# Position of Cart attached to Wall by Spring under Damping

## Theorem

### Problem Definition Consider a cart $C$ of mass $m$ attached to a wall by means of a spring $S$.

Let $C$ be free to move along a straight line in a medium which applies a damping force $F_d$ whose magnitude is proportional to the speed of $C$.

Let the force constant of $S$ be $k$.

Let the constant of proportion of the damping force $F_d$ be $c$.

Let the displacement of $C$ at time $t$ from the equilibrium position be $\mathbf x$.

Let:

$a^2 = \dfrac k m$
$2 b = \dfrac c m$

Then the horizontal position of $C$ at time $t$ can be expressed as:

$x = \begin{cases} C_1 e^{m_1 t} + C_2 e^{m_1 t} & : b > a \\ & \\ C_1 e^{-a t} + C_2 t e^{-a t} & : b = a \\ & \\ e^{-b t} \left({C_1 \cos \alpha t + C_2 \sin \alpha t}\right) & : b < a \end{cases}$

where:

$C_1$ and $C_2$ depend upon the conditions of $C$ at time $t = 0$
$m_1$ and $m_2$ are the roots of the auxiliary equation $m^2 + 2 b + a^2 = 0$:
$m_1 = -b + \sqrt {b^2 - a^2}$
$m_2 = -b - \sqrt {b^2 - a^2}$
$\alpha = \sqrt {a^2 - b^2}$

## Proof

From Motion of Cart attached to Wall by Spring under Damping, the horizontal position of $C$ is given as:

$\dfrac {\d^2 \mathbf x} {\d t^2} + \dfrac c m \dfrac {\d \mathbf x} {\d t} + \dfrac k m \mathbf x = 0$

With the given substitutions $a$ and $b$, this resolves to:

$\dfrac {\d^2 \mathbf x} {\d t^2} + 2 b \dfrac {\d \mathbf x} {\d t} + a^2 \mathbf x = 0$

Recall that $m_1$ and $m_2$ are the roots of the auxiliary equation:

$m^2 + 2 b + a^2 = 0$
 $\displaystyle m_1, m_2$ $=$ $\displaystyle \dfrac {- 2 b \pm \sqrt {\paren {2 b}^2 - 4 a^2} } 2$ $\displaystyle$ $=$ $\displaystyle -b \pm \sqrt {b^2 - a^2}$

From the initial problem definition, we have that $k, m, c \in \R_{>0}$.

Hence $a, b \in \R_{>0}$.

Hence the nature of $m_1$ and $m_2$ is dependent upon whether $b > a$, $b = a$ or $b < a$.

### Overdamped

When $b > a$, we have $b^2 - a^2 > 0$ and so $m_1$ and $m_2$ are real and distinct.

$\mathbf x = C_1 e^{m_1 t} + C_2 e^{m_2 t}$

where

 $\displaystyle m_1$ $=$ $\displaystyle -b + \sqrt {b^2 - a^2}$ $\displaystyle m_2$ $=$ $\displaystyle -b - \sqrt {b^2 - a^2}$

$\Box$

### Critically Damped

When $b = a$, we have $b^2 - a^2 = 0$ and so:

$m_1 = m_2 = -b = -a$
$C_1 e^{-a x} + C_2 x e^{-a x}$

$\Box$

### Underdamped

When $b < a$, we have $b^2 - a^2 < 0$ and so:

 $\displaystyle m_1$ $=$ $\displaystyle -b + i \sqrt {a^2 - b^2}$ $\displaystyle m_2$ $=$ $\displaystyle -b - i \sqrt {a^2 - b^2}$
$\mathbf x = e^{-b t} \left({C_1 \cos \alpha t + C_2 \sin \alpha t}\right)$

where:

$\alpha = \sqrt {a^2 - b^2}$

$\blacksquare$