revised free vibrations (Finally!) and added lec 21

content/Dirak δ-function (lec 21).md

@@ -48,4 +48,96 @@ $0.2I_{1}+0.1I_{3}'+2I_{1}=g(t) \qquad \text{where } g(t)=\begin{cases}6, & 0\le
 $-I_{2}+0.1I_{3}'=0$
 $I_{1}=I_{2}+I_{3}$
 voila, a system of three equations.
-#end of lec 21
+#end of lec 21
+#start of lec 22
+The prof shared some really cool story of George Green and how he lived on his father's grain mill and derived Green's Theorem and how this theorem eventually came to Stokes desk and Stoke decided to put this new, never before seen theorem on his exam and how it's really not like nowadays where everything is Mcdonalds style, Mcdonalds style education.
+Another cool story of Ramanujan, he was very poor, but he managed to enter a school, an accomplishment in and of it self. His teacher saw his magical ability and told him to write to Hardy, a British mathematician. Hardy was impressed and he flew Ramanujan there and worked with him and he saw they things Ramanujan showed and it was inhuman, but he had few mathematical skills so he taught him how to make proofs and now Hardy said that if we were to rate all the mathematicians in the world he would give:
+myself 20/100
+Hilbert 40/100
+Ramanujan 80/100
+
+OKAY get back to the lecture, that's 6 minutes worth of stories. (I really live for Petar's stories. Idk why but they just feel like jewels and I love saving them. I hope he's okay with me doing that.)
+We continue on the last lecture problem:
+$0.2I_{1}+0.1I_{3}'+2I_{1}=g(t) \qquad \text{where } g(t)=\begin{cases}6, & 0\leq t\leq 1 \\0, & 1<t\end{cases}$
+$-I_{2}+0.1I_{3}'=0$
+$I_{1}=I_{2}+I_{3}$
+two differential equations, one algebraic equation.
+express g in terms of unit step funtion:
+$g(t)=6-6u(t-1)$
+applying this on the second equation:
+$-I_{1}+I_{3}+0.1I_{3}'=0$
+
+now we have two equations:
+$0.2I_{1}+0.1I_{3}'+2I_{1}=6-6u(t-1)$
+$-I_{1}+I_{3}+0.1I_{3}'=0$
+where $I_1(0)=I_{2}(0)=I_{3}=0$
+multiply eqations by 10:
+$2I_{1}+1I_{3}'+20I_{1}=60(1-u(t-1))$
+$-10I_{1}+10I_{3}+I_{3}'=0$
+hit it with the LT!
+$J_{1}=\mathcal{L}\{I_{1}(t)\}(s)$
+$J_3=\mathcal{L}\{I_{3}\}$
+$2sJ_{1}+sJ_{3}+20J_{1}=\mathcal{L}\{g(t)\}$
+$2(s+10)J_{1}+sJ_{3}=\dots$
+$-10J_{1}+(s+10)J_{3}=0$
+where $\dots$ is:
+using:
+$\mathcal{L}\{f(t-a)u(t-a)\}=e^{-as}F(s)$
+$\mathcal{L}^{-1}\{e^{-as}F(s)\}=f(t-a)u(t-a)$
