# Position of Cart attached to Wall by Spring under Damping/Critically Damped/x = x0 at t = 0

## 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 $b = a$.

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:

$x = x_0 e^{-a t} \left({1 + a t}\right)$

Such a system is defined as being critically damped.

## Proof

$x = C_1 e^{-a t} + C_2 t e^{-a t}$

where $C_1$ and $C_2$ depend upon the conditions of $C$ at time $t = 0$.

It remains to evaluate $C_1$ and $C_2$ under the given conditions.

Differentiating $(1)$ with respect to $t$ gives:

$(2): \quad x' = -a C_1 e^{-a t} + C_2 e^{-a t} - a C_2 t e^{-a t}$

Setting the initial condition $x = x_0$ when $t = 0$ in $(1)$:

 $\displaystyle x_0$ $=$ $\displaystyle C_1 e^0 + C_2 e^0 \times 0$ $\displaystyle \leadsto \ \$ $\displaystyle x_0$ $=$ $\displaystyle C_1$

Setting the initial condition $x' = 0$ when $t = 0$ in $(1)$:

 $\displaystyle 0$ $=$ $\displaystyle -a C_1 e^0 + C_2 e^0 - a C_2 e^0 \times 0$ $\displaystyle \leadsto \ \$ $\displaystyle 0$ $=$ $\displaystyle -a C_1 + C_2$ $\displaystyle \leadsto \ \$ $\displaystyle C_2$ $=$ $\displaystyle a C_1$ $\displaystyle$ $=$ $\displaystyle a x_0$

Hence:

 $\displaystyle x$ $=$ $\displaystyle x_0 e^{-a t} + a x_0 t e^{-a t}$ $\displaystyle$ $=$ $\displaystyle x_0 e^{-a t} \left({1 + a t}\right)$

$\blacksquare$