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

$x = \dfrac {x_0} \alpha e^{-b t} \left({\alpha \cos \alpha t + b \sin \alpha t}\right)$

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

Such a system is defined as being underdamped.

## Proof

$(2): \quad x = e^{-b t} \left({C_1 \cos \alpha t + C_2 \sin \alpha t}\right)$

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:

$(2): \quad x' = -b e^{-b t} \left({C_1 \cos \alpha t + C_2 \sin \alpha t}\right) + e^{-b t} \left({-\alpha C_1 \sin \alpha t + \alpha C_2 \cos \alpha t}\right)$

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

 $\displaystyle x_0$ $=$ $\displaystyle e^0 \left({C_1 \cos 0 + C_2 \sin 0}\right)$ $\displaystyle$ $=$ $\displaystyle C_1$

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

 $\displaystyle 0$ $=$ $\displaystyle -b e^0 \left({C_1 \cos 0 + C_2 \sin 0}\right) + e^0 \left({-\alpha C_1 \sin 0 + \alpha C_2 \cos 0}\right)$ $\displaystyle$ $=$ $\displaystyle -b C_1 + \alpha C_2$ $\displaystyle \leadsto \ \$ $\displaystyle C_2$ $=$ $\displaystyle \frac {b C_1} \alpha$ $\displaystyle$ $=$ $\displaystyle \frac {b x_0} \alpha$

Hence:

 $\displaystyle x$ $=$ $\displaystyle e^{-b t} \left({x_0 \cos \alpha t + \frac {b x_0} \alpha \sin \alpha t}\right)$ $\displaystyle$ $=$ $\displaystyle \dfrac {x_0} \alpha e^{-b t} \left({\alpha \cos \alpha t + b \sin \alpha t}\right)$

$\blacksquare$