Motion 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 motion of $C$ is described by the second order ODE:

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


Proof

By Newton's Second Law of Motion, the force on $C$ equals its mass times its acceleration:

$\mathbf F = m \mathbf a$

By Acceleration is Second Derivative of Displacement with respect to Time:

$\mathbf a = \dfrac {\mathrm d^2 \mathbf x}{\mathrm d t^2}$

By Hooke's Law:

$\mathbf F = -k \mathbf x$

So:

\(\displaystyle m \mathbf a\) \(=\) \(\displaystyle -k \mathbf x\)
\(\displaystyle \implies \ \ \) \(\displaystyle m \dfrac {\mathrm d^2 \mathbf x}{\mathrm d t^2}\) \(=\) \(\displaystyle -k \mathbf x\)
\(\displaystyle \implies \ \ \) \(\displaystyle \dfrac {\mathrm d^2 \mathbf x} {\mathrm d t^2} + \dfrac k m \mathbf x\) \(=\) \(\displaystyle 0\)

$\blacksquare$


Sources