# 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)

#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 beautifully 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 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)
$b^2-4mk$
$y(t)=c_{1}\cos 5t+c_{2}\sin 5t$
$y(0)=c_1=1$
$y'(0)=0=c_{2}$
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 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
$r_{1,2}=-5$
$y(t)=(c_{1}+c_{2}t)e^{-5t}$
$y(0)=1=c_{1}$
$y'(0)=c_{2}-5c_{1}=0$
$c_{2}=5$
$y(t)=(1+5t)e^{-5t}\rightarrow0_{as\ t\to\infty}$
$y(t)=(1+5t)e^{-5t}>0$
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