MATH201/content/Free vibrations (lec 11-12).md

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