added lec 13

content/Resonance in free vibrations (lec 13).md

@@ -0,0 +1,41 @@
+#start of lec 13
+He has good news. he's excited to tell us about electric currents!
+$my''+by'+ky=F_{o}\cos(\gamma t)$ the system has an applied force with a forcing frequency of $\gamma$ and an amplitude of $F_{o}$ (a constant)
+$b^2-4mk<0$ (complex roots)
+$y_{h}(t)=e^{-bt/2m}\left( c_{1}\cos\left( \frac{\sqrt{ 4mk-b^2 }}{2m} t\right)+c_{2}\sin\left( \frac{\sqrt{ 4mk-b^2 }}{2m} t\right) \right)$
+$y_{h}(t)=Ae^{-bt/2m}\sin(\omega t+\phi)$ where $\omega$ is angular frequency and equals $\frac{\sqrt{ 4mk-b^2 }}{2m}$
+where $Ae^{-bt/2m}$ is called the transient part of the equation (makes equation go to 0 as t->$\infty$)
+We guess: $y_p=A_{1}\cos(\gamma t)+A_{2}\sin(\gamma t)$ where $\gamma\ne \omega$ because if $\gamma=\omega$ and b=0 then we would have to multiply our guess by t.
+$A_{1}=\frac{F_{o}(k-m\gamma)}{(k-m\gamma^2)^2+b^2\gamma^2}$
+$A_{2}=\frac{F_{o}b\gamma}{(k-m\gamma^2)^2+b^2\gamma^2}$
+$y_{p}(t)=\frac{F_{o}}{(k-m\gamma^2)^2+b^2\gamma^2}((k-m\gamma^2 )\cos(\gamma t)+b\gamma \sin \gamma t)$
+$\sin(\alpha+\beta)=\sin \alpha \cos \beta+\cos \alpha \sin \beta$
+$k-m\gamma^2=A\sin \theta$
+$br=A\cos \theta$
+$A=\sqrt{ (k-m\gamma^2)^2+b^2\gamma^2 }$
+$\tan \theta=\frac{k-m\gamma^2}{b\gamma}$
+$y_{p}(t)=\frac{F_{o}}{(k-m\gamma^2)^2+b^2\gamma^2}((k-m\gamma^2 )\cos(\gamma t)+b\gamma \sin \gamma t)=\frac{F_{o}\sqrt{ (k-m\gamma^2)^2+b^2\gamma^2 }}{(k-m\gamma^2)^2+\beta^2\gamma^2}\sin(\gamma t+\theta)$
+$=\frac{F_{o}}{\sqrt{ (k-m\gamma^2)^2+b^2\gamma^2 }}\sin(\gamma t+\theta)$
+where we define $\mu(\gamma)=\frac{1}{(k-m\gamma^2)^2+b^2\gamma^2}$ called the gain factor
+and the general solution is $y(t)=y_{h}(t)+y_{p}(t)$
+
+see how y_h goes to zero but y_p stays? y_p is the steady state part of the solution. if you were to graph it you would see a wah wah effect that decays to zero.
+
+if we make the value in the denominator of the gain factor small the amplitude goes to a very high value, higher than $F_{o}$ this is equivalent to tuning a radio circuit to resonate to a certain transmitted frequency, the amplitude grows to a great value in the receiver when excited with the right frequency. This is resonance! as b (friction) decreases to zero we get stronger and stronger resonance.
+lets find the maximum of the amplitude (resonance point)
+$\mu'(\gamma)=-\frac{{2m^2\gamma\left( \gamma^2-\left( \frac{k}{m}-\frac{b^2}{2m^2} \right) \right)}}{[(k-m\gamma^2)^2+b^2\gamma^2]^{3/2}}=0$
+cases: $\gamma=0$ not interesting, beacuse then the force applied is a constant force.
+$\gamma_{r}=\sqrt{ \frac{k}{m}-\frac{b^2}{2m^2} }$ where the r means resonance
+by taking second derivative$\mu_{max}(\gamma_{r})=\frac{2m}{b\sqrt{ 4mk-b^2 }}$
+if $b^2\geq 4mk>2mk$ (overly damped/critically damped case) (no resonance, imaginary numbers) $\gamma=0$
+if $b^2<2mk<4mk$ (assumed from the beginning above) then we get a resonant frequency $\gamma_{r}=\sqrt{\frac{  2mk-b^2}{2m^2}}$
+
+what if b=0? (no resistance):
+$my''+ky=F_{o}\cos(\gamma t)$
+$y_{h}(t)=c_{1}\cos \omega t+c_{2}\sin \omega t$ , $\omega=\sqrt{ \frac{k}{m} }$
+$=A\sin(\omega t+\phi)$
+
+$y_{p}(t)=A_{1}\cos(\gamma t)+A_{2}\sin(\gamma t)$
+assume $\gamma=\omega$ with zero resistance we get:
+$y_{p}(t)=\frac{F_{o}}{2m\omega}t\sin \omega t \underset{ t\to \infty }{ \to }\infty$ (your circuit blows up! Or equivalently, your bridge collapses.)
+#end of lec 13

content/_index.md

@@ -11,6 +11,7 @@ I have written these notes for myself, I thought it would be cool to share them.
 [Method of undetermined coefficients (lec 8-9)](method-of-undetermined-coefficients-lec-8-9.html)
 [Variation of parameters (lec 9-10)](variation-of-parameters-lec-9-10.html) (raw notes, not reviewed or revised yet.)
 [Free vibrations (lec 11-12)](free-vibrations-lec-11-12.html) (raw notes, not reviewed or revised yet.)
+[Resonance in free vibrations](resonance-in-free-vibrations-lec-13.html) (raw notes, not reviewed or revised yet.)
 </br>
 [How to solve any DE, a flow chart](Solve-any-DE.png)
 </br>