Motion of Cart attached to Wall by Spring under Damping

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Theorem

Problem Definition

CartOnSpringWithDamping.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 in a medium which applies a damping force $F_d$ whose magnitude is proportional to the speed of $C$.

Let the force constant of $S$ be $k$.

Let the constant of proportion of the damping force $F_d$ be $c$.

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:

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


Proof

Let $\mathbf F_s$ be the force on $C$ exerted by the spring $S$.

Let $\mathbf F_d$ be the damping force.

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

$\mathbf F_s + \mathbf F_d = m \mathbf a$

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

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

By definition, the velocity $\mathbf v$ is defined as:

$\mathbf v = \dfrac {\mathrm d \mathbf x}{\mathrm d t}$

By Hooke's Law:

$\mathbf F = -k \mathbf x$

By definition of the damping force:

$\mathbf F_d = - c \mathbf v$

So:

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

$\blacksquare$


Sources