Position of Cart attached to Wall by Spring under Damping

Problem Definition
Let:
 * $a^2 = \dfrac k m$
 * $2 b = \dfrac c m$

Then the horizontal position of $C$ at time $t$ can be expressed as:
 * $x = \begin{cases}

C_1 e^{m_1 t} + C_2 e^{m_1 t} & : b > a \\ & \\ C_1 e^{-a t} + C_2 t e^{-a t} & : b = a \\ & \\ e^{-b t} \left({C_1 \cos \alpha t + C_2 \sin \alpha t}\right) & : b < a \end{cases}$

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


 * $m_1$ and $m_2$ are the roots of the auxiliary equation $m^2 + 2 b + a^2 = 0$:
 * $m_1 = -b + \sqrt {b^2 - a^2}$
 * $m_2 = -b - \sqrt {b^2 - a^2}$


 * $\alpha = \sqrt {a^2 - b^2}$

Proof
From Motion of Cart attached to Wall by Spring under Damping, the horizontal position of $C$ is given as:
 * $\dfrac {\mathrm d^2 \mathbf x} {\mathrm d t^2} + \dfrac c m \dfrac {\mathrm d \mathbf x} {\mathrm d t} + \dfrac k m \mathbf x = 0$

With the given substitutions $a$ and $b$, this resolves to:


 * $\dfrac {\mathrm d^2 \mathbf x} {\mathrm d t^2} + 2 b \dfrac {\mathrm d \mathbf x} {\mathrm d t} + a^2 \mathbf x = 0$

This is a homogeneous linear second order ODE with constant coefficients.

Recall that $m_1$ and $m_2$ are the roots of the auxiliary equation:
 * $m^2 + 2 b + a^2 = 0$

By Solution to Quadratic Equation: Real Coefficients:

From the initial problem definition, we have that $k, m, c \in \R_{>0}$.

Hence $a, b \in \R_{>0}$.

Overdamped
When $b > a$, we have $b^2 - a^2 > 0$ and so $m_1$ and $m_2$ are real and distinct.

So from Solution of Constant Coefficient Homogeneous LSOODE: Real Roots of Auxiliary Equation:
 * $\mathbf x = C_1 e^{m_1 t} + C_2 e^{m_2 t}$

where

Critically Damped
When $b = a$, we have $b^2 - a^2 = 0$ and so:
 * $m_1 = m_2 = -b = -a$

So from Solution of Constant Coefficient Homogeneous LSOODE: Equal Real Roots of Auxiliary Equation:
 * $C_1 e^{-a x} + C_2 x e^{-a x}$

Underdamped
When $b < a$, we have $b^2 - a^2 < 0$ and so:

So from Solution of Constant Coefficient Homogeneous LSOODE: Complex Roots of Auxiliary Equation:


 * $\mathbf x = e^{-b t} \left({C_1 \cos \alpha t + C_2 \sin \alpha t}\right)$

where:
 * $\alpha = \sqrt {a^2 - b^2}$