物理

❴栖岸计划❵有趣的推导

$m \ddot{ξ}_i = k ξ$

或

$m_1 \ddot{x}_1 + \cdots m_n \ddot{x}_n = [k_1 x_1 + \cdots k_n x_n]$

上式加点上，可变为矩阵平衡形式，即转动双自由度常用

写出列向量 $ \vec{x} = \begin{bmatrix} x_1 \\ \vdots \\ x_n \end{bmatrix} ，\vec{k} = \begin{bmatrix} k_1 \\ \vdots \\ k_n \end{bmatrix} $

$\vec{M} \cdot \ddot{\vec{x}} = -\vec{K} \cdot \vec{x}$

猜解 $ \vec{x} = \vec{a} e^{-i\omega t} $（阶跃符合简振形式）

$-\vec{a} \omega^2 \vec{M} e^{-i\omega t} = -\vec{a} \vec{K} \cdot \vec{a} e^{-i\omega t}$

$\vec{a} \cdot (\omega^2 \vec{M} - \vec{K}) = 0$

$\vec{a} \ne 0 \Rightarrow |\omega^2 \vec{M} - \vec{K}| = 0$

带入实例

(1.1)

$\ddot{x}_1 = a_{11} x_1 + a_{12} x_2 + a_{13} x_3$

$\ddot{x}_2 = a_{21} x_1 + a_{22} x_2 + a_{23} x_3$

$\ddot{x}_3 = a_{31} x_1 + a_{32} x_2 + a_{33} x_3$

$\quad(1.1)$

有简振模 $ξ _1,ξ _2, ξ_3 $

$ξ_1 = b_{11} x_1 + b_{12} x_2 + b_{13} x_3 $

$ξ_2 = b_{21} x_1 + b_{22} x_2 + b_{23} x_3 $

$ξ_3 = b_{31} x_1 + b_{32} x_2 + b_{33} x_3$

$\quad (1.2)$

$\ddot{ξ}_1 = \omega_1^2 ξ_1, \quad \ddot{ξ}_2 = \omega_2^2 ξ_2, \quad \ddot{ξ}_3 = \omega_3^2 ξ_3 \quad (1.3)$

代入 (1.1) 代入 (1.2)

$\ddot{ξ}_1=[b_{11}a_{11} +b_{12} a_{21}+b_{13}a_{31}]x_1$

$+[b_{11}a_{12} +b_{12}a_{22}+b_{13}a_{32}]x_2$

$+[b_{11}a_{13}+b_{12}a_{23}+b_{13}a_{33}] x_3 \quad (1.8)$

(1.3) (1.2) 联立

$\ddot{\xi}_1 = \omega_1^2 b_{11} x_1 + \omega_1^2 b_{12} x_2 + \omega_1^2 b_{13} x_3 \quad (1.5)$

(1.5) (1.8) 联立

$[(a_{11} - \omega^2) b_{11} + \cdots] x_1 + [\cdots + (a_{22} - \omega^2) b_{22} + \cdots] x_2 + [\cdots + (a_{33} - \omega^2) b_{33}] x_3 = 0$

有

$\begin{bmatrix}a_{11} - \omega^2, & a_{21}, & a_{31} \\a_{12}, & a_{22} - \omega^2, & a_{32} \\a_{13}, & a_{23}, & a_{33} - \omega^2\end{bmatrix}$

$\begin{bmatrix}b_{11} \\ b_{21} \\ b_{31}\end{bmatrix}$

$= 0 \quad (1.7)$

以此类推

$\begin{bmatrix}a_{11} - \omega^2, & a_{21}, & a_{31} \\a_{12}, & a_{22} - \omega^2, & a_{32} \\a_{13}, & a_{23}, & a_{33} - \omega^2\end{bmatrix}$

$\begin{bmatrix}b_{21} \\ b_{22} \\ b_{23}\end{bmatrix}$

$= 0$