谐和激励

自由振动

\[ \begin{equation} \ddot{x} + 2 \xi \omega_0 \dot{x} + \omega_0^2 x = 0, \, \omega_0 = \sqrt{\frac{k}{m}}, \, \xi = \frac{c}{2 m \omega_0} \in \mathbb{R}^+ \end{equation} \]

代入 \(x(t) = e^{\lambda t}\)，得到特征方程

\[ \lambda^2 + 2 \xi \omega_0 \lambda + \omega_0^2 = 0 \]

1. \(\lambda_1, \lambda_2 \in \mathbb{R}\)（\(\xi > 1\)）：过阻尼 $$ x(t) = C_1 e^{\lambda_1 t} + C_2 e^{\lambda_2 t} $$
2. \(\lambda_1 = \lambda_2\)（\(\xi = 1\)）：临界阻尼 $$ x(t) = (C_1 + C_2 t) e^{-\omega_0 t} $$

3. \(\lambda_1, \lambda_2 \in \mathbb{C}\)（\(0 \leq \xi < 1\)）：欠阻尼 $$ \begin{equation} \lambda_{1,2} = -\xi \omega_0 \pm i \omega_0 \sqrt{1 - \xi^2} = -\xi \omega_0 \pm i \omega_{\mathrm{d}} \end{equation} $$

   \[ \begin{equation} \begin{aligned} x(t) &= e^{-\xi \omega_0 t} (C_1 \cos \omega_{\mathrm{d}} t + C_2 \sin \omega_{\mathrm{d}} t) \\ &= A e^{-\xi \omega_0 t} \sin(\omega_{\mathrm{d}} t + \varphi) \end{aligned} \end{equation} \]

   代入初始条件 \(x(0) = x_0, \dot{x}(0) = \dot{x}_0\) 可解出

   \[ \begin{equation} A = \sqrt{x_0^2 + \left(\frac{\dot{x}_0 + \xi \omega_0 x_0}{\omega_{\mathrm{d}}}\right)^2}, \quad \varphi = \arctan \frac{x_0 \omega_{\mathrm{d}}}{\dot{x}_0 + \xi \omega_0 x_0} \end{equation} \]

   对于单位脉冲响应 \(x_0 = 0, \dot{x}_0 = 1\)，有

   \[ \begin{equation} h(t) = x(t) = \frac{1}{\omega_{\mathrm{d}}} e^{-\xi \omega_0 t} \sin \omega_{\mathrm{d}} t \quad (t \geq 0) \end{equation} \]

测量阻尼比 \(\xi\) 的实验方法

• 测量两个振动幅值（这里以相邻的为例）\(A_1, A_2\)
• \(\eta = \ln \frac{A_2}{A_1} = \ln \frac{A e^{-\xi \omega_0 (t + T_{\mathrm{d}})}}{A e^{-\xi \omega_0 t}} = - \xi \omega_0 T_{\mathrm{d}}, \, T_{\mathrm{d}} = \frac{2\pi}{\omega_0 \sqrt{1 - \xi^2}}\)

\[ \xi = - \frac{\eta}{\omega_0 T_{\mathrm{d}}} = \frac{1}{\omega_0 T_{\mathrm{d}}} \ln \frac{A_1}{A_2} \approx \frac{1}{2\pi} \ln \frac{A_1}{A_2} \]

提高测量精度：测量 \(A_1, A_{n+1}\)，则

\[\xi = \frac{1}{n \omega_0 T_{\mathrm{d}}} \ln \frac{A_1}{A_{n+1}} \approx \frac{1}{2\pi n} \ln \frac{A_1}{A_{n+1}}\]

谐和激励

\[ \begin{equation} \ddot{x} + 2 \xi \omega_0 \dot{x} + \omega_0^2 x = \frac{F_0}{m} \cos(\omega_0 t + \varphi_0) \end{equation} \]

通解的振幅项指数衰减，很快消失，以特解为主导。

代入 \(\bar{x} = A \cos \omega t + B \sin \omega t\)，即可求得特解。也可使用复变方法。

\[ \begin{equation} \ddot{x} + 2 \xi \omega_0 \dot{x} + \omega_0^2 x = \frac{F_0}{m} e^{\mathrm{i} \omega t} \end{equation} \]

代入 \(\tilde{x} = \tilde{A} e^{\mathrm{i} \omega t}\)，解得

\[ \begin{equation} \begin{aligned} \tilde{A} &= \frac{F_0 / m}{\omega_0^2 - \omega^2 + 2 \mathrm{i} \xi \omega_0 \omega} && := H(\omega) \frac{F_0}{m} \\ \xrightarrow{\frac{F_0}{m} = \frac{F_0}{k} \omega_0^2} &= \frac{1}{1 - \eta^2 + 2 \mathrm{i} \xi \eta} \frac{F_0}{k} && := \beta(\eta) e^{- \mathrm{i} \varphi_0(\eta)} \frac{F_0}{k}, \quad \eta = \frac{\omega}{\omega_0} \end{aligned} \end{equation} \]

其中

\[ \begin{aligned} \beta(\eta) &= \frac{1}{\sqrt{(1 - \eta^2)^2 + (2 \xi \eta)^2}} \\ \varphi_0(\eta) &= \arctan \frac{2 \xi \eta}{1 - \eta^2} \end{aligned} \]

得到振动解

\[ \begin{equation} \tilde{x}(t) = \beta(\eta) \frac{F_0}{k} e^{\mathrm{i} (\omega t - \varphi_0(\eta))} \end{equation} \]

半带宽、品质因数

考虑地面运动的振动

1. 绝对坐标

\[ m \ddot{x} = -k (x - x_{\text{g}}) - c (\dot{x} - \dot{x}_{\text{g}}) \]

1. 相对坐标

\[ m (\ddot{\tilde{x}} + \ddot{x}_{\text{g}}) = -k \tilde{x} - c \dot{\tilde{x}} \]