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

Problem Definition
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:
 * $(2): \quad x = e^{-b t} \left({C_1 \cos \alpha t + C_2 \sin \alpha t}\right)$

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 $(1)$ $t$ gives:
 * $(2): \quad x' = -b e^{-b t} \left({C_1 \cos \alpha t + C_2 \sin \alpha t}\right) + e^{-b t} \left({-\alpha C_1 \sin \alpha t + \alpha C_2 \cos \alpha t}\right)$

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

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

Hence: