Difference between revisions of "Forest UCM Osc Damped"
Jump to navigation
Jump to search
Line 38: | Line 38: | ||
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^ | + | : <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^ | + | :<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^ | + | :<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^ | + | ::<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^ | + | ::<math> x= e^{\left (- \beta - \sqrt{\beta^2 -\omega^2_0} \right )t} </math> |
− | :<math> \left (\frac{d}{dt} + \beta - \sqrt{\beta^2 -\omega^ | + | :<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^ | + | ::<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^ | + | ::<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^ | + | :<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=== | ||
Line 64: | Line 64: | ||
Then | Then | ||
− | :<math> x= \left ( C_1 e^{ \sqrt{\beta^2 -\omega^ | + | :<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\ | + | ::<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.
- : in 1-D
or
or
let
- undamped oscillation frequency
- damping constant
then
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:
Setting the term in parentheses to zero and using the quadratic formula
You have change the second order differential equation into two first order differential equations
constructing a complete solution from the two solutions (orthogonal functions) above.
Undamped oscillator
If
= 0Then
- Forest_UCM_Osc_SHM#Equation_of_motion the SHM solution derived before at
Under damped Oscillator
In this case the term