Difference between revisions of "Forest UCM Osc Damped"

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Setting the term in parentheses to zero and using the quadratic formula  
 
Setting the term in parentheses to zero and using the quadratic formula  
  
: <math>O = \frac{- 2\beta \pm \sqrt{(2\beta)^2 -4\omega}}{2} =  - \beta \pm \sqrt{\beta^2 -\omega}</math>
+
: <math>O = \frac{- 2\beta \pm \sqrt{(2\beta)^2 -4\omega^2}}{2} =  - \beta \pm \sqrt{\beta^2 -\omega^2}</math>
  
:<math>  \left ( O - \beta + \sqrt{(2\beta)^2 -4\omega\right )  \left ( O - \beta - \sqrt{(2\beta)^2 -4\omega\right )  x = 0 </math>
+
:<math>  \left ( O - \beta + \sqrt{\beta^2 -\omega^2} \right )  \left ( O - \beta - \sqrt{\beta^2 -\omega^2}\right )  x = 0 </math>
  
  
 
[[Forest_UCM_Osc#Damped_Oscillations]]
 
[[Forest_UCM_Osc#Damped_Oscillations]]

Revision as of 12:55, 5 October 2014

1-D Damped Oscillaions

Equation of Motion

As in the case of air resistance, assume there is frictional force proportional to the velocity of the oscillation body.


[math] \sum \vec{F}_{ext} = -k\vec r - b \vec \dot v = m \vec \ddot r[/math]
[math] \sum F_{ext} = -kx - b \dot x = m \ddot x[/math]: in 1-D

or

[math] m \ddot x + kx + b \dot x = 0[/math]

or

[math] \ddot x + \frac{k}{m}x + \frac{b}{m} \dot x = 0[/math]


let

[math]\frac{k}{m} = \omega^2_0 =[/math] undamped oscillation frequency
[math]\frac{b}{m} \equiv 2 \beta =[/math] damping constant

then

[math] \ddot x + 2 \beta \dot x + \omega^2_0x = 0[/math]


As see in section Forest_UCM_Osc_SHM#Equation_of_motion, you can determine solutions to the above by writing the analogous auxilary equation:

[math] \left ( O^2 + 2 \beta O + \omega^2_0 \right ) x = 0 \;\;\;\;\;\; O \equiv \frac{d}{dt}[/math]

Setting the term in parentheses to zero and using the quadratic formula

[math]O = \frac{- 2\beta \pm \sqrt{(2\beta)^2 -4\omega^2}}{2} = - \beta \pm \sqrt{\beta^2 -\omega^2}[/math]
[math] \left ( O - \beta + \sqrt{\beta^2 -\omega^2} \right ) \left ( O - \beta - \sqrt{\beta^2 -\omega^2}\right ) x = 0 [/math]


Forest_UCM_Osc#Damped_Oscillations