6.4 KiB
#start of lec 13 He has good news. he's excited to tell us about electric currents! In particular, how a radio uses resonance to selectively listen to a particular frequency:
Resonance
Let's imagine a mass spring system which has an applied force with a forcing frequency of \gamma and an amplitude of F_{o} (a constant):
my''+by'+ky=F_{o}\cos(\gamma t)
(Think of the driving force being the radio transmitter, and the mass-spring system is an LC tank circuit in a old-style radio)
In order to study the phenomenon of resonance, we need an underdamped system.
so we let: b^2-4mk<0 (ie: complex roots)
then the homogenous solution becomes:
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 called the angular frequency and equals \frac{\sqrt{ 4mk-b^2 }}{2m}
and where Ae^{-bt/2m} is called the transient part of the equation (goes to 0 as t->\infty).
For particular solution, we use method of undetermined coefficients #mouc
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+b^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}{\sqrt{ (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 as time progresses but y_p stays? y_p is the steady state part of the solution. If you were to graph y(t) you would see a "beating" effect due to the sum of the two sins that eventually decays off.
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 the equivalent of tuning a radio receiving 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) is decreased to zero we get stronger and stronger resonance.
Lets find the maximum of the amplitude (resonance point)
take the derivative of \mu wrt to \gamma:
\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
case one: \gamma=0 not interesting, because then the force applied would just a constant force.
case two:
\gamma_{r}=\sqrt{ \frac{k}{m}-\frac{b^2}{2m^2} } where the r means resonance
By plugging in \gamma_{r} into \mu() we can know the maximum amplitude at resonance: \mu_{max}(\gamma_{r})=\mu(\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 very 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
#start of lecture 14
Amplitude modulation
Last lecture we showed we can selectively listen to a specific signal by using resonance.
He has something else he's excited to show us; amplitude modulation! Lets recall from last lecture:
my''+ky=F_{o}\cos(\gamma t) (undamped driven mass-spring system)
solving characteristic equation:
\omega=\sqrt{ \frac{k}{m} }
y_{h}(t)=c_{1}\cos(\omega t)+c_{2}\sin(\omega t) ; \gamma\ne\omega
y_{p}=A\cos(\gamma t) we can guess the particular solution is a constant times \cos. There will be no \sin term on the LHS as there's no first derivative (aka no friction)
\Rightarrow A=\frac{F_{o}}{m(\omega^2-\gamma^2)}\cos(\gamma t)
y(t)=c_{1}\cos(\omega t)+c_{2}\sin(\omega t)+\frac{F_{o}}{m(\omega^2-\gamma^2)}\cos(\gamma t)
assume y(0)=0=y'(0)
c_{2}=0 c_{1}=-\frac{F_{o}}{m(\omega^2-\gamma)}
y(t)=\frac{F_{o}}{m(\omega^2-\gamma^2)}(\cos(\gamma t)-\cos(\omega t))
Hmm, It's not very easy to visualize a cos minus cos term.
Use trig identity to make equation easier to visualize: 2\sin(\alpha)\sin(\beta)=\cos\left( \frac{{\alpha-\beta}}{2} \right)-\cos\left( \frac{{\alpha+\beta}}{2} \right)
2\gamma t=\alpha-\beta
2\omega t=\alpha+\beta
\Rightarrow \alpha=\frac{(\omega+\gamma)t}{2} \beta=\frac{(\omega-\gamma)t}{2}
y(t)=\frac{F_{o}}{m(\omega^2-\gamma^2)}\sin\frac{(\omega-\gamma)t}{2}\sin\frac{(\omega+\gamma)t}{2}This is an amplitude modulated signal! also can be seen as a "beating frequency".
To recap, taking a frictionless mass spring system which is initially at rest, and applying a driving frequency that is strictly different from the system's natural frequency, results in a displacement that follows an AM modulated signal. How cool is that!
Sometimes I feel this effect when I'm sitting in the back of the bus, where the bus is stopped at a red light. I wonder if this modulated vibration pattern is attributed to this exact phenomenon.
Shortcut for solving DE of a mass spring system
!Drawing 2023-10-06 13.24.11.excalidraw
my''+by'+ky=mg+F
move mg to LHS and replace y with y_{new} (remember, the \frac{mg}{k} is a constant, its derivative is 0):
m\left( y-\frac{mg}{k} \right)''+b\left( y-\frac{mg}{k} \right)'+k\left( y-\frac{mg}{k} \right)=F
my_{new}''+by_{new}'+ky_{new}=F
This simplifies our approach to solving a mass spring system. We could do it without this rearrangement, but it's more complex as the RHS has a sum of two terms. Either way works though, pick what you like.