Physical example for this kind of "resonance" ? - Page 1
Forum section: electronics / general

author: vlad777
author_rank: Frequent Contributor
timestamp: February 07, 2017, 11:53:29 pm
content: HiIn this video Gilbert Strang talks in part about resonance (as a special case for differential equation).What kind of system is described with this equation?y'=a* y(t) + e^(s*t)And what resonance in this system physically is?("Resonance" occurs when a==s.)Edit:Is there a circuit that would be representative of this phenomenon?Edit2:What I don't understand, is that there are no frequencies here , so how can we have resonance?Many thanks.

author: ataradov
author_rank: Super Contributor
timestamp: February 08, 2017, 12:45:58 am
content: It is not a real resonance, just a removable discontinuity in the solution. It is removed by introducing that special form of the solution at the point of discontinuity. No perceivable physical change happens at that point.And the system - he told what it is. The bank account with interest and additional investments. When the interest rate and additional investment rate are the same, some math breaks down a bit. In real life noting particular will happen.

author: vlad777
author_rank: Frequent Contributor
timestamp: February 08, 2017, 12:53:31 am
content: I don't know, hard core scientists chose the word "resonance" and I would like to know why.(I get what you are saying BTW, the removing of discontinuity.)

author: ataradov
author_rank: Super Contributor
timestamp: February 08, 2017, 12:58:23 am
content: I think it is just a misnomer. Even hardcore scientist use words loosely sometimes. In this case this is clearly not a classical resonance, since function is pretty much smooth except for this one point.

author: vlad777
author_rank: Frequent Contributor
timestamp: February 08, 2017, 01:26:50 am
content: Could it be something about amplitude magnitude as, for example generator freq, approaches resonant frequency of a series LC circuit?(Or something like that)Or is it really a misnomer in a sense that in resonance frequencies are same, but here rates are the same?

author: ataradov
author_rank: Super Contributor
timestamp: February 08, 2017, 01:50:05 am
content: Quote from: vlad777 on February 08, 2017, 01:26:50 amNo. In a real resonance function behavior changes around the resonance point. Here function is continuously smooth. There is no way to show this physically, as this is not a physical phenomenon, it is pure math problem.

author: rstofer
author_rank: Super Contributor
timestamp: February 08, 2017, 06:24:33 am
content: Here is a document describing resonance in differential equations:http://www.math.utah.edu/~gustafso/2250resonance.pdf

author: ataradov
author_rank: Super Contributor
timestamp: February 08, 2017, 06:25:47 am
content: Quote from: rstofer on February 08, 2017, 06:24:33 amThis is a classical resonance. What's in the video is a very-very different thing.

author: JacquesBBB
author_rank: Frequent Contributor
timestamp: February 08, 2017, 09:13:16 am
content: Usually, we speak of resonances when you have a motion with a proper frequency that is excited with the same frequency.If you want to think of the present problem in these terms, just assume that \(s\) and \(a\) are pure imaginary. Just replace them by\( i s\) and \(i a\).Then the differential equation without second member ,$$ {dy \over dt} = i a y$$is the differential equation of a rotator of frequency \(a\), and the full problem is$$ {dy \over dt} = i a y + e^{ist}$$.This is the equation for the rotator of frequency \(a\), excited by an oscillation of frequency \(s\).If the two frequencies are different, the final motion is a sum of two oscillations of frequencies \(a\) and \(s\), and the motion isbounded$$ y(t) = {{e^{ist} - e^{iat}}\over {s-a}} + y(0) e^{iat } $$But if the exiting frequency \(s\) is the same as the proper frequency \(a\) , you end up with an unbounded motionwith solution$$ y(t) = (y(0) + t) e^{iat} $$for a mechanical problem, this can lead to the rupture of the system. In practice, the system does not go to infinity, as often, there aresome additional non linear components that will change the proper frequency as the amplitude (of \(y\) ) changes, so you thenescape of the resonance., but you still have a motion of very large amplitude.In the present case, the situation is somewhat different, as the coefficients \(a\) and \(s\) are real (although it is not mentioned), butthe problem inherits the denomination of the oscillator case.Sorry, I do not have the proper syntax for inline latex expression.

author: vlad777
author_rank: Frequent Contributor
timestamp: February 09, 2017, 01:46:31 am
content: JacquesBBB, thank you very much, can you please give a link or an image of the mechanical rotator?Is it like a rotor in a motor? What is the natural frequency of a rotator in a mechanical example?Because I imagine that a rotor can spin at any rate?

