旋转失衡
$m \ddot{x} + c \dot{x} + kx = m e \omega^2 \sin \omega t \qquad \Longleftrightarrow \qquad$ 简谐运动 $\frac{X_0 k}{F_0} = \frac{1}{\sqrt{(1 - \tilde{\omega}^2)^2 + (2\xi \tilde{\omega})^2}}$
$\Rightarrow$ 稳态解: $x(t) = X_0 \sin(\omega t - \varphi)$
$X_0 = \frac{me \omega^2}{\sqrt{(k - m \omega^2)^2 + (c \omega)^2}} = \frac{m e \omega^2 / k}{\sqrt{(1 - \tilde{\omega}^2)^2 + (2\xi \tilde{\omega})^2}}$
$\Theta = \arctan \frac{c \omega}{k - m \omega^2} = \arctan \frac{2 \xi \tilde{\omega}}{1 - \tilde{\omega}^2}$
加速度幅值放大因子 $\mu_0 = \frac{\tilde{\omega}^2}{\sqrt{(1 - \tilde{\omega}^2)^2 + (2\xi \tilde{\omega})^2}} = \mu \tilde{\omega}^2 = \frac{mX_0 \omega^2}{me}$
粘性阻尼系统
频率比 $\tilde{\omega} = \frac{\omega}{\omega_n} > \frac{1}{\sqrt{1-2\xi^2}} > 1$
放大率 $\frac{M X_0}{me} = \frac{1}{2 \xi \sqrt{1 - \xi^2}}$
转子旋转与临界转速
$R = \frac{\omega \tilde{\omega}^2}{\sqrt{(1 - \tilde{\omega}^2)^2 + (2\xi \tilde{\omega})^2}}$
临界转速: $\omega_{cr} = \frac{\omega_n}{\sqrt{1 - 2\xi^2}}$ => 小阻尼 $\omega_{cr} = \omega_n = \sqrt{\frac{k}{m}}$
基础激励与隔振
幅值放大因子 $\Lambda® = \frac{r^2}{\sqrt{(1 - r^2)^2 + (2\xi r)^2}}$
相位差 $\Theta® = \arctan \frac{2 \xi r}{1 - r^2}$
$r = \frac{\omega}{\omega_n}$, $\rho_1 e^{j \Theta_1} = \rho_2 e^{j \Theta_2}$
$\beta_1 = \frac{1 + (2\xi r)^2}{(1 - r^2)^2 + (2\xi r)^2}$, $\Theta_2 = \arctan(2\xi r)$
力传递率 $S = \frac{F_T}{F_0} = \sqrt{\frac{1 + (2\xi r)^2}{(1 - r^2)^2 + (2\xi r)^2}}$
运动方程: $m \ddot{x} + c \dot{x} + kx = F_0 \cos \omega t$
稳态响应: $x(t) = X \cos(\omega t - \varphi)$
$X = \frac{F_0}{\sqrt{(k - m \omega^2)^2 + (c \omega)^2}}$ $\varphi = \tan^{-1} (\frac{\omega c}{k - m \omega^2})$
传递到基础的力：$F_T(t) = kx(t) + cx(t) = k X \cos(\omega t - \varphi) - c \omega X \sin(\omega t - \varphi)$
=> 其幅值 $F_T = [(kx)^2 + (c \dot{x})^2]^{\frac{1}{2}} = X \sqrt{k^2 + \omega^2 c^2}$
位移传递率：$(k - m \omega^2 + jc \omega) X e^{j\omega t} = (k + jc\omega) Y e^{j\omega t}$
$T_d = \frac{Y}{X} = \sqrt{\frac{1 + 4 \xi^2 \tilde{\omega}^2}{(1 - \tilde{\omega}^2)^2 + (2 \xi \tilde{\omega})^2}}$
隔振效率 $\nu$
分贝表示：$\Psi = 10 \log_{10}{\frac{1}{S^2}} (dB)$
百分比表示: $\Psi = (1 - S) \times 100%$
测振器原理
把振动物体的测量工作, 转换为测量惯性质量元件相对于外壳的
强迫振动的工作，可以测量位移、速度或加速度