+
+so:
+$2(s+10)J_{1}+sJ_{3}=60 \frac{1-e^{-s}}{s}$
+multiply 2 by s+10
+multiply 1 by s:
+
+$J_{1}(s)=30 \frac{s+10}{s(s+5)(s+20)}(1-e^{-s})$ 
+use partial fraction:
+$30 \frac{s+10}{s(s+5)(s+20)}=\frac{A}{s}+\frac{B}{s+5}+\frac{C}{s+20}$
+$=\frac{{A(s+5)(s+20)+B(s^2+20s)+C(s^2+5s)}}{s(s+5)(s+20)}$
+then: $A+B+C=0$
+$25A+20B+5C=30$
+$100A=300 \implies A=3$
+$\implies B=-2 \qquad C=-1$
+$J_{1}(s)=\left( \frac{3}{s}-\frac{2}{s+5}- \frac{1}{s+20} \right)(1-e^{-s})$ 
+invert the LT: (use $\mathcal{L}^{-1}\{e^{-as}F(s)\}=f(t-a)u(t-a)$)
+$I_{1}=3-2e^{-5t}-e^{-20t}-u(t-1)(3-2e^{-5(t-1)}-e^{-20(t-1)})$
+whats $I_{3}$? use: $-10I_{1}+10I_{3}+I_{3}'=0$ (i think)
+$J_{3}=\frac{10}{s+10}J_{1}$
+$J_{3}=300 \frac{1}{s(s+5)(s+20)}(1-e^{ -s })$
+partial fraction it so we can eventuall take the inverse LT:
+skip some steps:
+$J_{3}=\left( \frac{3}{s}-\frac{4}{s+5}+\frac{1}{s+20} \right)(1-e^{ -s })$
+$I_{3}=3-4e^{ -5t }+e^{ -20t }-(3-4e^{ -5(t-1) }+e^{ -20(t-1) })u(t-1)$
+$I_{2}=I_{1}-I_{3}$
+
+$ay''+by'+cy=f$
+$y'=z$
+$\begin{cases}y'-z=0 \\az'+bz+cy=f\end{cases}$
+
+last example of the chapter:
+#ex
+$x'+y=0, \qquad x(0)=0$
+$x+y'=1-u(t-z) \qquad y(0)=0$
+this is review problem in chapter 7 of the textbook.
+hit equation 1 and 2 with the LT:
+$sX+Y=0$
+$X+sY=\frac{{1-e^{ -2s }}}{s}$
+multiply equation 1 by s:
+$-(s^2-1)X=\frac{{1-e^{ -2s }}}{2}$
+$X(s)=e^{=2s} \frac{1}{s(s-1)(s+1)}-\frac{1}{s(s-1)(s+1)}$
+use partial fractions:
+$\frac{A}{s}+\frac{B}{s-1}+\frac{C}{s+1}=\frac{1}{s(s-1)(s+1)}$
+$A(s^2-1)+B(s^2+s)+C(s^2-s)=1$
+$A+B+C=0$
+$B-C=0$
+$-A=1\implies A=-1$
+$\implies B=\frac{1}{2} \qquad C=\frac{1}{2}$
+$X(s)=\left( -\frac{1}{s}+\frac{1}{2} \frac{1}{s-1}+\frac{1}{2} \frac{1}{s+1} \right)(e^{ -2s }-1)$
+inverse laplace:
+$x(t)=1-\frac{1}{2}e^t-\frac{1}{2}e^{-t}-u(t-2)\left( 1-\frac{1}{2}e^{t-2} -\frac{1}{2}e^{-(t-2)}\right)$
+$Y=-sX=-s (\frac{-1}{s(s-1)(s+1)}+ \frac{e^{ -2s }}{s(s-1)(s+1)})$
+$=\frac{1}{(s-1)(s+1)}- \frac{e^{-2s}}{(s-1)(s+1)}$
+$=\frac{1}{2}(1-e^{-2s})\left( \frac{1}{s-1}-\frac{1}{s+1} \right)$
+$\implies y(t)=\frac{1}{2}(e^t-e^{-t})-\frac{1}{2}u(t-2)(e^{t-2}-e^{-(t-2)})$
+we are done
+#end of lec 22

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

@@ -1,74 +1,109 @@
 # Free vibrations
-Free vibrations are when there are no externally applied forces acting upon an oscillatory system. RHS=0.
