Question:Verify that the exponentially damped sinusoid $\mathbf{y}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}{\mathbf{e}}^{\mathbf{-}\mathbf{3}\mathbf{t}}\mathbf{sin}\mathbf{(}\sqrt{\mathbf{3}\mathbf{t}}\mathbf{)}$is a solution to the equation if ${\mathbf{F}}_{\mathbf{ext}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}\mathbf{0}\mathbf{,}\mathbf{m}\mathbf{=}\mathbf{1}\mathbf{,}\mathbf{b}\mathbf{=}\mathbf{6}\mathbf{}$and k= 12. What is the limit of this solution as role="math" localid="1655378672841" $\mathbf{}\mathbf{\infty }$?

Hooke’s law states that the strain of the material is proportional to the applied stress within the elastic limit of that material.

$\mathbf{F}\mathbf{=}\mathbf{-}\mathbf{kx}$

In the equation, is the force, is the extension length, and is the constant of proportionality known as spring constant in $\mathbf{N}\mathbf{/}\mathbf{m}$.

The differential equation for the mass-spring oscillator is $\mathbf{my}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{by}\mathbf{\text{'}}\mathbf{+}\mathbf{ky}\mathbf{=}{\mathbf{F}}_{\mathbf{ext}}$.

Substitute the given values of

$F_{ext}\left(t\right)=0,m=1,b=6$ and k =12

The differential equation becomes
.$F_{ext}\left(t\right)=0,m=1,b=6$

The auxiliary equation of the differential is;

${\mathbf{m}}^{\mathbf{2}}\mathbf{+}\mathbf{6}\mathbf{m}\mathbf{+}\mathbf{12}\mathbf{=}\mathbf{0}$

Find the roots of the auxiliary equation using the formula;

$\mathbf{m}\mathbf{=}\frac{\mathbf{-}\mathbf{b}\mathbf{\pm }\sqrt{{\mathbf{b}}^{\mathbf{2}}\mathbf{-}\mathbf{4}\mathbf{ac}}}{\mathbf{2}\mathbf{a}}$

Here and .

$$\begin{gathered} \mathbf{m}\mathbf{=}\frac{\mathbf{-}\mathbf{6}\mathbf{\pm }\sqrt{{\mathbf{6}}^{\mathbf{2}}\mathbf{-}\mathbf{4}\mathbf{\left(}\mathbf{1}\mathbf{\right)}\mathbf{\left(}\mathbf{12}\mathbf{\right)}}}{\mathbf{2}\mathbf{\left(}\mathbf{1}\mathbf{\right)}} \\ \mathbf{=}\frac{\mathbf{-}\mathbf{6}\mathbf{\pm }\mathbf{2}\mathbf{i}\sqrt{\mathbf{3}}}{\mathbf{2}} \\ \mathbf{=}\mathbf{-}\mathbf{3}\mathbf{\pm }\sqrt{\mathbf{3}}\mathbf{i} \end{gathered}$$

Therefore, the solution is $\mathbf{y}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}{\mathbf{e}}^{\mathbf{(}\mathbf{-}\mathbf{3}\mathbf{t}\mathbf{)}}\mathbf{sin}\mathbf{(}\sqrt{\mathbf{3}}\mathbf{t}\mathbf{)}$.

If $\mathbf{t}\to \infty$, then ${\mathbf{e}}^{\mathbf{(}\mathbf{-}\mathbf{3}\mathbf{t}\mathbf{)}}\mathbf{=}\mathbf{0}$.

Therefore,

$\underset{l\to a}{\lim }y\left(t\right))=\underset{l\to a}{\lim }{e}^3\sin \begin{array}{cc}\left(\sqrt{3)}\right)& \\ =e^a& \sin \left(\sqrt{3}\times 0\right))=0\end{array}$

Hence, if $\mathbf{t}\to \infty$then and therefore it is a solution.