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

## Contents

## 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 $\mathbf x$.

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:

- $x = x_0 \cos \alpha t$

where $\alpha = \sqrt {\dfrac k m}$.

## Proof

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

- $(1): \quad x = C_1 \cos \alpha t + C_2 \sin \alpha t$

where:

- $C_1$ and $C_2$ depend upon the conditions of $C$ at time $t = 0$
- $\alpha = \sqrt {\dfrac k m}$

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

- $(2): \quad x' = -C_1 \alpha \sin \alpha t + C_2 \alpha \cos \alpha t$

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

\(\displaystyle x_0\) | \(=\) | \(\displaystyle C_1 \cos 0 + C_2 \sin 0\) | |||||||||||

\(\displaystyle \implies \ \ \) | \(\displaystyle x_0\) | \(=\) | \(\displaystyle C_1\) |

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

\(\displaystyle 0\) | \(=\) | \(\displaystyle -C_1 \alpha \sin 0 + C_2 \alpha \cos 0\) | |||||||||||

\(\displaystyle \implies \ \ \) | \(\displaystyle 0\) | \(=\) | \(\displaystyle C_2\) |

Hence:

- $x = x_0 \cos \alpha t$

where $\alpha = \sqrt {\dfrac k m}$.

$\blacksquare$

## Sources

- 1972: George F. Simmons:
*Differential Equations*... (previous) ... (next): $\S 3.20$: Vibrations in Mechanical Systems: $(6)$