-$mr^2+br+k=0$ 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)
+## Definition
+This is a mass-spring system with friction: (imagine the $\beta$ was $b$ instead)
+![mbk.png](drawings/mbk.png)
+A mass-spring system with friction can be modelled by the following differential equation:
+$$my''+by'+ky=F_{ext}(t)$$
+Where: 
+$$\begin{matrix}F_{ext}(t) & \text{any external forces applied }(N) \\m & \text{mass } (kg) \\b & \text{damping constant } \left( \frac{Ns}{m} \right) \\k  & \text{spring constant } \left( \frac{N}{m} \right)\end{matrix}$$
+Free vibrations are when there are <u>no external forces</u> applied upon an oscillatory system. ie:
+$$F_{ext}(t)=0$$
+Free vibrations is really another word for #second_order_homogenous But it puts emphasis that we are modelling a friction-mass-spring system and that we can describe the solution—the position of the mass $y(t)$—in terms of frequency, phase shift, and amplitude: $\omega,\ \phi,\ A$.
+## System derivation
+Let's start solving the DE!
+$mr^2+br+k=0$ is the characteristic polynomial.
+we can solve this polynomial equation using the quadratic formula:
+$$r_{1,2}=\frac{-b\pm\sqrt{ b^2-4mk }}{2m}$$
+Our solution $y_{h}$ will depend on what the value inside the $\sqrt{\quad}$ is.
+This nested value, $b^2-4mk$ is called the discriminant.
+We have three cases:
+</br>
+over damped case, $b^2-4mk>0$:
+$$\begin{align}r_{1}&\ne r_{2}\quad (\text{non-repeated roots})\\y_{h}(t)&=c_{1}e^{r_{1}t}+c_{2}e^{r_{2}t}\xrightarrow[\text{as }t\to \infty]{}0 \\r_{1,2}&=-\frac{b}{2m}\pm \frac{\sqrt{ b^2-4mk }}{2m}<0\end{align}$$
 
+</br>
+critically damped case, $b^2-4mk=0$:
+
+$$\begin{align}r_{1}&=r_{2}\quad \text{(repeated roots)} \\y_{h}(t)&= c_{1}e^{rt}+c_{2}te^{rt}\xrightarrow[\text{as }t\to \infty]{}0 \\r_{1,2}&= -\frac{b}{2m}<0\end{align}$$
+</br>
+under damped case, $b^2-4mk<0$:
+$$\begin{align}r_{1,2}&=\alpha\pm i\beta \quad \text{(imaginary roots)} \\ y_{h}(t)&=e^{\alpha t}(c_{1}\cos (\beta t)+c_{2}\sin(\beta t))\xrightarrow[\text{as }t\to \infty]{}0 \\ r_{1,2}&=-\frac{b}{2m}\pm i\frac{\sqrt{ 4mk-b^2 }}{2m}\end{align}$$
+</br>
 #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:
+
+Recall the solution for the under damped case, $y_{h}(t)=e^{\alpha t}(c_{1}\cos (\beta t)+c_{2}\sin(\beta t))$
+It's hard to visualize a sum of two trig functions. It would be nice if we could define $y_{h}$ in terms of phase shift, amplitude, and frequency.
+The frequency part is easy: let $\omega=\beta$ (called the angular frequency)
+Now for phase:
+Which trig identity could help us here? This one looks handy:
 $\sin(\alpha+\beta)=\sin \alpha\cos \beta+\cos \alpha \sin \beta$
-cant make c_1 c_2 sin or cos so what we do?
-do a power transform to convert cartesian into cylindrical coordinates
+Ah but $c_1$ $c_2$ isn't in terms of $\sin$ or $\cos$, so what we do?
+Do a substitution! Convert cartesian into polar 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 beautifully the world of electronic circuits of R L C also has these equations. Biology too. Nature is beautiful and harmonic.
+$Ae^{\alpha}(\sin \phi \cos \omega t+\cos \phi \sin \omega t)$
+$=Ae^{\alpha}\sin(\omega t+\phi)$ Much neater looking!
+$\phi$ is the phase shift in radians.
+$\frac{\omega}{2\pi}$ is the natural frequency (oscillations/second).
