3.2 KiB
#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