91Ó°ÊÓ

The differential equation is,

$\mathbf{my}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{by}\mathbf{\text{'}}\mathbf{+}\mathbf{ky}\mathbf{=}\mathbf{f}\left(\mathbf{t}\right)                             ......\left(\mathbf{1}\right)$

From the given information,

$$\begin{gathered} \mathbf{m}\mathbf{=}\frac{\mathbf{32}}{\mathbf{32}}\mathbf{=}\mathbf{1} \\ \mathbf{32}\mathbf{=}\mathbf{k}\frac{\mathbf{6}}{\mathbf{12}} \\ \mathbf{k}\mathbf{=}\mathbf{64} \\ \mathbf{b}\mathbf{=}\mathbf{2} \end{gathered}$$

And $\mathbf{f}\left(\mathbf{t}\right)\mathbf{=}\mathbf{4}\mathbf{cos}\mathbf{8}\mathbf{t}$

Substitute the all value of m, k, b and f(t) in the equation (1),

$\begin{array}{rcl}\mathbf{my}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{by}\mathbf{\text{'}}\mathbf{+}\mathbf{ky}& \mathbf{=}& \mathbf{f}\left(\mathbf{t}\right)\\ \left(\mathbf{1}\right)\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\left(\mathbf{2}\right)\mathbf{y}\mathbf{\text{'}}\mathbf{+}\left(\mathbf{64}\right)\mathbf{y}& \mathbf{=}& \mathbf{4}\mathbf{cos}\mathbf{8}\mathbf{t}\\ \mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{2}\mathbf{y}\mathbf{\text{'}}\mathbf{+}\mathbf{64}\mathbf{y}& \mathbf{=}& \mathbf{4}\mathbf{cos}\mathbf{8}\mathbf{t}                           ......\left(\mathbf{2}\right)\\ & & \end{array}$

The auxiliary equation for the above equation,

$\begin{array}{rcl}{\mathbf{m}}^{\mathbf{2}}\mathbf{+}\mathbf{2}\mathbf{m}\mathbf{+}\mathbf{64}& \mathbf{=}& \mathbf{0}\\ \mathbf{m}& \mathbf{=}& \frac{\mathbf{-}\mathbf{2}\mathbf{Â\pm }\sqrt{\mathbf{4}\mathbf{-}\mathbf{4}\mathbf{×}\mathbf{64}}}{\mathbf{2}}\\ \mathbf{m}& \mathbf{=}& \mathbf{-}\mathbf{1}\mathbf{Â\pm }\mathbf{i}\sqrt{\mathbf{63}}\\ & & \end{array}$

The root of auxiliary equation is,

${\mathbf{m}}_{\mathbf{1}}\mathbf{=}\mathbf{-}\mathbf{1}\mathbf{+}\mathbf{i}\sqrt{\mathbf{63}}\mathbf{,}   {\mathbf{m}}_{\mathbf{2}}\mathbf{=}\mathbf{-}\mathbf{1}\mathbf{-}\mathbf{i}\sqrt{\mathbf{63}}$

The complimentary solution of the given equation is,

${\mathbf{y}}_{\mathbf{c}}\mathbf{=}{\mathbf{e}}^{\mathbf{-}\mathbf{t}}\left[\mathbf{Acos}\sqrt{\mathbf{63}}\mathbf{t}\mathbf{+}\mathbf{Bsin}\sqrt{\mathbf{63}}\mathbf{t}\right]                          ......\left(\mathbf{3}\right)$

Assume, the particular solution of equation (1),

${\mathbf{y}}_{\mathbf{p}}\left(\mathbf{t}\right)\mathbf{=}\frac{\mathbf{1}}{{\mathbf{D}}^{\mathbf{2}}\mathbf{+}\mathbf{2}\mathbf{D}\mathbf{+}\mathbf{64}}\left(\mathbf{4}\mathbf{cos}\mathbf{8}\mathbf{t}\right)                          ......\left(\mathbf{4}\right)$