+Taking the inverse, $\frac{2\pi}{\omega}$ is the period (seconds/oscillation).
+This behavior is all a result of classical mechanics, but rather beautifully, electronic circuits composed of resistors, inductors, and capacitors can also be described by this equation. You'll see it in biology too.
+*"... Nature is beautiful and harmonic."* -Prof (not an exact quotation, I didn't catch what he precisely said.)
 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 vibration (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)
-$b^2-4mk$
-$y(t)=c_{1}\cos 5t+c_{2}\sin 5t$
+so we can compute $A$ and $\phi$ from $c_1$ and $c_2$.
+## Examples
+#ex #second_order_homogenous #IVP
+Solve the following IVP's when $b=0$,  $b=6$, $b=10$, and $b=12$
+(This is more of a study than an example problem. We have already solved equations like this so treat it as an exploration. Note the big takeaways at the end of each case)
+$$y''+by'+25y=0 \qquad y(0)=1\quad y'(0)=0$$
+1.) b=0 -> no friction in the system (undamped)
+$b^2-4mk<0$ (under damped)
+$\alpha=-\frac{b}{2m}=0$
+$\beta=\frac{\sqrt{ 4mk-b^2 }}{2m}=\frac{\sqrt{ 4(1)(25)-0^2 }}{2(1)}=5$
+$y(t)=e^{0t}(c_{1}\cos 5t+c_{2}\sin 5t)=c_{1}\cos 5t+c_{2}\sin 5t$
 $y(0)=c_1=1$
+$y'(0)=-5\sin 5t+5c_{2}\cos 5t$
 $y'(0)=0=c_{2}$
-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 $b^2-4mk=36-4*25=-64$
-$r_{1,2}=-\frac{6}{2}\pm4i$
+$y(t)=\cos(5t)=\sin\left( 5t+\frac{\pi}{2} \right)$ (by trig identity)
+$$\phi=\frac{\pi}{2} \qquad \omega=5 \qquad A=1$$
+Important take away from undamped case: amplitude is constant 1, oscillates forever.
+2.) b=6
+compute $b^2-4mk=36-4*25=-64<0$
+Still under damped situation.
+$r_{1,2}=-\frac{b}{2m}\pm \frac{\sqrt{ 64 }}{2m}i=-3\pm4i$
 $y(t)=e^{-3t}(c_{1}\cos4t+c_{2}\sin4t)$
-still under damped situation.
 $y(0)=1=c_{1}$
-$y'(0)=0=-3c_{1}+4c_{2}\Rightarrow c_{2}=\frac{3}{4}$
-$A=\frac{5}{4}$
-$\tan \phi=\frac{4}{3}$
+$y'(t)=e^{-3t}(-4c_{1}\sin 4t+4c_{2}\cos 4t)+ -{3}e^{-3t}(c_{1}\cos 4t+c_{2}\sin 4t)$
+$y'(0)=0=4c_{2}-3c_{1}\Rightarrow c_{2}=\frac{3}{4}$
+$A=\sqrt{ c_{1}^2+c_{2}^2 }=\frac{5}{4}$
+$\tan \phi=\frac{c_{1}}{c_{2}}=\frac{4}{3}$
 $\phi \approx 0.9273\dots$
 $$y(t)=\frac{5}{4}e^{-3t}\sin(4t+\phi)$$
-"I know engineers love calculators, I know mathematicians hate 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
-$r_{1,2}=-5$
+Important take away: We computed $\phi$ and $A$ in this example. We found a way to know the envelope of the amplitude of the oscillating system and it's phase shift.
