Position of Cart attached to Wall by Spring/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 horizontal 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 $(1)$:
| \(\ds x_0\) | \(=\) | \(\ds C_1 \cos 0 + C_2 \sin 0\) | ||||||||||||
| \(\ds \leadsto \ \ \) | \(\ds x_0\) | \(=\) | \(\ds C_1\) |
Setting the initial condition $x' = 0$ when $t = 0$ in $(1)$:
| \(\ds 0\) | \(=\) | \(\ds -C_1 \alpha \sin 0 + C_2 \alpha \cos 0\) | ||||||||||||
| \(\ds \leadsto \ \ \) | \(\ds 0\) | \(=\) | \(\ds C_2\) |
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: $(6)$