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?

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

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,$\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{4}\mathbf{y}\mathbf{=}\mathbf{2}\mathbf{cos}\mathbf{2}\mathbf{t}$

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.