+
+"I know engineers love calculators, I know mathematicians hate calculators, and that's probably the only difference between mathematicians and engineers." -Prof (referring to calculating arctan(4/3) on an exam)
+3.) b=10
+$r_{1,2}=-\frac{10}{2}\pm \frac{\sqrt{ 10^2-4*25 }}{2}=-5$ (repeated root, critically damped)
 $y(t)=(c_{1}+c_{2}t)e^{-5t}$
 $y(0)=1=c_{1}$
-$y'(0)=c_{2}-5c_{1}=0$
+$y'(t)=-5c_{1}e^{-5t}+c_{2}e^{-5t}-5c_{2}te^{-5t}$
+$y'(0)=-5c_{1}+c_{2}=0$
 $c_{2}=5$
 $y(t)=(1+5t)e^{-5t}\rightarrow0_{as\ t\to\infty}$
-$y(t)=(1+5t)e^{-5t}>0$
-4) b=12
- $r_{1,2}=-6\pm \sqrt{ 11 }$
-$y(t)=c_{1}e^{(-6\pm \sqrt{ 11 })t}+c_{2}e^{(-6-\sqrt{11 })t}$
+Big take away: notice that $y(t)$ here is always strictly positive ($y(t)$>0). It doesn't oscillate at all. (Also if you recall from PHYS130, critically damped is the optimal amount of damping to get the mass to settle to the equilibrium position as fast as possible.)
+4.) b=12
+ $r_{1,2}=-\frac{12}{2}\pm \frac{\sqrt{ 144-100 }}{2}-6\pm \sqrt{ 11 }$
+$y(t)=c_{1}e^{(-6+ \sqrt{ 11 })t}+c_{2}e^{(-6-\sqrt{11 })t}$
 $y(0)=c_{1}+c_{2}=1$
 $y'(0)=(-6+\sqrt{ 11 })c_{1}+(-6-\sqrt{ 11 })c_{2}=0$
+solving the linear system of equations gives:
 $c_{1}=\frac{11+6\sqrt{ 11 }}{22}$
 $c_{2}=\frac{{11-6\sqrt{ 11 }}}{22}$
-this is an over damped case.
+Big takeaway: this is an over damped case.
 
-lets look at the graphs: (graphs featuring the three cases shown on projector.)
+Final big takeaway:
+Lets look at the graphs, we are assuming $t>0$ to keep things realistic. Notice how the black line ($b=10$) decays faster than the $b=12$ and the $b=6$. Also notice how there can be a local max/min in an over damped system. It's possible to have local max/min in a critically damped system as well. This is a cause of giving the mass some initial speed when you released it. An under damped system has infinitely many local max/mins. (psst, A possible question you may come across is to ask what $t$ an oscillatory system reaches it's max/min value, and what value it attains). An over damped and critically damped system will never cross the x axis unless you give it sufficient speed when you release it and that you're "throwing it" in the direction of the equilibrium position. Finally, notice how the under damped system will cross the x axis infinitely many times, and has a constant period, and can be viewed as a sin wave with an exponentially decaying envelope and a bit of phase. 
+![mbk-graphs.png](drawings/mbk-graphs.png)
+Hey, we are all done! See, free vibrations is more of a case study rather than new content.
 #end of lec 12

content/_index.md

@@ -14,7 +14,7 @@ Good luck on midterms! <3 -Oct 18 2023
 [Variation of parameters (lec 9-10)](variation-of-parameters-lec-9-10.html) 
 [Cauchy-Euler equations (lec 10-11)](cauchy-euler-equations-lec-10-11.html)
 [Reduction of order (lec 11)](reduction-of-order-lec-11.html)
-[Free vibrations (lec 11-12)](free-vibrations-lec-11-12.html) (raw notes, not reviewed or revised yet.)
+[Free vibrations (lec 11-12)](free-vibrations-lec-11-12.html) 
 [Resonance & AM (lec 13-14)](resonance-am-lec-13-14.html)
 [Laplace transform (lec 14-16)](laplace-transform-lec-14-16.html) (raw notes, not reviewed or revised yet.)
 [Solving IVP's using Laplace transform (lec 17-18)](solving-ivps-using-laplace-transform-lec-17-18.html) (raw notes, not reviewed or revised yet.)