MATH201/content/Free vibrations (lec 11-12).md

#start of lecture 11
last lecture we did cauchy euler equations:
$ax^2{\frac{d^2y}{dx^2}+bx{\frac{ dy }{ dx }}+cy=f(x)},\ x>0$
where $a,\ b,\ c$ are still constants and $\in \mathbb{R}$
1) $x=e^t$
$a{\frac{d^2y}{dt^2}}+(b-a){\frac{dy}{dt}}+cy=f(e^t)$ <- lousy notation, the y here isnt quite the same as in the above definition.
2) $y=x^r$
$ar^2+(b-a)r+C=0$
three cases:
(i) $r_1\ne r_{2}$
then: $y_{h}(x)=c_{1}x^{r_{1}}+c_{2}x^{r_{2}}$
(ii) $r_{1}=r_{2}=r$
then: $y_{h}(x)=c_{1}x^r+c_{2}x^r\ln(x)$
(iii) $r_{1,2}=\alpha+i\beta$
then: $y_{h}(x)=x^2(c_{1}\cos(\ln \beta x)+c_{2}\sin \ln(\beta x))$
now find one particular solution for a non homogenous soultion, using variation of parameters, combine the y_h and y_p to get y(x).

not all equations can fall into cauchy euler type.
$y''+p(x)y'+q(x)y=f(x)$ (1) <- no general solution procudure always
but, if $y_{1}(x)$ solves $y''+p(x)y'+q(x)y=0$
then we can find the general solution to the non homogenous equation (1) by guessing it in the form $y(x)=v(x)y_{1}(x)$
$y'=v'y_{1}+vy_{1}'$
$y''=v''y_{1}+2v'y_{1}'+vy_{1}''$
$(v''y_{1}+2v_{1}'y_{1}'+y_{1}''v)+p(x)(v'y_{1}+vy_{1}')+q(x)vy_{1}=f(x)$
$v\cancelto{ 0 }{ (y_{1}''+p(x)y_{1}'+q(x)y_{1}) }+v''y_{1}+(2y_{1}'+p(x)y_{1})v'=f$
$y_{1}v''+()$
$v''+\left( \frac{2y_{1}'}{y_{1}}+p \right)v'=\frac{f}{y_{1}}$
$v'=u$
$u'+\left( \frac{2y_{1}'}{y_{1}}+p \right)u=\frac{f}{y_{1}}$<- this is a linear first order equation
how to solve linear first order equation? we  compute the integrating factor $\mu$
$\mu=e^{\int(2y_{1}'/y_{1}+p)dx}=e^{\ln(y_{1})^2}e^{\int P(x) \, dx}=y_{1}^2e^{\int p(x) \, dx}$
isnt this nice? some kind of magic. We made some guesses and we arrived somewhere.

#ex find the general solution to the equation:
$y''+4xy'+(4x^2+2)y=8e^{-x(x+2)}$
if $y_{1}(x)=e^{-x^2}$ is one solution.
therefore were finding the solution of the form: $y(x)=v(x)y_{1}=v(x)e^{-x^2}$
$v'=u$
$u'+\left( \frac{2y_{1}'}{y_{1}}+4x \right)u=\frac{8{e^{-x^2}e^{-2x}}}{e^{-x^2}}$ <-(p(x)=4x)
$u'+\left( \frac{2{e^{-x^2}(-2x)}}{e^{-x^2}}+4x \right)u=8e^{-2x}$
$u'=8e^{-2x}$
$u=-4e^{-2x}+c_{1}$
$v'=u=-4e^{-2x}+c_{1}$
$v(x)=2e^{-2x}+c_{1}x+c_{2}$
general solution:
$$y(x)=v(x)y_{1}(x)=(2e^{-2x}+c_{1}x+c_{2})e^{-x^2}$$

## Free vibrations
$mr^2+br+k=0$ characteristic polynomial
(i) $r_{1}\ne r_{2}$ $b^2-4mk>0$
$y_{h}(t)=c_{1}e^{r_{1}t}+c_{2}e^{r_{2}t}$
$r_{1,2}=-\frac{b}{2m}\pm \frac{\sqrt{ b^2-4mk }}{2m}<0$
then the limit of the homogenous solution is 0 as t->$\infty$ (over damped case)
(ii) $r_{1}=r_{2}=-\frac{b}{2m}$
$r_{1}=r_{2}=-\frac{b}{2m}$
$y_{h}(t)=e^-\frac{b}{2m}+c_{2}te^{-b/2m}t$ limit =0 as t approaches inf (critically damped)

