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