Questions and answers for Phys 2210, Sp 11

Please send us your questions about anything class-related - this week's homework assignment, course content, course administration, or whatever. We will either email an answer to you directly, or if we think the question might be of general interest, we will instead post the question and answer on this page instead, with your name removed. So check this page periodically to see what questions your classmates have asked!

Please email us at Steven.Pollock at colorado.edu or Rachel.Pepper at colorado.edu or arey at jilau1.colorado.edu with "Phys2210" in the subject line of your email!

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Prof. Pollock I found this really cool oscillation video that I think you would appreciate.

http://www.subliminalmessages.com/0076_music.htm

It seems to complement the first problem on the current homework pretty well, with the multiple driving forces. I especially like how you can see the transients dieing out on the pendulum second from the left.

See you in class,

Thanks, this was indeed cool. We haven't covered "coupled oscillators" yet (Taylor Ch 11) but I agree, you can still connect this to what we're doing in class (and the transients superposing with the long term behaviour was pretty dramatic. )

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Physics challenges: A student from our class asked me if I had some fun/interesting challenges, like physics puzzles. Here's what I came up with:

There was a sweet little extra credit problem on hw set#4 that almost nobody tried: "An (ideal, unstretchable, massless) string is wrapped tightly around the earth. Then, 1 extra meter is added to its length. You pull up at *one point* of the string. Approximately how high above the ground can you pull this point? (You may be quite surprised by the answer, I was!)" Did you ever try solving that one? I enjoyed it, myself.

If you want more -

1) The Physics Teacher has a monthly challenge problem "for students and teachers", written by Boris Korsunksy. You can find hard copies of the magazine in the physics library, but it's also all online. (Nice thing about these is that solutions are available :-)

2) I just spotted this site recently, it's a bunch of rather amusing challenge puzzles from Duke University - a little less "calculation" focused than the above, more conceptual -

3) I told you about Fermi problems (Wikipedia has a good introduction to the idea) U. Maryland built up a collection of them -

Cheers, Steve

P.S. Mike Dubson runs a once-a-week informal "Physics X" course for undergrads every fall, it's aimed at prepping for the physics GRE, but he throws lots of good puzzles out for people to work on. Keep your eyes open for it, I recommend it highly, both as GRE prep but also as a way to improve your general physics problem solving skills!

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Preflight #15:
And I read the notes. What does the second to last paragraph say about cc in norther hemisphere and clockwise in southern hemisphere? I couldn't read one of the words in this line: "The coreolis force is the reason why _______ ....." What was the blank?

It says "The Coriolis force is the reason why hurricanes tend to be CClockwise in the northern hemisphere and Clockwise in the southern hemisphere". (See the new main webpage for a little more commentary on this. (It's really very complicated and I have a bit of a hard time wrapping my head around it. There are other elements too, like the motion of the center of the hurricane itself, rather than the relative motion of the winds within the hurricane) -Steve

This is more a clarification question about fourier transform theory in general. The transform helps us to break up a certain function so that we can look at the power of its various frequency components. In practical application I guess this aids in being able to eliminate known causes of noise (as long as you 'know' them by their frequency') so that you could say, eliminate a 60kHz signal that happens to be a microwave that got turned on during an observing run using a telescope or something...My question is regards to moving backwards, the transform of the transform is the original function (to a factor of some constants), which makes sense. So suppose we have a function f and its transform g. Then we edit function g, removing some type of unwanted noise and get the edited transform, g', can we simply take the transform of g' to go back to a trimmed version of f, or am I oversimplifying something?

Yup, you can clean up the transform, get a trimmed version, and then inverse transform to get a cleaned up version of what you started with. This is done all the time, in fact, it's a means to process both visual signals (images) and auditory signals. Some of the cleanup procedures in standard software packages are built around this method.

Here's a cool thing I just learned while prepping for lecture last week. In the old days (like the '60's and '70's!) the Fourier transform was often taken, not by digital software, but optically. (Running light through slits takes the Fourier transform of the slit pattern) And the "cleanup" was physical too, the light was simply BLOCKED in certain spatial regions (which corresponded to undesired frequencies) by a physical mask, and then the image was reconstructed - cleaned up! NASA did this with images that had certain kinds of rastering noise, and the cleanup is dramatic.

When I fired up Mathematica and went to their "demonstrations" page to look for software for class last week, I searched for "Fourier" and found a ton of demos, including this one:
http://www.demonstrations.wolfram.com/ImageDeconvolution/
Seemed a little beyond what we could get to in class, but kind of mathemagical, IMHO.

Steve

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Preflight #14:

is there an easy way to look at and answer a question like number 1? I had to do brief calculations, and i am still much less then convinced it is right.

Yes, there's a quick trick. *IF* two "opposite walls" are both held at T=0, then the Tutorial we did in class last Thursday tells us there's no possible way to get that to work with exponential functions (or linear functions), so you're stuck, you MUST set your separation constant's sign to give you sinusoidal solutions in that dimension.

