Position of Cart attached to Wall by Spring under Damping/Overdamped/x = x0 at t = 0
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 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:
- $x = \dfrac {x_0} {m_1 - m_2} \paren {m_1 e^{m_2 t} - m_2 e^{m_1 t} }$
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 $(1)$:
\(\ds x_0\) | \(=\) | \(\ds C_1 e^0 + C_2 e^0\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds x_0\) | \(=\) | \(\ds C_1 + C_2\) |
Setting the initial condition $x' = 0$ when $t = 0$ in $(1)$:
\(\ds 0\) | \(=\) | \(\ds C_1 m_1 e^0 + C_2 m_2 e^0\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds 0\) | \(=\) | \(\ds C_1 m_1 + C_2 m_2\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds 0\) | \(=\) | \(\ds C_1 m_1 + \paren {x_0 - C_1} m_2\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds C_1 \paren {m_1 - m_2}\) | \(=\) | \(\ds -x_0 m_2\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds C_1 \paren {m_1 - m_2}\) | \(=\) | \(\ds -\dfrac {x_0 m_2} {m_1 - m_2}\) |
Then we have:
\(\ds x_0\) | \(=\) | \(\ds C_1 + C_2\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds C_2\) | \(=\) | \(\ds x_0 - C_1\) | |||||||||||
\(\ds \) | \(=\) | \(\ds x_0 + \dfrac {x_0 m_2} {m_1 - m_2}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds x_0 \paren {1 + \dfrac {m_2} {m_1 - m_2} }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds x_0 \paren {\dfrac {\paren {m_1 - m_2} + m_2} {m_1 - m_2} }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac {x_0 m_1} {m_1 - m_2}\) |
Hence:
\(\ds x\) | \(=\) | \(\ds -\dfrac {x_0 m_2} {m_1 - m_2} e^{m_1 t} + \dfrac {x_0 m_1} {m_1 - m_2} e^{m_2 t}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac {x_0} {m_1 - m_2} \paren {m_1 e^{m_2 t} - m_2 e^{m_1 t} }\) |
$\blacksquare$
Sources
- 1972: George F. Simmons: Differential Equations ... (previous) ... (next): $\S 3.20$: Vibrations in Mechanical Systems: $(15)$