MATH201/content/Resonance & AM (lec 13-14).md

#start of lec 13 He has good news. he's excited to tell us about electric currents! In particular, how a radios use 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} which can be calculated as shown above.

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.