Position of Cart attached to Wall by Spring under Damping/Overdamped/x = x0 at t = 0

From ProofWiki
Jump to navigation Jump to search

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 $\mathbf x$.


Let:

$a^2 = \dfrac k m$
$2 b = \dfrac c m$

Let $b > a$.


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:

$x = \dfrac {x_0} {m_1 - m_2} \left({m_1 e^{m_2 t} - m_2 e^{m_1 t} }\right)$

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

$m_1 = -b + \sqrt {b^2 - a^2}$
$m_2 = -b - \sqrt {b^2 - a^2}$


Such a system is defined as being overdamped.


Proof

From Position of Cart attached to Wall by Spring under Damping: Overdamped:

$(1): \quad x = C_1 e^{m_1 t} + C_2 e^{m_2 t}$

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

It remains to evaluate $C_1$ and $C_2$ under the given conditions.


Differentiating $(1)$ with respect to $t$ gives:

$(2): \quad x' = C_1 m_1 e^{m_1 t} + C_2 m_2 e^{m_2 t}$


Setting the initial condition $x = x_0$ when $t = 0$ in $(1)$:

\(\displaystyle x_0\) \(=\) \(\displaystyle C_1 e^0 + C_2 e^0\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle x_0\) \(=\) \(\displaystyle C_1 + C_2\)


Setting the initial condition $x' = 0$ when $t = 0$ in $(1)$:

\(\displaystyle 0\) \(=\) \(\displaystyle C_1 m_1 e^0 + C_2 m_2 e^0\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle 0\) \(=\) \(\displaystyle C_1 m_1 + C_2 m_2\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle 0\) \(=\) \(\displaystyle C_1 m_1 + \left({x_0 - C_1}\right) m_2\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle C_1 \left({m_1 - m_2}\right)\) \(=\) \(\displaystyle -x_0 m_2\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle C_1 \left({m_1 - m_2}\right)\) \(=\) \(\displaystyle -\dfrac {x_0 m_2} {m_1 - m_2}\)


Then we have:

\(\displaystyle x_0\) \(=\) \(\displaystyle C_1 + C_2\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle C_2\) \(=\) \(\displaystyle x_0 - C_1\)
\(\displaystyle \) \(=\) \(\displaystyle x_0 + \dfrac {x_0 m_2} {m_1 - m_2}\)
\(\displaystyle \) \(=\) \(\displaystyle x_0 \left({1 + \dfrac {m_2} {m_1 - m_2} }\right)\)
\(\displaystyle \) \(=\) \(\displaystyle x_0 \left({\dfrac {\left({m_1 - m_2}\right) + m_2} {m_1 - m_2} }\right)\)
\(\displaystyle \) \(=\) \(\displaystyle \dfrac {x_0 m_1} {m_1 - m_2}\)


Hence:

\(\displaystyle x\) \(=\) \(\displaystyle -\dfrac {x_0 m_2} {m_1 - m_2} e^{m_1 t} + \dfrac {x_0 m_1} {m_1 - m_2} e^{m_2 t}\)
\(\displaystyle \) \(=\) \(\displaystyle \dfrac {x_0} {m_1 - m_2} \left({m_1 e^{m_2 t} - m_2 e^{m_1 t} }\right)\)

$\blacksquare$


Sources