Question

A model of a spring/mass system is $\mathbf{4}\mathit{x}\mathbf{"}\mathbf{+}\mathit{t}\mathit{x}\mathbf{=}\mathbf{0}$. By inspection of the differential equation only, discuss the behavior of the system over a long period of time

Short answer

It implies that the velocity of the mass is constant since its acceleration is zero and that the mass never returns to the equilibrium position as the spring stretches continuously.

Step-by-step solution

Definition:

Spring Constant or force constant is defined as the applied force if the displacement in the spring is unity.

Behavior in long term:

Consider a model of spring as below.

$4x^{\text{'}\text{'}}+e^{-0.1t}x=0$

For large values of$t$,localid="1668407530818" $e^{-0.1t}\to 0$(that is,localid="1668407536140" $\underset{t\to \infty }{\lim }{e}^{-0.1t}=0$).

Therefore, the above differential equation becomes,

localid="1668407545419" $$\begin{gathered} 4x^{\text{'}\text{'}}=0 \\ {x}^{\text{'}\text{'}}=0 \end{gathered}$$

This implies that, the speed of the system remains constant after a long period of time as the system acceleration becomes zero.

Integrate on both sides to obtain that,

localid="1668407552803" $\int x^{\text{'}\text{'}}dt=\int 0dt$

Here, localid="1668407559232" $x^{\text{'}}=c$andlocalid="1668407565141" $c$is constant of integration.

Again, integrate on both sides to obtain that,

localid="1668407572616" $\begin{array}{rcl}\int x^{\text{'}}dt& =& \int cdtx\\ & =& ct+k\end{array}$

Here,localid="1668407581421" $k$is an integrating constant.

Thus, for long period of time, the restoring force will have decayed to the point where the spring is incapable of returning the mass, and the spring will simply keep on stretching.