#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