# Period of Oscillation of Underdamped Cart attached to Wall by Spring

## 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 $C$ be underdamped.

Let $C$ be pulled aside to $x = x_0$ and released from stationary at time $t = 0$.

Then the period of oscillation of $C$ can be expressed as:

$T = \dfrac {2 \pi} {\sqrt {\dfrac k m - \dfrac {c^2} {4 m^2} } }$

## Proof

Let:

$a^2 = \dfrac k m$
$2 b = \dfrac c m$
$x = \dfrac {x_0} \alpha e^{-b t} \left({\alpha \cos \alpha t + b \sin \alpha t}\right)$

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

Let $T$ be the period of oscillation of $C$.

Then:

 $\displaystyle T$ $=$ $\displaystyle \dfrac {2 \pi} {\sqrt {a^2 - b^2} }$ Period of Oscillation of Underdamped System is Regular $\displaystyle$ $=$ $\displaystyle \dfrac {2 \pi} {\sqrt {\dfrac k m - \left({\dfrac c {2 m} }\right)^2} }$ $\displaystyle$ $=$ $\displaystyle \dfrac {2 \pi} {\sqrt {\dfrac k m - \dfrac {c^2} {4 m^2} } }$

$\blacksquare$