# Motion of Cart attached to Wall by Spring

## 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 with zero friction.

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

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

Then the motion of $C$ is described by the second order ODE:

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

## Proof

By Newton's Second Law of Motion, the force on $C$ equals its mass times its acceleration:

$\mathbf F = m \mathbf a$
$\mathbf a = \dfrac {\mathrm d^2 \mathbf x}{\mathrm d t^2}$

By Hooke's Law:

$\mathbf F = -k \mathbf x$

So:

 $\displaystyle m \mathbf a$ $=$ $\displaystyle -k \mathbf x$ $\displaystyle \implies \ \$ $\displaystyle m \dfrac {\mathrm d^2 \mathbf x}{\mathrm d t^2}$ $=$ $\displaystyle -k \mathbf x$ $\displaystyle \implies \ \$ $\displaystyle \dfrac {\mathrm d^2 \mathbf x} {\mathrm d t^2} + \dfrac k m \mathbf x$ $=$ $\displaystyle 0$

$\blacksquare$