Natural Frequency 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 natural frequency of $C$ can be expressed as:

$\nu = \dfrac 1 {2 \pi} \sqrt {\dfrac k m - \dfrac {c^2} {4 m^2} }$

Proof

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

where $T$ is the period of oscillation of $C$.

Then by definition of the natural frequency of $C$:

 $\displaystyle \nu$ $=$ $\displaystyle \dfrac 1 T$ $\displaystyle$ $=$ $\displaystyle \dfrac 1 {\left({\dfrac {2 \pi} {\sqrt {\frac k m - \frac {c^2} {4 m^2} } } }\right)}$ $\displaystyle$ $=$ $\displaystyle \dfrac 1 {2 \pi} \sqrt {\dfrac k m - \dfrac {c^2} {4 m^2} }$

$\blacksquare$