Motion of Cart attached to Wall by Spring

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 $\mathbf x$.

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

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

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:

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

By Hooke's Law:

$\mathbf F = -k \mathbf x$

So:

\(\ds m \mathbf a\)

\(=\)

\(\ds -k \mathbf x\)

\(\ds \leadsto \ \ \)

\(\ds m \dfrac {\d^2 \mathbf x} {\d t^2}\)

\(=\)

\(\ds -k \mathbf x\)

\(\ds \leadsto \ \ \)

\(\ds \dfrac {\d^2 \mathbf x} {\d t^2} + \dfrac k m \mathbf x\)

\(=\)

\(\ds 0\)

$\blacksquare$

Sources

• 1972: George F. Simmons: Differential Equations ... (previous) ... (next): $\S 3.20$: Vibrations in Mechanical Systems: $(2)$