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

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

---

ラプラス変換と計算

$$\begin{eqnarray} \mathfrak{L}\left[ \frac{\mathrm{d^2}x}{\mathrm{d^2}t} \right.&+&\left.\frac{c}{m}\frac{\mathrm{d}x}{\mathrm{d}t} \right.&+&\left.\frac{k}{m}x \right]&=0 \\\mathfrak{L}\left[ \frac{\mathrm{d^2}x}{\mathrm{d^2}t} \right] &+&\mathfrak{L}\left[ \frac{c}{m}\frac{\mathrm{d}x}{\mathrm{d}t} \right] &+&\mathfrak{L}\left[ \frac{k}{m}x\right] &=0 \;\ldots\; \href{https://shikitenkai.blogspot.com/2021/04/gtb-ht.html}{ \mathfrak{L}\left[ a g(t)+b h(t) \right]=a\mathfrak{L}\left[ {g\left( t \right)} \right] + b\mathfrak{L}\left[ {h\left( t \right)} \right] } \\ s^2X-sx_0 -v_0 &+&\frac{c}{m}\left(sX-x_0 \right) &+&\frac{k}{m}X &=0 \\ \\&&&&&\;\ldots\;\mathfrak{L}\left[x\right]=X \\&&&&&\;\ldots\;\href{https://shikitenkai.blogspot.com/2021/04/fracmathrmdfmathrmdt.html}{\mathfrak{L}\left[ \frac{\mathrm{d}x}{\mathrm{d}t}\right] =s^2X-x_0,\;x_0=x(0)} \\&&&&&\;\ldots\;\href{https://shikitenkai.blogspot.com/2021/04/blog-post_62.html}{\mathfrak{L}\left[ \frac{\mathrm{d^2}x}{\mathrm{d^2}t}\right] =s^2X-sx_0 -v_0,\;v_0=x'(0)} \\ s^2X-sx_0 -v_0 &+&s\frac{c}{m}X-\frac{c}{m}x_0 &+&\frac{k}{m}X &=0 \end{eqnarray}$$ $$\begin{eqnarray} s^2X+s\frac{c}{m}X+\frac{k}{m}X &=& sx_0 +v_0 +\frac{c}{m}x_0 \\ \left(s^2+s\frac{c}{m}+\frac{k}{m}\right)X &=& sx_0 +v_0 +\frac{c}{m}x_0 \\ X&=&\frac{sx_0 +v_0 +\frac{c}{m}x_0 }{s^2+s\frac{c}{m}+\frac{k}{m}} \\&=&\frac{sx_0 +v_0 +\frac{c}{m}x_0 }{\left(s-\lambda_1\right)\left(s-\lambda_2\right)} \;\ldots\;\lambda_{1,2}=s^2+s\frac{c}{m}+\frac{k}{m}の解 \\&=&\frac{C_1 }{s-\lambda_1}+\frac{C_2 }{s-\lambda_2} \\&=&\frac{C_1 \left(s-\lambda_2\right)+C_2 \left(s-\lambda_1\right)}{\left(s-\lambda_1\right)\left(s-\lambda_2\right)} \\&=&\frac{(C_1 +C_2 )s-(C_1 \lambda_2+C_2 \lambda_1)}{\left(s-\lambda_1\right)\left(s-\lambda_2\right)} \end{eqnarray}$$

---

逆ラプラス変換

$$\begin{eqnarray} \mathfrak{L}^{-1}\left[X\right]&=&\mathfrak{L}^{-1}\left[C_1 \frac{1}{s-\lambda_1}+C_2 \frac{1}{s-\lambda_2}\right] \\&=&C_1 \mathfrak{L}^{-1}\left[\frac{1}{s-\lambda_1}\right]+C_2 \mathfrak{L}^{-1}\left[\frac{1}{s-\lambda_2}\right] \\x(t)&=&C_1 e^{\lambda_1 t}+C_2 e^{\lambda_2 t} \\&&\;\ldots\;\mathfrak{L}^{-1}\left[X\right]=x(t) \\&&\;\ldots\;\mathfrak{L}^{-1}\left[ \frac{1}{s+a} \right]=e^{-at} \end{eqnarray}$$