# Position of Cart attached to Wall by Spring under Damping/Overdamped

## 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:

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

Let $b > a$.

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

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

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 $m^2 + 2 b + a^2 = 0$:
$m_1 = -b + \sqrt {b^2 - a^2}$
$m_2 = -b - \sqrt {b^2 - a^2}$

Such a system is defined as being overdamped.

### Initial Conditions: $x = x_0$ at $t = 0$

Let $C$ be pulled aside to $x = x_0$ and released from stationary at time $t = 0$.

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

$x = \dfrac {x_0} {m_1 - m_2} \left({m_1 e^{m_2 t} - m_2 e^{m_1 t} }\right)$

## 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 $b < a$.

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

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

$\blacksquare$