バネマスダンパー系，運動方程式，ラプラス変換，逆ラプラス変換，入力なし

バネマスダンパー系

運動方程式

$$\begin{array}{rclrclr}\ddot{x}& +& \frac{c}{m}\dot{x}& +& \frac{k}{m}x& =& 0\\ \frac{{\mathrm{d}}^{2}x}{{\mathrm{d}}^{2}t}& +& \frac{c}{m}\frac{\mathrm{d}x}{\mathrm{d}t}& +& \frac{k}{m}x& =& 0\end{array}$$

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ラプラス変換と計算

$$\begin{array}{rclrcl}\mathfrak{L}[\frac{{\mathrm{d}}^{2}x}{{\mathrm{d}}^{2}t}& +& \frac{c}{m}\frac{\mathrm{d}x}{\mathrm{d}t}& +& \frac{k}{m}x]& =0\\ \mathfrak{L}\left[\frac{{\mathrm{d}}^{2}x}{{\mathrm{d}}^{2}t}\right]& +& \mathfrak{L}\left[\frac{c}{m}\frac{\mathrm{d}x}{\mathrm{d}t}\right]& +& \mathfrak{L}\left[\frac{k}{m}x\right]& =0\dots \mathfrak{L}[ag(t)+bh(t)]=a\mathfrak{L}\left[g\left(t\right)\right]+b\mathfrak{L}\left[h\left(t\right)\right]\\ s^{2}X-s{x}_0-v_0& +& \frac{c}{m}(sX-x_0)& +& \frac{k}{m}X& =0\\ \\ & & & & & \dots \mathfrak{L}\left[x\right]=X\\ & & & & & \dots \mathfrak{L}\left[\frac{\mathrm{d}x}{\mathrm{d}t}\right]=s^{2}X-x_0,x_0=x(0)\\ & & & & & \dots \mathfrak{L}\left[\frac{{\mathrm{d}}^{2}x}{{\mathrm{d}}^{2}t}\right]=s^{2}X-s{x}_0-v_0,v_0=x^{\prime }(0)\\ s^{2}X-s{x}_0-v_0& +& s\frac{c}{m}X-\frac{c}{m}{x}_0& +& \frac{k}{m}X& =0\end{array}$$ $$\begin{array}{rcl}{s}^{2}X+s\frac{c}{m}X+\frac{k}{m}X& =& s{x}_0+v_0+\frac{c}{m}{x}_0\\ (s^2+s\frac{c}{m}+\frac{k}{m})X& =& s{x}_0+v_0+\frac{c}{m}{x}_0\\ X& =& \frac{s{x}_0+v_0+\frac{c}{m}{x}_0}{s^2+s\frac{c}{m}+\frac{k}{m}}\\ & =& \frac{s{x}_0+v_0+\frac{c}{m}{x}_0}{(s-{\lambda }_1)(s-{\lambda }_2)}\dots {\lambda }_{1,2}=s^2+s\frac{c}{m}+\frac{k}{m}\mathrm{の}\mathrm{解}\\ & =& \frac{C_1}{s-{\lambda }_1}+\frac{C_2}{s-{\lambda }_2}\\ & =& \frac{C_1(s-{\lambda }_2)+C_2(s-{\lambda }_1)}{(s-{\lambda }_1)(s-{\lambda }_2)}\\ & =& \frac{(C_1+C_2)s-(C_1{\lambda }_2+C_2{\lambda }_1)}{(s-{\lambda }_1)(s-{\lambda }_2)}\end{array}$$

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逆ラプラス変換

$$\begin{array}{rcl}{\mathfrak{L}}^{-1}\left[X\right]& =& {\mathfrak{L}}^{-1}[C_1\frac{1}{s-{\lambda }_1}+C_2\frac{1}{s-{\lambda }_2}]\\ & =& C_1{\mathfrak{L}}^{-1}\left[\frac{1}{s-{\lambda }_1}\right]+C_2{\mathfrak{L}}^{-1}\left[\frac{1}{s-{\lambda }_2}\right]\\ x(t)& =& C_1{e}^{{\lambda }_{1}t}+C_2{e}^{{\lambda }_{2}t}\\ & & \dots {\mathfrak{L}}^{-1}\left[X\right]=x(t)\\ & & \dots {\mathfrak{L}}^{-1}\left[\frac{1}{s+a}\right]=e^{-at}\end{array}$$