91Ó°ÊÓ

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

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

We will transform the given equation:

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

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

$$\begin{gathered} y\text{'}\left(t\right)=\left(cos\mathrm{Ó\neg }t\right)\text{'} \\ =-sin\mathrm{Ó\neg }t×\left(\mathrm{Ó\neg }t\right)\text{'} \\ =-\mathrm{Ó\neg }sin\mathrm{Ó\neg }t \\ y\text{'}\text{'}\left(t\right)=\left(y\text{'}\left(t\right)\right)\text{'} \\ =(-\mathrm{Ó\neg }sin\mathrm{Ó\neg }t)\text{'} \\ =-\mathrm{Ó\neg }cos\mathrm{Ó\neg }t×\left(\mathrm{Ó\neg }t\right)\text{'} \\ =-{\mathrm{Ó\neg }}^{2}cos\mathrm{Ó\neg }t \end{gathered}$$

Now substituting this into equation, we have that:

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

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

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