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

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Theorem

Problem Definition

CartOnSpring.png

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

$\blacksquare$


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