双自由度系统

无阻尼自由振动

柔度法振动方程

$$\begin{array}{r}{y}_1=-m_1{\ddot{y}}_1{\delta }_{11}-m_2{\ddot{y}}_2{\delta }_{12}\\ y_2=-m_1{\ddot{y}}_1{\delta }_{21}-m_2{\ddot{y}}_2{\delta }_{22}\end{array}$$

设 $y_1$ , $y_2$ 为如下形式：

$$\begin{array}{r}{y}_1={\rho }_1\sin (\omega t+\alpha )\\ y_2={\rho }_2\sin (\omega t+\alpha )\end{array}$$

代入并消去共同项 $\sin (\omega t+\alpha )$ 后得到：

$$\begin{array}{r}{\rho }_1=m_1{\delta }_{11}{\omega }^2{\rho }_1+m_2{\delta }_{12}{\omega }^2{\rho }_2\\ {\rho }_2=m_1{\delta }_{21}{\omega }^2{\rho }_1+m_2{\delta }_{22}{\omega }^2{\rho }_2\end{array}$$

为使方程有解，则要求行列式为 $0$ ：

$$\begin{array}{r}\left|\begin{array}{cc}{\displaystyle m_1{\delta }_{11}-\frac{1}{{\omega }^2}}& m_2{\delta }_{12}\\ m_1{\delta }_{21}& {\displaystyle m_2{\delta }_{22}-\frac{1}{{\omega }^2}}\end{array}\right|=0\end{array}$$

设 ${\displaystyle \lambda =\frac{1}{{\omega }^2}}$ ，可解得 ${\lambda }_1$ ， ${\lambda }_2$ ，并可分别求出一个 ${\displaystyle \frac{{\rho }_1}{{\rho }_2}}$ ，对应不同的振型。

无阻尼受迫振动

$$\begin{array}{r}{y}_1=[p_1(t)-m_1{\ddot{y}}_1]{\delta }_{11}+[p_2(t)-m_2{\ddot{y}}_2]{\delta }_{12}\\ y_2=[p_1(t)-m_1{\ddot{y}}_1]{\delta }_{21}+[p_2(t)-m_2{\ddot{y}}_2]{\delta }_{22}\end{array}$$

这里考虑同频率简谐振动

$$\begin{array}{r}{p}_1=P_1\sin (\theta t)\\ p_2=P_2\sin (\theta t)\end{array}$$

再考虑到

$$\begin{array}{r}{\mathrm{\Delta }}_1=P_1{\delta }_{11}+P_2{\delta }_{12}\\ {\mathrm{\Delta }}_2=P_1{\delta }_{21}+P_2{\delta }_{22}\end{array}$$

则原方程可化为

$$\begin{array}{r}{\rho }_1=m_1{\delta }_{11}{\omega }^2{\rho }_1+m_2{\delta }_{12}{\omega }^2{\rho }_2+{\mathrm{\Delta }}_1\\ {\rho }_2=m_1{\delta }_{21}{\omega }^2{\rho }_1+m_2{\delta }_{22}{\omega }^2{\rho }_2+{\mathrm{\Delta }}_2\end{array}$$

由线性代数克莱默法则得出方程的解。

同频同相位

外力和惯性力同时取到最大值。由此可计算结构最大的弯矩，并得出动力放大系数 $\text{MF}$ 。