一般激励

一般激励

\[ \ddot{x} + 2 \xi \omega_0 \dot{x} + \omega_0^2 x = f(t) \]

最好的办法：Laplace 变换 \(F(s) = \int_0^{\infty} f(t) e^{-st} \, \mathrm{d}t\)

\[ \begin{aligned} \mathcal{L}[\ddot{x}] &= s^2 X(s) - s x(0) - \dot{x}(0) \\ \mathcal{L}[\dot{x}] &= s X(s) - x(0) \\ \mathcal{L}[x] &= X(s) \\ \end{aligned} \]

初始条件取 \(x(0) = 0, \, \dot{x}(0) = 0\)，则

\[ X(s) = \frac{F(s)}{s^2 + 2 \xi \omega_0 s + \omega_0^2} \]

逆变换得到

\[ \begin{aligned} x(t) = \mathcal{L}^{-1}[X(s)] &= f(t) * h(t) \\ &= \int_0^t f(t - \tau) h(\tau) \, \mathrm{d}\tau \\ \end{aligned} \]

此式称为 Duhamel 积分。其中 \(h(t)\) 为脉冲响应函数，满足

\[ \mathcal{L}[h(t)] = \frac{1}{s^2 + 2 \xi \omega_0 s + \omega_0^2} \]

这等价于

\[ \left \{ \begin{aligned} & \ddot{h} + 2 \xi \omega_0 \dot{h} + \omega_0^2 h = 0 \\ & h(0) = 0, \quad \dot{h}(0) = 1/m \end{aligned} \right. \]

解析解为

\[ h(t) = \frac{1}{m \sqrt{1 - \xi^2}} e^{-\xi \omega_0 t} \sin(\sqrt{1 - \xi^2} \omega_0 t) \quad (t \geq 0) \]

物理上理解：

\[ \left \{ \begin{aligned} & \ddot{\tilde{x}} + 2 \xi \omega_0 \dot{\tilde{x}} + \omega_0^2 \tilde{x} = f(\tau) \mathrm{d}\tau \, \delta(t - \tau) \\ & t > \tau \end{aligned} \right. \]

响应谱 \(\frac{|X|_{\max}}{\delta} \sim \frac{T}{T_0}, \delta = F_0/k\) 弹簧静伸缩量

---

广义单自由度系统

能量等效

• 动能：等效质量 \(\bar{M}\)
• 势能：等效刚度 \(\bar{K}\)

以连续梁为例，任意一点的位移为 \(u(x,t) = \psi(x) y(t)\)

\[ \begin{aligned} T &= \int_0^L \frac{1}{2} \rho(x) A(x) \, \mathrm{d}x \, \left( \psi(x) \dot{y}(t) \right)^2 \\ &= \frac{1}{2} \left[\int_0^L \rho(x) A(x) \psi^2(x) \, \mathrm{d}x \right] \dot{y}^2(t) \\ &= \frac{1}{2} \bar{M} \dot{y}^2(t) \\[2ex] U &= \int_0^L \frac{1}{2} E(x) I(x) \left( \frac{\partial^2}{\partial x^2} \left( \psi(x) y(t) \right) \right)^2 \, \mathrm{d}x \\ &= \frac{1}{2} \left[ \int_0^L E(x) I(x) \left( \frac{\mathrm{d}^2 \psi}{\mathrm{d} x^2} \right)^2 \, \mathrm{d}x \right] y^2(t) \\ &= \frac{1}{2} \bar{K} y^2(t) \end{aligned} \]

\(\psi(x)\) 满足的边界条件（固定端）：\(\begin{cases} \psi(0) = 0 \\ \psi'(0) = 0 \\ \psi(L) = 1 s\end{cases}\)

若加一集中质量 \(m\)，则动能多一项，势能不变

拉格朗日方程

如果

\[ \begin{aligned} & L = T - V \\ & \frac{\mathrm{d}}{\mathrm{d}t} \left( \frac{\partial L}{\partial \dot{y}} \right) - \frac{\partial L}{\partial y} = Q \end{aligned} \]