Q6E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

Question: An external force $\mathbf{F}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}\mathbf{2}\mathbf{cos}\mathbf{2}\mathbf{t}$ is applied to a mass-spring system with,$\mathbf{m}\mathbf{=}\mathbf{1}\mathbf{,}\mathbf{b}\mathbf{=}\mathbf{0}$ and, $\mathbf{k}\mathbf{=}\mathbf{4}$which is initially at rest; i.e., $\mathbf{y}\left(\mathbf{0}\right)\mathbf{=}\mathbf{0}\mathbf{,}\mathbf{y}\mathbf{\text{'}}\left(\mathbf{0}\right)\mathbf{=}\mathbf{0}$. Verify that $\mathbf{y}\left(t\right)\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{tsin}\mathbf{2}\mathbf{t}\mathbf{ }\mathbf{,}\mathbf{y}\left(\mathbf{0}\right)\mathbf{=}\mathbf{0}\mathbf{,}\mathbf{y}\mathbf{\text{'}}\left(\mathbf{0}\right)\mathbf{=}\mathbf{0}$gives the motion of this spring. What will eventually (asincreases) happen to the spring?

Step-by-Step Solution

Answer

Answer

 

If keeps on increasing then the spring will break.

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;

 

, F = -kx

 

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

Step 2: Finding the values of the mass-spring oscillator.

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

  

Substitute the given values of $\mathbf{F}\left(\mathbf{t}\right)\mathbf{=}\mathbf{2}\mathbf{cos}\mathbf{2}\mathbf{t}$ and $\mathbf{m}\mathbf{=}\mathbf{1}\mathbf{,}\mathbf{b}\mathbf{=}\mathbf{0}\mathbf{,}\mathit{k}\mathbf{=}\mathbf{4}$ &  in the differential equation then we get,$\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{4}\mathbf{y}\mathbf{=}\mathbf{2}\mathbf{cos}\mathbf{2}\mathbf{t}$

 

Step 3: Verify y = 1 2 tsin 2 t

Let $\mathbf{y}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{tsin}\mathbf{2}\mathbf{t}.$

Differentiateand get; 

$$\begin{gathered} \mathbf{y}\text{'}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{sin}\mathbf{2}\mathbf{t}\mathbf{+}\mathbf{tcos}\mathbf{2}\mathbf{t}\mathbf{\dots }\mathbf{\left(}\mathbf{1}\mathbf{\right)} \\ \mathbf{y}\text{'}\text{'}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}\mathbf{cos}\mathbf{2}\mathbf{t}\mathbf{+}\mathbf{cos}\mathbf{2}\mathbf{t}\mathbf{-}\mathbf{2}\mathbf{tsin}\mathbf{2}\mathbf{t}\mathbf{\dots }\mathbf{(}2\mathbf{)} \end{gathered}$$

 Substitute (1) and (2) in the differential equation,

$$\begin{gathered} \mathbf{2}\mathbf{cos}\mathbf{2}\mathbf{t}\mathbf{-}\mathbf{2}\mathbf{tsin}\mathbf{2}\mathbf{t}\mathbf{+}\mathbf{4}\left(\frac{\mathbf{1}}{\mathbf{2}}\mathbf{tsin}\mathbf{2}\mathbf{t}\right)\mathbf{=}\mathbf{2}\mathbf{cos}\mathbf{2}\mathbf{t} \\ \mathbf{2}\mathbf{cos}\mathbf{2}\mathbf{t}\mathbf{-}\mathbf{2}\mathbf{tsin}\mathbf{2}\mathbf{t}\mathbf{+}\mathbf{2}\mathbf{tsin}\mathbf{2}\mathbf{t}\mathbf{=}\mathbf{2}\mathbf{cos}\mathbf{2}\mathbf{t} \\ \mathbf{2}\mathbf{cos}\mathbf{2}\mathbf{t}\mathbf{=}\mathbf{2}\mathbf{cos}\mathbf{2}\mathbf{t} \end{gathered}$$

 

Therefore, RHS = LHS.

 

Therefore $\mathbf{y}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{tsin}\mathbf{2}\mathbf{t}$ is the solution for the given differential equation.

 

Now from the solution, it can be observed that  increases ifincreases. So, if  keeps on increasing then the spring will break.