バネマスダンパー系，λ1,λ2を求める

バネマスダンパー系

${\lambda }_{1,2}\mathrm{を}\mathrm{求}\mathrm{め}\mathrm{る}$

$$\begin{array}{r}\frac{s{x}_0+v_0+\frac{c}{m}{x}_0}{s^2+s\frac{c}{m}+\frac{k}{m}}=\frac{(C_1+C_2)s-(C_1{\lambda }_2+C_2{\lambda }_1)}{(s-{\lambda }_1)(s-{\lambda }_2)}\end{array}$$ $$\begin{array}{r}\\ s^2+s\frac{c}{m}+\frac{k}{m}& =& (s-{\lambda }_1)(s-{\lambda }_2)\cdots \mathrm{分}\mathrm{母}\mathrm{に}\mathrm{つ}\mathrm{い}\mathrm{て}\mathrm{考}\mathrm{え}\mathrm{る}\\ {\lambda }_{1,2}& =& \frac{-\frac{c}{m}\pm \sqrt{{\left(\frac{c}{m}\right)}^2-4\cdot 1\cdot \frac{k}{m}}}{2\cdot 1}\\ & =& -\frac{c}{2m}\pm \frac{1}{2}\sqrt{{\left(\frac{c}{m}\right)}^2-4\frac{k}{m}}\\ & =& -\frac{c}{2m}\pm \sqrt{\frac{1}{4}({\left(\frac{c}{m}\right)}^2-4\frac{k}{m})}\\ & =& -\frac{c}{2m}\pm \sqrt{{\left(\frac{c}{2m}\right)}^2-\frac{k}{m}}\\ & =& -\gamma \pm \sqrt{{\gamma }^2-{\omega }_0^2}\\ & & \dots \gamma =\frac{c}{2m},{\omega }_0^2=\frac{k}{m}\end{array}$$