author: T3sl4co1l
author_rank: Super Contributor
timestamp: February 09, 2017, 04:07:36 am
content: A flexural resonator is a good example. A rod sticking up out of an infinite block, say. The resonant frequency is determined by the equivalent* spring rate and mass of the rod, and the rod is able to flex in the Re and Im directions. If you apply a rotating acceleration**, it simply wobbles around, the tip tracing out a circular path. (There are two degrees of freedom, so rotation is not a unique solution. We get this simplification by constraining the input to circular motion, and simplifying the result to show just the circular motion.)*Because, as I am so fond of emphasizing: real elements are transmission line segments***, but if we're okay with analyzing the system at its first resonant frequency, it can be approximated as a lumped equivalent model.**Which is pretty stupendously challenging, what with that infinite block weighing everything down! Which means to actually test this in the lab, we need a very much finite reaction mass, which means it will have a finite mechanical impedance, and will exchange energy with the driving source, and all that. Nothing unexpected, because passive linear systems must obey reciprocity, but it makes it harder to actually implement.***Which is even worse in this mechanical analog, because there are many different kinds of real motion: longitudinal (which carries the bulk of the bending moment in this example), translational, shear, surface acoustic... And many of these modes are dispersive, giving rise to anharmonic resonances.****Properly, accurately explaining real physical processes, in common spoken language, is horrible... It would be nice to simply say, "go to college and learn about these things in depth so you know what I'm talking about", but that's hardly practical...Tim

author: vlad777
author_rank: Frequent Contributor
timestamp: February 09, 2017, 04:37:50 pm
content: Oh so tip of the rod is moving vertically and horizontally , and the position of the rod tip is a complex numberwhere Re is horizontal and Im is vertical "coordinate".The speed of the tip of the rod is equal to i*a*position .i*a is like a two dimensional spring constant.And e^(i*s*t) is cos(s*t)+i*sin(s*t) by Euler's formula, and in our example it is just a cos force in Re direction and a sin force in Im direction.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: February 09, 2017, 08:09:51 pm
content: Yes.I don't think I'm missing anything about the problem, and that's an actually representative example...So here's a riddle...Suppose you have such a setup: you have an extremely massive flywheel, attached to a gimbal so that it can be spun on its natural axis, and that axis can be rotated. (Ignore its gyroscopic inertia*: we'll assume the gimbal can be turned hard enough to get it where it needs to go for the problem.) The flywheel axis can be pointed straight vertical (i.e., parallel to gravity), all the way down to horizontal (perpendicular to gravity). A thin rod sticks out from the axis of the flywheel, so that the rod is our 2D SHO (Simple Harmonic Oscillator).Thus, when the axis is pointed vertical, there is no acceleration pushing the rod around, and there is no vibration, no matter what frequency the flywheel revolves at.When the axis is off vertical, gravitational acceleration has a component perpendicular to the rod, and therefore is able to deflect it.The meat of the riddle is this:1. If the axis is tilted off vertical, will the rod deflect at all?2. Will the rod deflection follow the flywheel's angle?3. Does the deflection vary with frequency? Is there a resonance? What is the peak amplitude (if finite), or does it grow without bound?4. Suppose we rotate the Earth, instead of rotating the flywheel and its axis. (Um, yeah, okay... good luck with that!) Or more theoretically speaking -- suppose we change the direction of acceleration (nevermind how that is achieved). Has the problem changed? If so, how? If not, how is it equivalent?*Though, I have a feeling this will still bear some significance for the part of the problem we're interested in. Namely: the orbital vibration carries angular momentum, and therefore will precess when tilted.You must be very careful with problems like this, because everything is in motion, and motion is continuous. You can't simply flip the angle from one way to the next -- it has to be done smoothly, and the torque on all components must be accounted for.You can often use a shortcut, like potentials and total energy, or thermodynamics, to arrive at the correct answer, but remember that these methods -- when used correctly -- do account for the dynamics, and if you put in conditions that require impossible dynamics, you'll probably get unreasonable answers!So, I conceived this as a means to test the system: the most straightforward excitation would be, for example, having the flywheel-and-rod part sitting on a table, and sliding it around a circular path while resting on the table. This delivers a circular acceleration, and therefore the rod will flex (relative to the flywheel, and whatever source is driving this motion), consistent with my earlier example.In what other ways can we apply a rotating acceleration? Well, suppose we rotated the whole system on an axis, so that gravity becomes our rotating acceleration!Great, but that makes things muddy, because now the axial rotation of the system is an additional force, that we must account for.So the conclusions should be:1. Yes. The rod will be statically deflected, according to its sprung mass, spring rate, and the perpendicular component of gravity. (All linear terms, given that we aren't talking about the angle of gravity, and as long as the angle of the rod's deflection is also small. Since, a very flexible rod could bend completely over, and then the acceleration becomes distributed and awkward. A stiff metal rod, with a deflection much less than its length, will be nice and linear.)2. No, at least at small angular velocities. As long as everything remains quasi-static, the rod will simply be pulled a little off axis, and that's fine. (Using more flexible materials, this can be turned into a rotary flex shaft. Handy!)3. Yes, but this is complicated. It is well known that flexible rods, attached to spinning flywheels, can bend and flop around, even without an applied static acceleration! And, once the deflection leaves the linear range, it very quickly (exponentially? hyperbolically?) diverges! The problem is, at sufficient angular velocity and distance from the rotational axis, the centripetal acceleration exceeds gravitational acceleration, changing the problem...So, this is not at all a good substitute for the simplest case (swirling the flywheel on a table, but not rotating it on axis)!4. Supposing we could rotate the Earth around the flywheel-and-rod system, this would work fine. The lack of equivalency is because rotation is invariant under transformation: when the flywheel is rotating, it "knows" it's rotating, and anything else watching it will reach the same conclusion.So, trying to excite this system by spinning it under gravity? Not a good idea!(And if this ALL seems horribly contrived and arbitrary? Consider it a lesson on brainstorming problems, and checking them for validity -- I first considered cases where acceleration is applied to such a system, then designed an experiment to test and verify those methods. The rotating case breaks, but the swirling case is just fine.)Tim

