Forest UCM Osc Driven

From New IAC Wiki
Jump to navigation Jump to search

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

¨x+2β˙x+ω20x=0


making it

¨x+2β˙x+ω20x=f(t)

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

O=ddt

to solve the second order linear differential equation.

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

D=d2dt2+2βddt+ω20

then

¨x+2β˙x+ω20x=f(t)

becomes

Dx=f(t)
Linear differential equations have coefficient that can constant or variable coefficients that can be transformed into constant coefficients.


Note
D is a linear operator

meaning

D(ax1+bx2)=D(ax1)+D(bx2)

the above is a property of differential calculus where

D(ax)=aD(x) and D(x1+x2)=D(x1)+D(x2)


Solving the Inhomogeneous Diff. Eq.

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

D(xh+xp)=D(xh)+D(xp)=0+f=f


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 xp that satisfies

D(xp)=f

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

Dxh=0

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

xh=C1er1t+C2er2t

where

r1=β+β2ω20
r2=β+β2+ω20


Case 1 : f(t) is sinusoidal

Consider the case where the driving force is a sinusoidal function

f(t)=f0cos(ωt)

We seek a solution to the particular equation

D(xp)=f0cos(ωt)
Remember
ω is not necessarily at the natural (resonant) frequency ω0


Trick
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

D(yp)=f0sin(ωt)


You can construct a complex solution now such that

zp=Ceiωt


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

D(zp)=(ω2+2iβω+ω20)Ceiωt=f0eiωt
C=f0(ω2+2iβω+ω20)

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

C=Aeiδ

where the real amplitude A is given by

A2=CC=f0(ω2+2iβω+ω20)(f0(ω2+2iβω+ω20))
=f0(ω2+2iβω+ω20)(f0(ω22iβω+ω20))
=f20(ω20ω2)2+4β2ω2
C=Aeiδ=f20(ω20ω2)2+4β2ω2eiδ
zp=Ceiωt=f20(ω20ω2)2+4β2ω2eiδeiωt=Ceiωt=f0(ω20ω2)2+4β2ω2ei(ωtδ)


What is delta?

C=f0(ω2+2iβω+ω20)=Aeiδ
f0(ω2+2iβω+ω20)=Aeiδ
f0eiδ=A((ω20ω2)+2iβω)

Since f0 and A are not imaginary variable

The phase and δ must be due to the remaining complex term ((ω20ω2)+2iβω)

To detmine the angle δ we construct a right trianble in the complex plane where the leg along the real axis is given by (ω20ω2) and the other leg of the triangle is along the imaginary axis with a length of 2βω

tanδ=2βω(ω20ω2)
< math>\delta = \tan^{-1}\left ( \frac{2 \beta \omega}{(\omega_0^2- \omega^2)} \right )</math>
The final complex solution si
zp=f0(ω20ω2)2+4β2ω2ei(ωtδ)


Forest_UCM_Osc#Damped_Oscillations_with_driving_source