Forest UCM Osc Driven

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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]


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


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

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


[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]


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

Case 1 : f(t) is sinusoidal

Consider the case where the driving force is a sinusoidal function

[math]f(t) = f_0 \cos ( \omega t)[/math]

We seek a solution to the particular equation

[math]D(x_p) = f_0 \cos ( \omega t)[/math]
[math] \omega[/math] is not necessarily at the natural (resonant) frequency [math]\omega_0[/math]

If a differential equation has the cosine function as a solution then the sine function may also be a solution since the difference between the two is only a phase shift.

It must also be try that

[math]D(y_p) = f_0 \sin ( \omega t)[/math]

You can construct a complex solution now such that

[math]z_p = C e^{i \omega t}[/math]

the constant C is determine by substituting the solution into the equation

[math]D(z_p) = \left ( - \omega^2 + 2 i \beta \omega + \omega_0 \right ) C e^{i \omega t} = f_0 e^{i \omega t}[/math]
[math]\Rightarrow C= \frac{f_0} {\left ( - \omega^2 + 2 i \beta \omega + \omega_0 \right ) } [/math]

The amplitude of the solution is a complex number that may be cast in terms of a real amplitude times the complex exponential such that

[math] C = A e^{-i\delta}[/math]

where the real amplitude A is given by

[math]A^2 = C C* = \frac{f_0} {\left ( - \omega^2 + 2 i \beta \omega + \omega_0 \right ) } \left ( \frac{f_0} {\left ( - \omega^2 + 2 i \beta \omega + \omega_0 \right ) }\right )^*[/math]
[math]= \frac{f_0} {\left ( - \omega^2 + 2 i \beta \omega + \omega_0 \right ) } \left ( \frac{f_0} {\left ( - \omega^2 - 2 i \beta \omega + \omega_0 \right ) }\right )[/math]
[math]= \frac{f_0^2} { (\omega^2 - \omega^2)^2 + 4 \beta^2 \omega^2 }[/math]