91Ó°ÊÓ

The general solution to (1) in the case $0<{\mathrm{b}}^2<4\mathrm{mk}$:

$\mathrm{y}\left(\mathrm{t}\right)={\mathrm{Ae}}^{-\left(\frac{\mathrm{b}}{2\mathrm{m}}\right)\mathrm{t}}\sin \left(\frac{\sqrt{4\mathrm{mk}-{\mathrm{b}}^2}}{2\mathrm{m}}\mathrm{t}+\mathrm{ϕ}\right)+\frac{{\mathrm{F}}_0}{\sqrt{{\left(\mathrm{k}-\mathrm{m}{\mathrm{γ}}^2\right)}^2+{\mathrm{b}}^2{\mathrm{γ}}^2}}\sin \left(\mathrm{γ}\mathrm{t}+\mathrm{θ}\right)$

The angular frequency:

The amplitude of the steady-state solution to equation (1) depends on the angular frequency of the forcing function and it is given by $\mathrm{A}\left(\mathrm{γ}\right)={\mathrm{F}}_0\mathrm{M}\left(\mathrm{γ}\right)$, where

$\mathrm{M}\left(\mathrm{γ}\right):=\frac{1}{\sqrt{{\left(\mathrm{k}-\mathrm{m}{\mathrm{γ}}^2\right)}^2+{\mathrm{b}}^2{\mathrm{γ}}^2}}                 …\left(1\right)$

The undamped system:

The system is governed by $\mathrm{m}\frac{{\mathrm{d}}^2\mathrm{y}}{{\mathrm{dt}}^2}+\mathrm{ky}={\mathrm{F}}_0\cos \mathrm{γ}\mathrm{t}$. And the homogenous solution of it is. And the corresponding homogeneous equation is ${\mathrm{y}}_{\mathrm{h}}\left(\mathrm{t}\right)=\mathrm{Asin}\left(\mathrm{Ó\neg }\mathrm{t}+\mathrm{Ï•}\right),\mathrm{Ó\neg }:=\sqrt{\frac{\mathrm{k}}{\mathrm{m}}}$

And the correct form is ${\mathrm{y}}_{\mathrm{p}}\left(\mathrm{t}\right)={\mathrm{A}}_1\mathrm{tcos}\mathrm{Ó\neg }\mathrm{t}+{\mathrm{A}}_2\mathrm{tsin}\mathrm{Ó\neg }\mathrm{t}$.

So, the general solution of the system is$\mathrm{y}\left(\mathrm{t}\right)=\mathrm{Asin}\left(\mathrm{Ó\neg }\mathrm{t}+\mathrm{Ï•}\right)+\frac{{\mathrm{F}}_0}{2\mathrm{m}\mathrm{Ó\neg }}\mathrm{tsin}\mathrm{Ó\neg }\mathrm{t}$

Referring to equation (8): one gets,

$\mathrm{y}\left(\mathrm{t}\right)={\mathrm{Ae}}^{-\left(\frac{\mathrm{b}}{2\mathrm{m}}\right)\mathrm{t}}\sin \left(\frac{\sqrt{4\mathrm{mk}-{\mathrm{b}}^2}}{2\mathrm{m}}\mathrm{t}+\mathrm{ϕ}\right)+\frac{{\mathrm{F}}_0}{\sqrt{{\left(\mathrm{k}-\mathrm{m}{\mathrm{γ}}^2\right)}^2+{\mathrm{b}}^2{\mathrm{γ}}^2}}\sin \left(\mathrm{γ}\mathrm{t}+\mathrm{θ}\right)$

In the overdamped motion case, the auxiliary equation has two real roots.

$$\begin{gathered} {\mathrm{r}}_1=-\frac{\mathrm{b}}{2\mathrm{m}}+\frac{1}{2\mathrm{m}}\sqrt{{\mathrm{b}}^2-4\mathrm{mk}} \\ {\mathrm{r}}_2=-\frac{\mathrm{b}}{2\mathrm{m}}-\frac{1}{2\mathrm{m}}\sqrt{{\mathrm{b}}^2-4\mathrm{mk}} \end{gathered}$$

Since ${\mathrm{b}}^2>4\mathrm{mk}$the homogeneous solution to $\mathrm{m}\frac{{\mathrm{d}}^2\mathrm{y}}{{\mathrm{dt}}^2}+\mathrm{b}\frac{\mathrm{dy}}{\mathrm{dt}}+\mathrm{ky}={\mathrm{F}}_0\cos \left(\mathrm{γ}\mathrm{t}\right)$and it becomes:

$$\begin{gathered} {\mathrm{y}}_{\mathrm{h}}\left(\mathrm{t}\right)={\mathrm{c}}_1{\mathrm{e}}^{{\mathrm{r}}_1\mathrm{t}}+{\mathrm{c}}_2{\mathrm{e}}^{{\mathrm{r}}_2\mathrm{t}} \\ {\mathrm{y}}_{\mathrm{p}}\left(\mathrm{t}\right)=\frac{{\mathrm{F}}_0}{{\left(\mathrm{k}-\mathrm{m}{\mathrm{γ}}^2\right)}^2+{\mathrm{b}}^2{\mathrm{γ}}^2}\left[\left(\mathrm{k}-\mathrm{m}{\mathrm{γ}}^2\right)\cos \left(\mathrm{γ}\mathrm{t}\right)+\mathrm{b}\mathrm{γ}\sin \left(\mathrm{γ}\mathrm{t}\right)\right] \end{gathered}$$

Then, the general solution is;

$\mathrm{y}\left(\mathrm{t}\right)={\mathrm{c}}_1{\mathrm{e}}^{\left(-\frac{\mathrm{b}}{2\mathrm{m}}+\frac{1}{2\mathrm{m}}\sqrt{{\mathrm{b}}^2-4\mathrm{mk}}\right)\mathrm{t}}+{\mathrm{c}}_2{\mathrm{e}}^{\left(-\frac{\mathrm{b}}{2\mathrm{m}}-\frac{1}{2\mathrm{m}}\sqrt{{\mathrm{b}}^2-4\mathrm{mk}}\right)\mathrm{t}}+\frac{{\mathrm{F}}_0}{{\left(\mathrm{k}-\mathrm{m}{\mathrm{γ}}^2\right)}^2+{\mathrm{b}}^2{\mathrm{γ}}^2}\left[\left(\mathrm{k}-\mathrm{m}{\mathrm{γ}}^2\right)\cos \left(\mathrm{γ}\mathrm{t}\right)+\mathrm{b}\mathrm{γ}\sin \left(\mathrm{γ}\mathrm{t}\right)\right]$