Consider,

$\begin{array}{rcl}\mathbf{X}& \mathbf{=}& \frac{\mathbf{1}}{{\mathbf{D}}^{\mathbf{2}}\mathbf{+}\mathbf{2}\mathbf{D}\mathbf{+}\mathbf{64}}\left(\mathbf{4}\mathbf{cos}\mathbf{8}\mathbf{t}\right)\\ \mathbf{Y}& \mathbf{=}& \frac{\mathbf{1}}{{\mathbf{D}}^{\mathbf{2}}\mathbf{+}\mathbf{2}\mathbf{D}\mathbf{+}\mathbf{64}}\left(\mathbf{4}\mathbf{sin}\mathbf{8}\mathbf{t}\right)\\ & & \\ \mathbf{X}\mathbf{+}\mathbf{iY}& \mathbf{=}& \frac{\mathbf{1}}{{\mathbf{D}}^{\mathbf{2}}\mathbf{+}\mathbf{2}\mathbf{D}\mathbf{+}\mathbf{64}}\left(\mathbf{4}\mathbf{cos}\mathbf{8}\mathbf{t}\mathbf{+}\mathbf{i}\mathbf{4}\mathbf{sin}\mathbf{8}\mathbf{t}\right)\\ & \mathbf{=}& \mathbf{4}\frac{\mathbf{1}}{{\mathbf{D}}^{\mathbf{2}}\mathbf{+}\mathbf{2}\mathbf{D}\mathbf{+}\mathbf{64}}\left({\mathbf{e}}^{\mathbf{8}\mathbf{it}}\right)\\ & \mathbf{=}& \mathbf{4}\frac{\mathbf{1}}{{\left(\mathbf{8}\mathbf{i}\right)}^{\mathbf{2}}\mathbf{+}\mathbf{2}\left(\mathbf{8}\mathbf{i}\right)\mathbf{+}\mathbf{64}}\left({\mathbf{e}}^{\mathbf{8}\mathbf{it}}\right)\\ & \mathbf{=}& \mathbf{4}\frac{\mathbf{1}}{\mathbf{-}\mathbf{64}\mathbf{+}\mathbf{16}\mathbf{i}\mathbf{+}\mathbf{64}}\left({\mathbf{e}}^{\mathbf{8}\mathbf{it}}\right)\\ & \mathbf{=}& \frac{\mathbf{4}}{\mathbf{16}\mathbf{i}}\left({\mathbf{e}}^{\mathbf{8}\mathbf{it}}\right)\\ & \mathbf{=}& \frac{\mathbf{1}}{\mathbf{4}}\left(\mathbf{sin}\mathbf{8}\mathbf{t}\mathbf{-}\mathbf{icos}\mathbf{8}\mathbf{t}\right)\\ & & \end{array}$

Therefore, the real and imaginary parts are;

$\mathbf{X}\mathbf{=}\frac{\mathbf{1}}{\mathbf{4}}\mathbf{sin}\mathbf{8}\mathbf{t}  and  \mathbf{Y}\mathbf{=}\mathbf{-}\frac{\mathbf{1}}{\mathbf{4}}\mathbf{cos}\mathbf{8}\mathbf{t}$

From the equation (4),

${\mathbf{y}}_{\mathbf{p}}\left(t\right)\mathbf{=}\frac{\mathbf{1}}{\mathbf{4}}\mathbf{sin}\mathbf{8}\mathbf{t}$

Therefore, the general solution is,

$$\begin{gathered} \mathbf{y}\mathbf{=}{\mathbf{y}}_{\mathbf{c}}\left(\mathbf{t}\right)\mathbf{+}{\mathbf{y}}_{\mathbf{p}}\left(\mathbf{t}\right) \\ \mathbf{y}\mathbf{=}{\mathbf{e}}^{\mathbf{-}\mathbf{t}}\left[\mathbf{Acos}\sqrt{\mathbf{63}}\mathbf{t}\mathbf{+}\mathbf{Bsin}\sqrt{\mathbf{63}}\mathbf{t}\right]\mathbf{+}\frac{\mathbf{1}}{\mathbf{4}}\mathbf{sin}\mathbf{8}\mathbf{t}               ......\left(\mathbf{5}\right) \end{gathered}$$

Thus, steady state solution of the system for $\mathbf{t}→∞$

$$\begin{gathered} \mathbf{y}\mathbf{=}{\mathbf{e}}^{\mathbf{-}\mathbf{t}}\left[\mathbf{Acos}\sqrt{\mathbf{63}}\mathbf{t}\mathbf{+}\mathbf{Bsin}\sqrt{\mathbf{63}}\mathbf{t}\right]\mathbf{+}\frac{\mathbf{1}}{\mathbf{4}}\mathbf{sin}\mathbf{8}\mathbf{t} \\ \mathbf{y}\mathbf{=}\frac{\mathbf{1}}{\mathbf{4}}\mathbf{sin}\mathbf{8}\mathbf{t}\mathbf{,}  {\mathbf{e}}^{\mathbf{-}\mathbf{t}}→\mathbf{0}  \mathbf{as} \mathbf{t}→∞ \end{gathered}$$

Therefore, the solution,

$\mathbf{y}\mathbf{=}\frac{\mathbf{1}}{\mathbf{4}}\mathbf{sin}\mathbf{8}\mathbf{t}$

Thus, the frequency is$\mathbf{f}\mathbf{=}\frac{\sqrt{\mathbf{63}}}{\mathbf{2}\mathbf{Ï€}}$.