Position of Cart attached to Wall by Spring

Theorem

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 $\mathbf x$.

Then the horizontal position of $C$ at time $t$ can be expressed as:

$x = C_1 \cos \alpha t + C_2 \sin \alpha t$

where:

$C_1$ and $C_2$ depend upon the conditions of $C$ at time $t = 0$

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

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:

$x = x_0 \cos \alpha t$

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

As both $k$ and $m$ are positive, $\dfrac k m$ can be expressed as:

$\dfrac k m = \alpha^2$

for some $\alpha \in \R_{>0}$.

Hence:

$\dfrac {\d^2 x} {\d t^2} + \alpha^2 x = 0$

$x = C_1 \cos \alpha t + C_2 \sin \alpha t$

Hence the result.

$\blacksquare$

1972: George F. Simmons: Differential Equations ... (previous) ... (next): $\S 3.20$: Vibrations in Mechanical Systems: $(4)$