The Mass-Spring System

Problem 1. An object with mass $m$ is loaded onto a mass-spring system. After loading the object that is at rest, the length of the spring is $s$ . Let $x = x(t)$ denote the displacement of the object at time $t$ . In Newtonian mechanics, we are given the following information:

Hooke’s law tells us that the force that the spring acts on the object is $F_{\mathrm{S}}= -k(s+x)$ , where $k > 0$ is a spring constant of proportionality.
Newton’s law of gravitation tells us that the gravitational force acting on the object is $F_{\mathrm{G}} = mg$ , where $g \approx 9.81$ is the gravitational constant of proportionality (at least on earth).
The damping force is given by $F_{\mathrm{D}} = -\gamma x'$ , where $\gamma >0$ is a damping constant of proportionality.

Let the damping force be characterised by the constant $\gamma > 0$ and the force arising from the spring be characterised by $k > 0$ . Newton’s second law states that at any point in time,

$mx'' = F_{\mathrm{S}} + F_{\mathrm{G}} + F_{\mathrm{D}}.$

When setting up the equations, we have the boundary conditions $mg = ks$ .

Derive a homogeneous second-order differential equation in terms of $m, \gamma, k$ and $x$ and its derivatives. Hence, find an expression for $x$ in terms of $t$ in the case:

$\gamma^2 - 4mk > 0$ ,
$\gamma^2 - 4mk = 0$ ,
$\gamma^2 - 4mk < 0$ .

What is the long-run behaviour of $x$ ? What would happen if there was no damping force (i.e. $\gamma = 0$ )?

(Click for Solution)

Solution. By Newton’s second law,

$mx'' = -k(s+x)+mg+(-\gamma x').$

Since $mg = ks$ ,

$mx'' = -kx-\gamma x' \quad \Rightarrow \quad mx'' + \gamma x' + kx = 0.$

Consider the auxiliary equation $mr^2 + \gamma r + k = 0$ , where we used $r$ to reduce ambiguity. By the quadratic equation, the roots $r_{\pm}$ are given by

$\displaystyle r_{\pm} = \frac{-\gamma \pm \sqrt{\gamma^2 - 4mk}}{2m}.$

In the case $\gamma^2 - 4mk > 0$ , the roots $r_{\pm}$ are real and distinct and negative. Thus,

$x(t) = C_1 e^{r_- t} + C_2 e^{r_+ t}.$

This is known as an overdamped system.

In the case $\gamma^2 - 4mk = 0$ , the roots $r_{\pm} = -\gamma/(2m)$ are real and repeated and negative. Thus,

$x(t) = (C_1 t + C_2)e^{rt}.$

This is known as a critically damped system.

In the case $\gamma^2 - 4mk < 0$ , the roots $r_{\pm} = p \pm iq$ are complex conjugates with $p = -\gamma/(2m) < 0$ . Thus,

$x(t) = e^{pt} (C_1 \sin(qt) + C_2 \cos(qt)).$

This is known as an underdamped system.

In all instances, the negative term $-r/(2m)$ induces all exponential terms to $\to 0$ , so that $x(t) \to 0$ as $t \to \infty$ . In the case $\gamma = 0$ , we only have the case $-4mk < 0$ , so that