2}{16})$$

4.13 二氧化碳分子经典模型如图 4.15 所示,平衡时原子在一直线上。只考虑原子在直线上运动的模式,求分子的简正频率和简正振动模式。

直接给出相对于平衡位置的坐标运动方程

$$\begin{array}{c}{\ddot{x}}_1=\frac{k}{m}(x_2-x_1)\\ {\ddot{x}}_2=\frac{k}{M}(x_1+x_3-2x_2)\\ {\ddot{x}}_3=\frac{k}{m}(x_2-x_3)\end{array}$$

令 ${\omega }_1^2=\frac{k}{m},{\omega }_2^2=\frac{k}{M}$，得到系数矩阵

$$\left(\begin{array}{ccc}-{\omega }_1^2& {\omega }_1^2& 0\\ {\omega }_2^2& -2{\omega }_2^2& {\omega }_2^2\\ 0& {\omega }_1^2& -{\omega }_1^2\end{array}\right)$$

对角化

$$A=\left(\begin{array}{ccc}1& 1& -1\\ 1& 0& 2\frac{{\omega }_2^2}{{\omega }_1^2}\\ 1& -1& -1\end{array}\right)\left(\begin{array}{ccc}0& 0& 0\\ 0& -{\omega }_1^2& 0\\ 0& 0& -{\omega }_1^2-2{\omega }_2^2\end{array}\right)\frac{1}{2(2{\omega }_2^2+{\omega }_1^2)}\left(\begin{array}{ccc}2{\omega }_2^2& 2{\omega }_1^2& 2{\omega }_2^2\\ 2{\omega }_2^2+{\omega }_1^2& 0& -(2{\omega }_2^2+{\omega }_1^2)\\ -{\omega }_1^2& 2{\omega }_1^2& -{\omega }_1^2\end{array}\right)$$

得到两个振动模式的频率和简正坐标为（特征值0对应了平动）

$$\begin{array}{c}\omega ={\omega }_1=\sqrt{\frac{k}{m}}\\ x=x_1-x_3\end{array}$$$$\begin{array}{c}\omega =\sqrt{{\omega }_1^2+2{\omega }_2^2}=\sqrt{\frac{k}{m}+2\frac{k}{M}}\\ x=-x_1+2x_2-x_3\end{array}$$

4.15 一耦合摆作小振动，如图 4.17所示，两个小球质量都是 $m$，系在长度都是 $l$ 的轻杆，小球之间用劲度系数为 $k$ 的弹簧相连，弹簧的自然长度与两个悬点 $O,O^{\prime }$ 之间的距离相同，系统限于两个轻杆自然下垂的铅直平面内运动。两球初始静止，一球在平衡位置，另一球则拉到 ${\theta }_2=\alpha$，求解系统。

首先写出运动方程

$$\left(\begin{array}{c}{\ddot{\theta }}_1\\ {\ddot{\theta }}_2\end{array}\right)=\left(\begin{array}{cc}-\frac{g}{l}-\frac{k}{m}& \frac{k}{m}\\ \frac{k}{m}& -\frac{g}{l}-\frac{k}{m}\end{array}\right)\left(\begin{array}{c}{\theta }_1\\ {\theta }_2\end{array}\right)$$

令 ${\omega }_1^2=\frac{k}{m},{\omega }_2^2=\frac{g}{l}+\frac{k}{m}$，系数矩阵化为

$$\left(\begin{array}{cc}-{\omega }_2^2& {\omega }_1^2\\ {\omega }_1^2& -{\omega }_2^2\end{array}\right)$$

所以

$$\left(\begin{array}{cc}-{\omega }_2^2& {\omega }_1^2\\ {\omega }_1^2& -{\omega }_2^2\end{array}\right)=\left(\begin{array}{cc}\frac{1}{\sqrt{2}}& \frac{1}{\sqrt{2}}\\ \frac{1}{\sqrt{2}}& -\frac{1}{\sqrt{2}}\end{array}\right)\left(\begin{array}{cc}{\omega }_1^2-{\omega }_2^2& 0\\ 0& -{\omega }_1^2-{\omega }_2^2\end{array}\right)\left(\begin{array}{cc}\frac{1}{\sqrt{2}}& \frac{1}{\sqrt{2}}\\ \frac{1}{\sqrt{2}}& -\frac{1}{\sqrt{2}}\end{array}\right)$$

得到

$$\left(\begin{array}{c}{\ddot{\theta }}_1\\ {\ddot{\theta }}_2\end{array}\right)=\left(\begin{array}{cc}\frac{1}{\sqrt{2}}& \frac{1}{\sqrt{2}}\\ \frac{1}{\sqrt{2}}& -\frac{1}{\sqrt{2}}\end{array}\right)\left(\begin{array}{cc}{\omega }_1^2-{\omega }_2^2& 0\\ 0& -{\omega }_1^2-{\omega }_2^2\end{array}\right)\left(\begin{array}{cc}\frac{1}{\sqrt{2}}& \frac{1}{\sqrt{2}}\\ \frac{1}{\sqrt{2}}& -\frac{1}{\sqrt{2}}\end{array}\right)\left(\begin{array}{c}{\theta }_1\\ {\theta }_2\end{array}\right)$$

重新定义坐标

$$\begin{array}{c}{q}_1=\frac{1}{\sqrt{2}}({\theta }_1+{\theta }_2)\\ q_2=\frac{1}{\sqrt{2}}({\theta }_1-{\theta }_2)\end{array}$$

所以

$$\left(\begin{array}{c}{\ddot{q}}_1\\ {\ddot{q}}_2\end{array}\right)=\left(\begin{array}{cc}{\omega }_1^2-{\omega }_2^2& 0\\ 0& -{\omega }_1^2-{\omega }_2^2\end{array}\right)\left(\begin{array}{c}{q}_1\\ q_2\end{array}\right)$$

结合初始条件

$$\begin{array}{c}{q}_1(0)=\frac{\alpha }{\sqrt{2}}{q}_2(0)=-\frac{\alpha }{\sqrt{2}}\\ {\dot{q}}_1(0)={\dot{q}}_2(0)=0\end{array}$$

所以

$$\begin{array}{c}{q}_1(t)=\frac{\alpha }{\sqrt{2}}\cos (\sqrt{-{\omega }_1^2+{\omega }_2^2}t)\\ q_2(t)=-\frac{\alpha }{\sqrt{2}}\cos (\sqrt{{\omega }_1^2+{\omega }_2^2}t)\end{array}$$

得到

$$\begin{array}{c}{\theta }_1(t)=\frac{\alpha }{2}(\cos (\sqrt{\frac{g}{l}}t)-\cos (\sqrt{\frac{g}{l}+2\frac{k}{m}}t))\\ {\theta }_2(t)=\frac{\alpha }{2}(\cos (\sqrt{\frac{g}{l}}t)+\cos (\sqrt{\frac{g}{l}+2\frac{k}{m}}t))\end{array}$$