83 lines
6.4 KiB
Markdown
83 lines
6.4 KiB
Markdown
#start of lec 13
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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:
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# Resonance
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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):
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$my''+by'+ky=F_{o}\cos(\gamma t)$
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(Think of the driving force being the radio transmitter, and the mass-spring system is an LC tank circuit in a old-style radio)
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In order to study the phenomenon of resonance, we need an underdamped system.
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so we let: $b^2-4mk<0$ (ie: complex roots)
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then the homogenous solution becomes:
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$y_{h}(t)=e^{-bt/2m}\left( c_{1}\cos\left( \frac{\sqrt{ 4mk-b^2 }}{2m} t\right)+c_{2}\sin\left( \frac{\sqrt{ 4mk-b^2 }}{2m} t\right) \right)$
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$y_{h}(t)=Ae^{-bt/2m}\sin(\omega t+\phi)$
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where $\omega$ is called the angular frequency and equals $\frac{\sqrt{ 4mk-b^2 }}{2m}$
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and where $Ae^{-bt/2m}$ is called the transient part of the equation (goes to 0 as t->$\infty$).
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For particular solution, we use method of undetermined coefficients #mouc
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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.
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$A_{1}=\frac{F_{o}(k-m\gamma)}{(k-m\gamma^2)^2+b^2\gamma^2}$
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$A_{2}=\frac{F_{o}b\gamma}{(k-m\gamma^2)^2+b^2\gamma^2}$
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$y_{p}(t)=\frac{F_{o}}{(k-m\gamma^2)^2+b^2\gamma^2}((k-m\gamma^2 )\cos(\gamma t)+b\gamma \sin \gamma t)$
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$\sin(\alpha+\beta)=\sin \alpha \cos \beta+\cos \alpha \sin \beta$
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$k-m\gamma^2=A\sin \theta$
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$br=A\cos \theta$
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$A=\sqrt{ (k-m\gamma^2)^2+b^2\gamma^2 }$
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$\tan \theta=\frac{k-m\gamma^2}{b\gamma}$
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$y_{p}(t)=\frac{F_{o}}{(k-m\gamma^2)^2+b^2\gamma^2}((k-m\gamma^2 )\cos(\gamma t)+b\gamma \sin \gamma t)=\frac{F_{o}\sqrt{ (k-m\gamma^2)^2+b^2\gamma^2 }}{(k-m\gamma^2)^2+b^2\gamma^2}\sin(\gamma t+\theta)$
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$=\frac{F_{o}}{\sqrt{ (k-m\gamma^2)^2+b^2\gamma^2 }}\sin(\gamma t+\theta)$
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where we define $\mu(\gamma)=\frac{1}{\sqrt{ (k-m\gamma^2)^2+b^2\gamma^2 }}$ called the gain factor
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and the general solution is $y(t)=y_{h}(t)+y_{p}(t)$
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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.
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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.
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Lets find the maximum of the amplitude (resonance point)
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take the derivative of $\mu$ wrt to $\gamma$:
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$\mu'(\gamma)=-\frac{{2m^2\gamma\left( \gamma^2-\left( \frac{k}{m}-\frac{b^2}{2m^2} \right) \right)}}{[(k-m\gamma^2)^2+b^2\gamma^2]^{3/2}}=0$
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case one: $\gamma=0$ not interesting, because then the force applied would just a constant force.
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case two:
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$$\gamma_{r}=\sqrt{ \frac{k}{m}-\frac{b^2}{2m^2} }$$
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where the r means resonance.
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By plugging in $\gamma_{r}$ into $\mu()$ we can know the maximum amplitude at resonance: $$\mu_{max}(\gamma_{r})=\mu(\gamma_{r})=\frac{2m}{b\sqrt{ 4mk-b^2 }}$$
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if $b^2\geq 4mk>2mk$ (overly damped/critically damped case) (no resonance, imaginary numbers) $\gamma=0$
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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.
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what if b=0? (no resistance):
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$my''+ky=F_{o}\cos(\gamma t)$
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$y_{h}(t)=c_{1}\cos \omega t+c_{2}\sin \omega t$ , $\omega=\sqrt{ \frac{k}{m} }$
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$=A\sin(\omega t+\phi)$
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$y_{p}(t)=A_{1}\cos(\gamma t)+A_{2}\sin(\gamma t)$
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assume $\gamma=\omega$ with zero resistance we get:
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$y_{p}(t)=\frac{F_{o}}{2m\omega}t\sin \omega t \underset{ t\to \infty }{ \to }\infty$ (your circuit blows up! Or equivalently, your bridge collapses.)
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#end of lec 13 #start of lecture 14
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# Amplitude modulation
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Last lecture we showed we can selectively listen to a specific signal by using resonance.
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He has something else he's excited to show us; amplitude modulation! Lets recall from last lecture:
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$my''+ky=F_{o}\cos(\gamma t)$ (undamped driven mass-spring system)
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solving characteristic equation:
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$\omega=\sqrt{ \frac{k}{m} }$
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$y_{h}(t)=c_{1}\cos(\omega t)+c_{2}\sin(\omega t)$ ; $\gamma\ne\omega$
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$y_{p}=A\cos(\gamma t)$ 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)
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$\Rightarrow A=\frac{F_{o}}{m(\omega^2-\gamma^2)}\cos(\gamma t)$
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$y(t)=c_{1}\cos(\omega t)+c_{2}\sin(\omega t)+\frac{F_{o}}{m(\omega^2-\gamma^2)}\cos(\gamma t)$
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assume $y(0)=0=y'(0)$
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$c_{2}=0$ $c_{1}=-\frac{F_{o}}{m(\omega^2-\gamma)}$
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$y(t)=\frac{F_{o}}{m(\omega^2-\gamma^2)}(\cos(\gamma t)-\cos(\omega t))$
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Hmm, It's not very easy to visualize a cos minus cos term.
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Use trig identity to make equation easier to visualize: $2\sin(\alpha)\sin(\beta)=\cos\left( \frac{{\alpha-\beta}}{2} \right)-\cos\left( \frac{{\alpha+\beta}}{2} \right)$
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$2\gamma t=\alpha-\beta$
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$2\omega t=\alpha+\beta$
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$\Rightarrow \alpha=\frac{(\omega+\gamma)t}{2}$ $\beta=\frac{(\omega-\gamma)t}{2}$
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$$y(t)=\frac{F_{o}}{m(\omega^2-\gamma^2)}\sin\frac{(\omega-\gamma)t}{2}\sin\frac{(\omega+\gamma)t}{2}$$
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This is an amplitude modulated signal! also can be seen as a "beating frequency".
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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!
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>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.
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# Shortcut for solving DE of a mass-spring system
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![[Drawing 2023-10-06 13.24.11.excalidraw]]
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$my''+by'+ky=mg+F$
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move $mg$ to LHS and replace $y$ with $y_{new}$ (remember, the $\frac{mg}{k}$ is a constant, its derivative is 0):
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$m\left( y-\frac{mg}{k} \right)''+b\left( y-\frac{mg}{k} \right)'+k\left( y-\frac{mg}{k} \right)=F$
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$my_{new}''+by_{new}'+ky_{new}=F$
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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. |