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}}{2} =  - \beta \pm \sqrt{\beta^2 -\omega^2}</math>
+
: <math>O = \frac{- 2\beta \pm \sqrt{(2\beta)^2 -4\omega^2_0}}{2} =  - \beta \pm \sqrt{\beta^2 -\omega^2_0}</math>
  
:<math>  \left ( O + \beta + \sqrt{\beta^2 -\omega^2} \right )  \left ( O + \beta - \sqrt{\beta^2 -\omega^2}\right )  x = 0 </math>
+
:<math>  \left ( O + \beta + \sqrt{\beta^2 -\omega^2_0} \right )  \left ( O + \beta - \sqrt{\beta^2 -\omega^2_0}\right )  x = 0 </math>
  
  
 
You have change the second order differential equation into two first order differential equations
 
You have change the second order differential equation into two first order differential equations
  
:<math>  \left ( \frac{d}{dt} + \beta + \sqrt{\beta^2 -\omega^2} \right )  x = 0 </math>
+
:<math>  \left ( \frac{d}{dt} + \beta + \sqrt{\beta^2 -\omega^2_0} \right )  x = 0 </math>
::<math>  \Rightarrow \frac{dx}{x} =  \left ( -\beta - \sqrt{\beta^2 -\omega^2} \right ) dt </math>
+
::<math>  \Rightarrow \frac{dx}{x} =  \left ( -\beta - \sqrt{\beta^2 -\omega^2_0} \right ) dt </math>
::<math>  x=  e^{\left (- \beta - \sqrt{\beta^2 -\omega^2} \right )t} </math>
+
::<math>  x=  e^{\left (- \beta - \sqrt{\beta^2 -\omega^2_0} \right )t} </math>
  
  
:<math> \left (\frac{d}{dt} + \beta - \sqrt{\beta^2 -\omega^2}\right )  x = 0 </math>
+
:<math> \left (\frac{d}{dt} + \beta - \sqrt{\beta^2 -\omega^2_0}\right )  x = 0 </math>
::<math>  \Rightarrow \frac{dx}{x} =  \left ( - \beta + \sqrt{\beta^2 -\omega^2} \right ) dt </math>
+
::<math>  \Rightarrow \frac{dx}{x} =  \left ( - \beta + \sqrt{\beta^2 -\omega^2_0} \right ) dt </math>
::<math>  x=  e^{\left ( - \beta + \sqrt{\beta^2 -\omega^2} \right )t} </math>
+
::<math>  x=  e^{\left ( - \beta + \sqrt{\beta^2 -\omega^2_0} \right )t} </math>
  
 
constructing a complete solution from the two solutions (orthogonal functions) above.
 
constructing a complete solution from the two solutions (orthogonal functions) above.
  
:<math>  x=  \left ( C_1 e^{ \sqrt{\beta^2 -\omega^2} t} + C_2 e^{ -\sqrt{\beta^2 -\omega^2} t} \right) e^{- \beta t} </math>
+
:<math>  x=  \left ( C_1 e^{ \sqrt{\beta^2 -\omega^2_0} t} + C_2 e^{ -\sqrt{\beta^2 -\omega^2_0} t} \right) e^{- \beta t} </math>
  
 
===Undamped oscillator===
 
===Undamped oscillator===
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Then
 
Then
  
:<math>  x=  \left ( C_1 e^{ \sqrt{\beta^2 -\omega^2} t} + C_2 e^{ -\sqrt{\beta^2 -\omega^2} t} \right) e^{- \beta t} </math>
+
:<math>  x=  \left ( C_1 e^{ \sqrt{\beta^2 -\omega^2_0} t} + C_2 e^{ -\sqrt{\beta^2 -\omega^2_0} t} \right) e^{- \beta t} </math>
::<math>  =  \left ( C_1 e^{ i\omega t} + C_2 e^{ -i\omega t} \right) </math> the SHM solution derived before at [[Forest_UCM_Osc_SHM#Equation_of_motion]]
+
::<math>  =  \left ( C_1 e^{ i\omega_0 t} + C_2 e^{ -i\omega_0 t} \right) </math> the SHM solution derived before at [[Forest_UCM_Osc_SHM#Equation_of_motion]]
 
 
  
 
==Under damped Oscillator==
 
==Under damped Oscillator==

Revision as of 13:28, 5 October 2014

1-D Damped Oscillations

Newton's 2nd Law

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]

Solve for the Equation of Motion

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_0}}{2} = - \beta \pm \sqrt{\beta^2 -\omega^2_0}[/math]
[math] \left ( O + \beta + \sqrt{\beta^2 -\omega^2_0} \right ) \left ( O + \beta - \sqrt{\beta^2 -\omega^2_0}\right ) x = 0 [/math]


You have change the second order differential equation into two first order differential equations

[math] \left ( \frac{d}{dt} + \beta + \sqrt{\beta^2 -\omega^2_0} \right ) x = 0 [/math]
[math] \Rightarrow \frac{dx}{x} = \left ( -\beta - \sqrt{\beta^2 -\omega^2_0} \right ) dt [/math]
[math] x= e^{\left (- \beta - \sqrt{\beta^2 -\omega^2_0} \right )t} [/math]


[math] \left (\frac{d}{dt} + \beta - \sqrt{\beta^2 -\omega^2_0}\right ) x = 0 [/math]
[math] \Rightarrow \frac{dx}{x} = \left ( - \beta + \sqrt{\beta^2 -\omega^2_0} \right ) dt [/math]
[math] x= e^{\left ( - \beta + \sqrt{\beta^2 -\omega^2_0} \right )t} [/math]

constructing a complete solution from the two solutions (orthogonal functions) above.

[math] x= \left ( C_1 e^{ \sqrt{\beta^2 -\omega^2_0} t} + C_2 e^{ -\sqrt{\beta^2 -\omega^2_0} t} \right) e^{- \beta t} [/math]

Undamped oscillator

If [math]\beta[/math] = 0

Then

[math] x= \left ( C_1 e^{ \sqrt{\beta^2 -\omega^2_0} t} + C_2 e^{ -\sqrt{\beta^2 -\omega^2_0} t} \right) e^{- \beta t} [/math]
[math] = \left ( C_1 e^{ i\omega_0 t} + C_2 e^{ -i\omega_0 t} \right) [/math] the SHM solution derived before at Forest_UCM_Osc_SHM#Equation_of_motion

Under damped Oscillator

[math]\beta \lt \omega_0[/math]

In this case the term

[math]\sqrt{\beta^2 -\omega^2} =\sqrt{(-1)(\omega^2- \beta^2 } = i \sqrt{\omega^2- \beta^2 }[/math]
[math]O = \frac{- 2\beta \pm \sqrt{(2\beta)^2 -4\omega^2}}{2} = - \beta \pm \sqrt{\beta^2 -\omega^2}[/math]

Over damped Oscillator

[math]\beta \gt \omega_0[/math]

Critically damped Oscillator

[math]\beta = \omega_0[/math]


Forest_UCM_Osc#Damped_Oscillations