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

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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 $\mathbf x$.


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:

$x = \dfrac {x_0} \alpha e^{-b t} \left({\alpha \cos \alpha t + b \sin \alpha t}\right)$

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


Such a system is defined as being underdamped.


Underdamped.png


Proof

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

$(2): \quad x = e^{-b t} \paren {C_1 \cos \alpha t + C_2 \sin \alpha t}$

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:

$(2): \quad x' = -b e^{-b t} \paren {C_1 \cos \alpha t + C_2 \sin \alpha t} + e^{-b t} \paren {-\alpha C_1 \sin \alpha t + \alpha C_2 \cos \alpha t}$


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

\(\ds x_0\) \(=\) \(\ds e^0 \paren {C_1 \cos 0 + C_2 \sin 0}\)
\(\ds \) \(=\) \(\ds C_1\)


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

\(\ds 0\) \(=\) \(\ds -b e^0 \paren {C_1 \cos 0 + C_2 \sin 0} + e^0 \paren {-\alpha C_1 \sin 0 + \alpha C_2 \cos 0}\)
\(\ds \) \(=\) \(\ds -b C_1 + \alpha C_2\)
\(\ds \leadsto \ \ \) \(\ds C_2\) \(=\) \(\ds \frac {b C_1} \alpha\)
\(\ds \) \(=\) \(\ds \frac {b x_0} \alpha\)


Hence:

\(\ds x\) \(=\) \(\ds e^{-b t} \paren {x_0 \cos \alpha t + \frac {b x_0} \alpha \sin \alpha t}\)
\(\ds \) \(=\) \(\ds \dfrac {x_0} \alpha e^{-b t} \paren {\alpha \cos \alpha t + b \sin \alpha t}\)

$\blacksquare$


Sources

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