Q1E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

Verify that for $\mathit{b}\mathbf{=}\mathbf{0}$ and ${\mathit{F}}_{\mathit{e}\mathit{x}\mathit{t}}\mathbf{\left(}\mathit{t}\mathbf{\right)}\mathbf{=}\mathbf{0}$, equation (3) has a solution of the form $\mathbf{y}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}\mathbf{cos\omega t}\mathbf{,}\mathbf{\omega }\mathbf{=}\sqrt{\mathbf{k}\mathbf{/}\mathbf{m}}$

Step-by-Step Solution

Answer

For $y\left(t\right)=cos\omega t$ find the second derivative and substitute in an equation $my\text{'}\text{'}+ky=0$. Since $\omega =\sqrt{\frac{k}{m}}$ , the function $y\left(t\right)=cos\omega t$ satisfies the given equation and therefore it is a solution.

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.

 $\mathit{F}\mathbf{=}\mathbf{-}\mathit{k}\mathit{x}$

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

Step 2: Transforming the equation

For $b=0$ and $F_{ext}\left(t\right)\equiv 0$ the equation (3) transforms into $my\text{'}\text{'}+ky=0$

We will transform the given equation:

$$\begin{gathered} my\text{'}\text{'}+ky=0 \\ my\text{'}\text{'}=-ky \\ \frac{ y\text{'}\text{'}}{y}=-\frac{k}{m} \dots \left(a\right) \end{gathered}$$

 

Step 3: Finding the derivatives

In order to verify that $y\left(t\right)=cos\omega t$, where $\omega =\sqrt{\frac{k}{m}}$ is a solution of previous equation, first you will find those derivatives appearing in the given equation.

$$\begin{gathered} y\text{'}\left(t\right)=\left(cos\omega t\right)\text{'} \\ =-sin\omega t\times \left(\omega t\right)\text{'} \\ =-\omega sin\omega t \\ y\text{'}\text{'}\left(t\right)=\left(y\text{'}\left(t\right)\right)\text{'} \\ =(-\omega sin\omega t)\text{'} \\ =-\omega cos\omega t\times \left(\omega t\right)\text{'} \\ =-{\omega }^{2}cos\omega t \end{gathered}$$

Step 4: Substituting the values

Now substituting this into equation , we have that:

$$\begin{gathered} \frac{y\text{'}\text{'}}{y\text{'}}=\frac{-{\omega }^{2}cos\omega t}{cos\omega t} \\ =-{\omega }^2 \\ =-{\left(\sqrt{\frac{k}{m}}\right)}^2 \\ =-\frac{k}{m} \end{gathered}$$

 

So, $y\left(t\right)=cos\omega t$ where $\omega =\sqrt{\frac{k}{m}}$ satisfies the given equation and therefore it is a solution of it.

Hence, $\omega =\sqrt{\frac{k}{m}}$, the function $y\left(t\right)=cos\omega t$ satisfies the given equation and therefore it is a solution.