Position of Underdamped Cart attached to Wall by Spring under Forced Vibration

Problem Definition
Then the horizontal position of $C$ at time $t$ can be expressed as:
 * $x = e^{-b t} \left({C_1 \cos \alpha t + C_2 \sin \alpha t}\right) + \dfrac {F_0} {\sqrt {\left({k^2 - \omega^2 m^2}\right)^2 + \omega^2 c^2} } \cos \left({\omega t - \phi}\right)$

where:
 * $C_1$ and $C_2$ depend upon the conditions of $C$ at time $t = 0$
 * $\alpha = \sqrt {\dfrac k m - \dfrac {c^2} {4 m^2} }$
 * $\phi = \arctan \left({\dfrac {\omega c} {k - \omega^2 m} }\right)$

Proof
From Forced Vibration of Cart attached to Wall by Spring, the motion of $C$ is described by the second order ODE:
 * $(1): \quad m \dfrac {\mathrm d^2 \mathbf x} {\mathrm d t^2} + c \dfrac {\mathrm d \mathbf x} {\mathrm d t} + k \mathbf x = \mathbf F_0 \cos \omega t$

Setting:
 * $a^2 = \dfrac k m$
 * $2 b = \dfrac c m$

this can be written as:
 * $(2): \quad \dfrac {\mathrm d^2 \mathbf x} {\mathrm d t^2} + 2 b \dfrac {\mathrm d \mathbf x} {\mathrm d t} + a^2 \mathbf x = \dfrac {\mathbf F_0} m \cos \omega t$

We are given that $C$ is underdamped, and so $b < a$.

From Second Order ODE: $y'' + 2 b y' + a^2 y = K \cos \omega x$ for $b < a$, $(2)$ has the general solution:


 * $y = e^{-b t} \left({C_1 \cos \alpha x + C_2 \sin \alpha x}\right) + e^{-b x} \left({C_1 \cos \alpha x + C_2 \sin \alpha x}\right) + \dfrac K {\left({a^2 - \omega^2}\right)^2 + 4 b^2 \omega^2} \left({2 b \omega \sin \omega x + \left({a^2 - \omega^2}\right) \cos \omega x}\right)$

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