#end of lec 11 #start of lec 12 (oct 2 2023)
![[Drawing 2023-10-02 13.02.06.excalidraw]]
let $\omega =\frac{\sqrt{ 4mk-b^2 }}{2m}$ (angular frequency)
then the underdamped case is:
$y(t)=(c_{1}\cos \omega t+c_{2}\sin \omega t)e^{\frac{-b}{2m}t}$
we know the trig identity:
$\sin(\alpha+\beta)=\sin \alpha\cos \beta+\cos \alpha \sin \beta$
cant make c_1 c_2 sin or cos so what we do?
do a power transform to convert cartesian into cylindrical coordinates
$c_{1}=A\sin \phi$
$c_{2}=A\cos \phi$
then:
$Ae^{-bt/2m}(\sin \phi \cos \omega t+\cos \phi \sin \omega t)$
$=Ae^{-bt/2m}\sin(\omega t+\phi)$ where $\phi$ is the phase shift.
and $\frac{\omega}{2\pi}$ is the natural frequency
$\frac{2\pi}{\omega}$ is the period
but this is all classical mechanics, but beatifully the world of electronic circuits of R L C also has these equations. Biology too. Nature is beautiful and harmonic.
btw we know $A=\sqrt{ c_{1}^2+c_{2}^2 }$
and $\tan \phi=\frac{c_{1}}{c_{2}}$
so we can get A and phi from c_1 and c_2.
this under damped case also reaches 0 as t->$\infty$

this system in the drawing is in free vibrattion (RHS=0 means no external force=free vibration.)
#ex 
$y''+by'+25y=0 \qquad y(0)=1\quad y'(0)=0$
1) b=0 -> no friction in the system (undamped)
$b^2-4mk$
$y(t)=c_{1}\cos 5t+c_{2}\sin 5t$
$y(0)=c_1=1$
$y'(0)=0=c_{2}$
then $\sin 5t\Rightarrow y(t)=\cos(5t)=\sin\left( 5t+\frac{\pi}{2} \right)$ (by trig identity)
important take away from undamped case: amplitude is constant 1, oscillates forever.
2) b=6
compute $b^2-4mk=36-4*25=-64$
$r_{1,2}=-\frac{6}{2}\pm4i$
$y(t)=e^{-3t}(c_{1}\cos4t+c_{2}\sin4t)$
still under damped situation.
$y(0)=1=c_{1}$
$y'(0)=0=-3c_{1}+4c_{2}\Rightarrow c_{2}=\frac{3}{4}$
$A=\frac{5}{4}$
$\tan \phi=\frac{4}{3}$
$\phi \approx 0.9273\dots$
$$y(t)=\frac{5}{4}e^{-3t}\sin(4t+\phi)$$
"I know engineers loves calculators, I know mathematicians hates calculators, and that's probably the only difference between mathematicians and engineers." -Peter (referring to calculating arctan(4/3) on an exam)
3) b=10
$r_{1,2}=-5$
$y(t)=(c_{1}+c_{2}t)e^{-5t}$
$y(0)=1=c_{1}$
$y'(0)=c_{2}-5c_{1}=0$
$c_{2}=5$
$y(t)=(1+5t)e^{-5t}\rightarrow0_{as\ t\to\infty}$
$y(t)=(1+5t)e^{-5t}>0$
4) b=12
 $r_{1,2}=-6\pm \sqrt{ 11 }$
$y(t)=c_{1}e^{(-6\pm \sqrt{ 11 })t}+c_{2}e^{(-6-\sqrt{11 })t}$
$y(0)=c_{1}+c_{2}=1$
$y'(0)=(-6+\sqrt{ 11 })c_{1}+(-6-\sqrt{ 11 })c_{2}=0$
$c_{1}=\frac{11+6\sqrt{ 11 }}{22}$
$c_{2}=\frac{{11-6\sqrt{ 11 }}}{22}$
this is an over damped case.

lets look at the graphs: (graphs featuring the three cases shown on projector.)
#end of lec 12