author: Vtile
author_rank: Super Contributor
timestamp: February 09, 2017, 09:08:52 pm
content: Quote from: T3sl4co1l on February 09, 2017, 04:07:36 amNope, nope. It just so difficult task that most do not ever get in that level. You need to know the thing you are explaining in the level that you don't need to think about it. Even that doesn't count if the explainer doesn't have ability to go to the shoes of the listener. I insist of saying that if you can not explain something without the math you do not know it really.

author: orolo
author_rank: Frequent Contributor
timestamp: February 09, 2017, 10:13:55 pm
content: As an electronic example, an inverting differentiator followed by an inverting adder could perhaps provide an example. Given a voltage \$ y \$, it is differentiated and then added a sinusoidal excitation. Ideally, we should get a differential equation as:\$ y = RC\frac{\mathrm{d}y}{\mathrm{d}t} - f \$where \$ f \$ is the excitation.Analyzing the Bode plot of the circuit, it resonates at 4.815kHz, approx. So, feeding a 10mV peak sinusoid of that frequency gets this response:If a 1kHz sinusoid is given, there is no resonance:I have built this circuit in a few minutes, but I think it roughly captures the idea.

author: vlad777
author_rank: Frequent Contributor
timestamp: February 10, 2017, 01:14:52 am
content: @oroloI think f needs to be quadrature although your schematic looks little bit like a quadrature osc.Can you please provide .asc file?@T3sl4co1lWell everybody goes numerically, Runge–Kutta right?You just need to see if the disturbance is transient.

author: orolo
author_rank: Frequent Contributor
timestamp: February 10, 2017, 03:06:21 pm
content: Quote from: vlad777 on February 10, 2017, 01:14:52 amI attached the schematic with the LT1001 op amp model, which comes with the standard LTSPICE. The resonance is at about 3.58kHz. At that frequency, the 10mV excitation will be amplified to about 4V peak. I think this circuit could be built more cleanly with an integrator, instead of a differentiator, but then we get an equivalent integral equation \$ y = RC\int y - f \$ which is not immediatly equivalent to the original equation.@T3sl4co1lVery interesting mechanical model. I think a first, basic resonance mode is easy to derive and explain. Let me dust my mechanics a bit. If we follow the coordinates of the top of the rod with respect to its rest point, in a polar reference frame rotating with the disc, and we disregard gravity (for now) and variations in height, we get:\$ \left. \begin{array}{ccc} \ddot{r} & = & \left(\omega^2 - k^2\right)\cdot r \\\ddot{\theta} & = & 0 \\ \end{array}\right\} \$In this very basic model, k^2 is related to the elastic constant of the top of the rod (the restoring force that tends to keep it straight), and omega is the angular velocity of the disk. At this very basic level, the elasticity of the rod will conflict with the centrifugal force (we are in a rotating frame), and we get small oscillations of the top of the rod with frequency \$\sqrt{k^2 - \omega^2}\$. Of course, \$k > \omega \$, otherwise the rod is more like a rope that would then to flatten over the disk. The top of the rod will oscillate with angular frequency modulated by the rotational speed of the disk.Also, \$ \theta\$ is a cyclic coordinate, that is, the top of the rod could rotate with whichever angular velocity. If the rod is rigid, this will be precluded by the fact that it cannot twist, so the natural behavior will be close to $\theta = 0$ (or any other constant): the top of the rod rotates with the disk.If we gradually add gravity, some acceleration 'a' between 0 and 9.81, that perturbation, in our rotating frame, will rotate in the opposite direction, so the equation becomes:\$ \left. \begin{array}{ccc} \ddot{r} & = & \left(\omega^2 - k^2\right)\cdot r \ + \ a\cos(\omega t) \\\ddot{\theta} & = & -a\sin(\omega t) \\ \end{array}\right\} \$Disregarding any other effects, looking at the equation for \$ r\$, there will be resonance when the centrifugal force modulates the frequency of radial oscillations until it reaches \$ \omega \$ (or one of its harmonics).That is, a simple condition for resonance is \$ \sqrt{k^2 - \omega^2} = \omega \$. In other words, \$ \omega = k/\sqrt{2}\$.This resonant mode does not depend on the amount of gravity. It derives from a centrifugal weakening of the elastic constant of the rod until it resonates with the periodic gravitational perturbation. Perhaps this resonance is practically unreachable if the rod is very rigid. I'm sure there are more convoluted modes of oscillation, but I would expect trouble at that specific mode.