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

## 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 $\mathbf F_d$ 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 $\mathbf F_d$ be $c$.

Let there be applied to $C$ an external force which varies as a function of time as:

$\mathbf F_e = \mathbf F_0 \cos \omega t$

where $\mathbf F_0$ is constant.

Let the displacement of $C$ at time $t$ from the equilibrium position be $\mathbf x$.

Then the horizontal position of $C$ at time $t$ can be expressed as:

$\mathbf x = e^{-b t} \left({C_1 \cos \alpha t + C_2 \sin \alpha t}\right) + \dfrac {\mathbf F_0} {\sqrt {\left({k^2 - \omega^2 m^2}\right)^2 + \omega^2 c^2} } \cos \left({\omega t - \phi}\right)$

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 $b < a$.

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

where:

$\alpha = \sqrt {a^2 - b^2}$
$\phi = \arctan \left({\dfrac {2 b \omega} {a^2 - \omega^2} }\right)$

Substituting back:

$a^2 = \dfrac k m$
$2 b = \dfrac c m$
$K = \dfrac {\mathbf F_0} m$

we obtain:

 $\displaystyle$  $\displaystyle \dfrac K {\sqrt {4 b^2 \omega^2 + \left({a^2 - \omega^2}\right)^2} } \cos \left({\omega t - \phi}\right)$ $\displaystyle$ $=$ $\displaystyle \dfrac {\mathbf F_0} m \frac 1 {\sqrt {\left({\dfrac c m}\right)^2 \omega^2 + \left({\dfrac k m - \omega^2}\right)^2} } \cos \left({\omega t - \phi}\right)$ $\displaystyle$ $=$ $\displaystyle \dfrac {\mathbf F_0} m \frac 1 {\sqrt {\dfrac 1 {m^2} c^2 \omega^2 + \dfrac 1 {m^2} \left({k - \omega^2 m}\right)^2} } \cos \left({\omega t - \phi}\right)$ $\displaystyle$ $=$ $\displaystyle \dfrac {\mathbf F_0} m \frac 1 {\dfrac 1 m \sqrt {c^2 \omega^2 + \left({k - \omega^2 m}\right)^2} } \cos \left({\omega t - \phi}\right)$ $\displaystyle$ $=$ $\displaystyle \dfrac {\mathbf F_0} {\sqrt {c^2 \omega^2 + \left({k - \omega^2 m}\right)^2} } \cos \left({\omega t - \phi}\right)$

and:

 $\displaystyle \phi$ $=$ $\displaystyle \arctan \left({\dfrac {2 b \omega} {a^2 - \omega^2} }\right)$ $\displaystyle$ $=$ $\displaystyle \arctan \left({\dfrac {\dfrac c m \omega} {\dfrac k m - \omega^2} }\right)$ $\displaystyle$ $=$ $\displaystyle \arctan \left({\dfrac {\omega c} {k - \omega^2 m} }\right)$ multiplying top and bottom by $m$

and:

 $\displaystyle \alpha$ $=$ $\displaystyle \sqrt {a^2 - b^2}$ $\displaystyle$ $=$ $\displaystyle \sqrt {\dfrac k m - \left({\dfrac c {2 m} }\right)^2}$ $\displaystyle$ $=$ $\displaystyle \sqrt {\dfrac k m - \dfrac {c^2} {4 m^2} }$

$\blacksquare$