Deriving the Damping Coefficient (part 2)

Condition 1: At time = 0 and position y = 0

condition1

y(0) = e^{- \xi *0} \{Acos( \tilde{ \omega} *0) \ + \ Bsin(\tilde{ \omega} *0 )\}

 Acos( \tilde{ \omega} *0) \ = \ A(1) \ = \ 0

Therefore: A = 0

To use the condition of V_0 \ = \ \dot{y} at t = 0, we need to take the first derivative of the position equation

y(t) = e^{- \xi t} \{Bsin( \tilde{ \omega} t)\}

\dot{y}(t) = e^{- \xi t} \frac{d}{dt} \{Bsin( \tilde{ \omega} t)\} + \{Bsin( \tilde{ \omega} t)\} \frac{d}{dt} e^{- \xi t}

\dot{y}(t) = e^{- \xi t} \{ \tilde{ \omega} Bcos( \tilde{ \omega} t )\} - \xi \{Bsin( \tilde{ \omega} t) \} e^{- \xi t}

\dot{y}(t) = e^{- \xi t} B \{ \tilde{ \omega} cos( \tilde{ \omega} t )- \xi sin( \tilde{ \omega} t) \}

V_o = e^{- \xi *0} B \{ \tilde{ \omega} cos( \tilde{ \omega} *0 )- \xi sin( \tilde{ \omega}*0) \}

V_0 = B \{ \tilde{\omega}\}

B = \frac{V_0}{\tilde{\omega}}

This brings our position and velocity equations to:

y(t) = \frac{V_0 e^{- \xi t}}{\tilde{\omega}}sin(\tilde{\omega}t)

\dot{y}(t) = \frac{V_0 e^{- \xi t}}{\tilde{\omega}}\{\tilde{\omega} cos( \tilde{\omega}t) - \xi sin(\tilde{\omega}t)\}

The goal of solving this system is to extract the damping coefficient. To do this, evaluate the system at the instant the particle rebounds from the ground.

condition2

These conditions are: y = 0, t = t_f, \dot{y} = V_f

First we need to determine t = t_f

condition2b

The equation for position is a sine wave of the form: y = C \ \ sin(\tilde{\omega}t)

sinewave

The sine wave has a period T and the time elapsed during that period is T/2, where

period, T = \frac{2 \pi}{\tilde{\omega}}

Therefore,

time, t = \frac{T}{2} = \frac{\pi}{\tilde{\omega}}

Solving the velocity equation at the determined conditions: t = \frac{\pi}{\tilde{\omega}}, \dot{y} = V_f

y(t) = \frac{V_0 e^{- \xi t}}{\tilde{\omega}}sin(\tilde{\omega}t)

V_f = \frac{V_0 e^{- \xi \frac{\pi}{\tilde{\omega}}}}{\tilde{\omega}}\{\tilde{\omega} cos( \tilde{\omega}\frac{\pi}{\tilde{\omega}}) - \xi sin(\tilde{\omega}\frac{\pi}{\tilde{\omega}})\}

V_f = \frac{V_0 e^{- \xi \frac{\pi}{\tilde{\omega}}}}{\tilde{\omega}}\{\tilde{\omega} cos( \pi) - \xi sin(\pi)\} = V_0 e^{- \xi \frac{\pi}{\tilde{\omega}}}

Using this information for V_f and V_0 combined with the linear definition of the coefficient of restitution, e, which is given by:

e = \frac{V_f}{V_o}

e = exp^{- \xi \frac{\pi}{\tilde{\omega}}}

Bringing out system to its initial terms: m, k, c

ln(e)=- \zeta \omega_0 \frac{\pi}{\omega_0 \sqrt{1 - \zeta^2}}

ln(e)=- \zeta  \frac{\pi}{\sqrt{1 - \zeta^2}}

\sqrt{1 - \zeta^2} ln(e)=- \zeta \pi

(\sqrt{1 - \zeta^2}ln(e) )^2= (- \zeta \pi)^2

(\sqrt{1-\zeta^2}) [ln(e)]^2 = \pi^2 \zeta^2

 [ln(e)]^2 - \zeta^2 [ln(e)]^2 = \pi^2 \zeta^2

\pi^2 \zeta^2 + \zeta^2 [ln(e)]^2 = [lkn(e)]^2

\zeta^2 (\pi^2 + [ln(e)]^2) = [ln(e)]^2

\zeta^2 = \frac{[ln(e)]^2}{\pi^2 + [ln(e)]^2}

Recall:

damping ratio, \zeta = \frac{c}{C_{crit}} \rightarrow \frac{c}{2 \sqrt{km}}

 (\frac{c}{2 \sqrt{km}})^2 = \frac{[ln(e)]^2}{\pi^2 + [ln(e)]^2}

 \sqrt{(\frac{c}{2 \sqrt{km}})^2 }= \sqrt{\frac{[ln(e)]^2}{\pi^2 + [ln(e)]^2}}

 \frac{c}{2 \sqrt{km}}= \frac{ln(e)}{\sqrt{\pi^2 + [ln(e)]^2}}

c = \frac{2\sqrt{km}ln(e)}{\sqrt{\pi^2 + [ln(e)]^2}} = \frac{2\sqrt{km}ln(\frac{1}{e})}{\sqrt{\pi^2 + [ln(\frac{1}{e})]^2}}

This defines the viscous damping coefficient.

Leave a comment