Position of Cart attached to Wall by Spring

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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 $\mathbf x$.


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 $t = 0$
$\alpha = \sqrt {\dfrac k m}$


$x = x_0$ at time $t = 0$

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:

$x = x_0 \cos \alpha t$


Proof

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

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

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:

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

and so from Second Order ODE: $y'' + k^2 y = 0$:

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

Hence the result.

$\blacksquare$


Sources