Difference between revisions of "Forest UCM Osc Driven"

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==Solving the Inhomogeneous Diff. Eq. ==
 
==Solving the Inhomogeneous Diff. Eq. ==
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 +
To solve the Inhomogeneous problem we take advantage of the linear operator such that
 +
 +
:<math>D(x_h + x_p) = D(x_h) + D(x_p) = 0 + f = f</math>
 +
 +
 +
Since the solution of the 2nd order differential equation requires exactly two arbitrary constant,
 +
 +
We can formulate the solution by tacking a particular solution onto the homogeneous solution that already has two arbitrary constants.
 +
 +
All we need to do is find the particular solution <math>x_p</math> that satisfies
 +
 +
:<math>D(x_p) =f</math>
 +
 +
and we will have a complete set of solutions.
  
 
Break the equation up into a Homogeneous solution and a Particular Solution
 
Break the equation up into a Homogeneous solution and a Particular Solution
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 +
  
 
===Homogeneous Solution===
 
===Homogeneous Solution===

Revision as of 18:33, 6 October 2014

Damped Oscillators driven by an external source

An external force must be supplied to do work on a damped oscillator in an amount that is equal to or greater than the work done by the dissipative force.


An external force (source) is added to the homogeneous differential equation making it inhomogenous

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


making it

[math] \ddot x + 2 \beta \dot x + \omega^2_0x = f(t)[/math]

where f(t) represents the external force (source) that depends on time divided by the objects mass.


Differential equations in Operator form

In the previous sections we used the definition

[math]O = \frac{d}{dt}[/math]

to solve the second order linear differential equation.

Let's take this a step further with the following operator definition

[math]D = \frac{d^2}{dt^2} + 2 \beta \frac{d}{dt} + \omega^2_0[/math]

then

[math] \ddot x + 2 \beta \dot x + \omega^2_0x = f(t)[/math]

becomes

[math] D x = f(t)[/math]
Linear differential equations have coefficient that can constant or variable coefficients that can be transformed into constant coefficients.


Note
[math]D[/math] is a linear operator

meaning

[math]D(ax_1 + bx_2) = D(ax_1) + D(bx_2)[/math]

the above is a property of differential calculus where

[math]D(ax) = aD(x)[/math] and [math]D(x_1 + x_2) = D(x_1) + D(x_2)[/math]


Solving the Inhomogeneous Diff. Eq.

To solve the Inhomogeneous problem we take advantage of the linear operator such that

[math]D(x_h + x_p) = D(x_h) + D(x_p) = 0 + f = f[/math]


Since the solution of the 2nd order differential equation requires exactly two arbitrary constant,

We can formulate the solution by tacking a particular solution onto the homogeneous solution that already has two arbitrary constants.

All we need to do is find the particular solution [math]x_p[/math] that satisfies

[math]D(x_p) =f[/math]

and we will have a complete set of solutions.

Break the equation up into a Homogeneous solution and a Particular Solution


Homogeneous Solution

Th Homogenous solution solved the equation

[math]Dx_h = 0[/math]

we know from the previous section that the homogenous solution has the form

[math]x_h = C_1 e^{r_1 t} + C_2 e^{r_2 t}[/math]

where

[math]r_1 = - \beta + \sqrt{\beta^2 - \omega_0^2}[/math]
[math]r_2 = - \beta + \sqrt{\beta^2 + \omega_0^2}[/math]


Forest_UCM_Osc#Damped_Oscillations_with_driving_source