Question

A model of a spring/mass system is $\mathbf{4}{\mathit{x}}^{\mathbf{\text{'}}\mathbf{\text{'}}}\mathbf{+}\mathit{t}\mathit{x}\mathbf{=}\mathbf{0}$. By inspection of the differential equation only, discuss the behavior of the system over a long period of time.

Short answer

The system approaches to equilibrium over the time.

Step-by-step solution

Definition:

A differential equation is an equation with one or more derivatives of a function.

Behavior:

Given a second-order differential equation that models a spring/mass system.

$4x^{\text{'}\text{'}}+tx=0$ ..... (1)

The period of the system is given by,

$T=\frac{2\pi }{\omega }$

Here,$\omega =\sqrt{\frac{k}{m}}$

By differential equation in equation (1), we have $k=t$and $m=4$.

So the period is,

$\begin{array}{rcl}T& =& \frac{2\pi }{\sqrt{\frac{t}{4}}}\\ & =& \frac{2\pi }{\sqrt{t}}\\ & =& \frac{2\pi \cdot 2}{\sqrt{4}}\\ & =& \frac{4\pi }{t}\end{array}$

As $t\to \infty$, the period slowly approaches zero which means that the mass slowly gets closer and closer to the equilibrium point over time.