Position of Cart attached to Wall by Spring under Damping/Critically Damped/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:
- $x = x_0 e^{-a t} \paren {1 + a t}$
Such a system is defined as being critically damped.
Proof
From Position of Cart attached to Wall by Spring under Damping: Critically Damped:
- $x = C_1 e^{-a t} + C_2 t e^{-a t}$
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:
- $(2): \quad x' = -a C_1 e^{-a t} + C_2 e^{-a t} - a C_2 t e^{-a t}$
Setting the initial condition $x = x_0$ when $t = 0$ in $(1)$:
\(\ds x_0\) | \(=\) | \(\ds C_1 e^0 + C_2 e^0 \times 0\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds x_0\) | \(=\) | \(\ds C_1\) |
Setting the initial condition $x' = 0$ when $t = 0$ in $(1)$:
\(\ds 0\) | \(=\) | \(\ds -a C_1 e^0 + C_2 e^0 - a C_2 e^0 \times 0\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds 0\) | \(=\) | \(\ds -a C_1 + C_2\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds C_2\) | \(=\) | \(\ds a C_1\) | |||||||||||
\(\ds \) | \(=\) | \(\ds a x_0\) |
Hence:
\(\ds x\) | \(=\) | \(\ds x_0 e^{-a t} + a x_0 t e^{-a t}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds x_0 e^{-a t} \paren {1 + a t}\) |
$\blacksquare$
Sources
- 1972: George F. Simmons: Differential Equations ... (previous) ... (next): $\S 3.20$: Vibrations in Mechanical Systems: $(17)$