Q5E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

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 $\mathbf{ }\mathbf{\infty }$ ?

Step-by-Step Solution

Answer

Answer

 

If we put   in the given equation then we will get 

Step 1: Definition of Hooke’s Law

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 $\mathbf{N}\mathbf{/}\mathbf{m}$.

Step 2: Finding the auxiliary equation.

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;

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

Step 3: Find the roots of the auxiliary equation.

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 $\mathbf{y}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}{\mathbf{e}}^{\mathbf{(}\mathbf{-}\mathbf{3}\mathbf{t}\mathbf{)}}\mathbf{sin}\mathbf{(}\sqrt{\mathbf{3}}\mathbf{t}\mathbf{)}$.

Step 4: Put t → ∞

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.