So for the preflight with T=0 at the top and bottom, you MUST have Y(y) be sinusoidal (and thus, X(x) will be exponential) Simple enough :-) Cheers, Steve

I understand why a pulse is made up of several frequencies, it doesn't make sense to me that a pulse of light be made up of more than one frequency though.

I agree with you that this seems a little magical, but light *is* an electromagnetic wave. It's nothing else, that's what it is! (And, "color" is nothing more than our physiological response to frequency.) If you want to produce a wave (of anything - sound, water, string, or... light) which is "localized" in time, then the only way to do that is to superpose many waves with a spectrum of frequencies. In the case of sound, it means that a short clap will consist of many pitches superposed, and in the case of light, a short laser pulse will consist of many frequencies (colors) superposed. It sounds like you "buy" the mathematics here, but just find the physics puzzling, and I agree, but here the math and the physics are so deeply connected there's no way around it!

I once had this explained to me in a practical way (that wasn't totally satisfying, but helped a little). If I show you a picture of a wave (i.e, a plot of amplitude versus time) and ask "what's the frequency?", you would look at the picture, find the period, and say "f = 1/T". But how, exactly, do you "find the period"? Well, you look at the wave (i.e. you look at *many* repetitions) and see that it repeats, and measure how frequently it repeats. It needs to repeat *over and over* to clearly determine the period. The more repeats, the more confident I am in my answer. But if the wave *doesn't* repeat, then it's not so clear what the frequency (or period) is. You can't "check it", you might not even be able to easily measure it. The shorter the pulse, the less confidence you have in your answer. (And indeed, there simply*isn't* a definite period associated with a short pulse!)

SJP

im still not comfortable with process of inspecting your boundry conditions in order to get our X(x)Y(y)

I've covered Fourier in depth before but I feel unsure of my answer to preflight 1 because of the discontinuous boundary conditions (at the corners). Going over it in class would be nice.

I thought I understood it in class but after doing this preflight I am not entirely sure how to tell whether X(x) or Y(y) is the sinusoidal solution to a partial differential equation. Going over this again/in more detail would be helpful.

All of this seems very confusing and I'm hoping that I'm still holding on...

I think we need a bit more time on solving PDE's for different situations than the steady-state temperature of a square plate. I just want to see the effect of boundary conditions in different problems.

Question 3 and 4, and solving Laplace's equation I general is something I have difficulty with.

It would be nice if we could do a few more class examples of solving laplace equations.

Yup, based on your (and others') feedback (and my general sense that solving PDE's with separation of variables is nontrivial) I plan to spend a good chunk of class time today continuing with this topic. If you still feel confused after today's lecture, though, let me know, we can pursue it further in office hours! SJP

What if we had a function that was sort of periodic, like for blips of equal length separated by equal amounts of time--would we use a Fourier series or a transform? My feeling is a transform, but I'm not sure.

I'm not 100% sure I understand the question, but I *think* the answer is "transform". Basically, if your signal is periodic then you can (and would) use a series, and if your signal is *not* periodic, then you must use a transform. If you have a "clear period" for only 4 periods, but then nothing, (before and after) and that's the *nature* of your signal, then it's really not periodic, and you have to use a Fourier transform.

Of course, in real life, nothing is ever perfectly periodic. (I mean, the universe has a finite lifetime, so nothing can/has repeated "forever") There is something of a blurring of the methods -but if a signal repeats for *many* periods, then the fourier transform will consist of well separated, sharply definied frequencies... i.e., it will effectively "look" like a fourier series! So e.g. to a musician, you can analyze sounds with the method of fourier *series* (even though, of course, in reality the instrument started, and stopped, so techinically it' cannot be perfectly periodic. But in even one or two seconds of music, there are many hundreds or even thousands of cycles, so the Fourier series method is a great approach. )

Steve

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Preflight #13:

I was a little confused by the multiplication of solutions in the Laplace's Eqn section (especially p. 623) because Boas explained it this way: "Any linear combination of solutions of (2.1) is a solution of (2.1) because the diff eq is linear." But the solutions are being multiplied, not added, and I thought the linear combination only worked if they were being added.

I agree, it's confusing. We are multiplying X(x) times Y(y) to get "candidate solutions" for T(x,y). What we are then linearly combining is these candidates!
In other words, if you find T1(x,y) = X1(x)Y1(y) as one candidate that solves Laplace's PDE, and you have another T2(x,y) = X2(x)Y2(y) that ALSO solves Laplace's PDE, then you can argue (by linearity of Laplace's equation) that
aT1(x,y) + bT2(x,y) is ALSO a solution of Laplace's equation.

This will be the topic of Thursday's lecture! Thanks for the question -Steve

I want to know more about the phenomenon that occurs at the functions discontinuities, it seems that this would screw with analytical results which depended partly on the boundary between the discontinuities.

It's called the "Gibbs Phenomenon", and I'm afraid I don't really know much more than the sort of glib comment made in Boas and Taylor about it, I've never studied it. There are some folks in our class taking a math course on Fourier Analysis, maybe their textbook has more details? And of course, there's always http://en.wikipedia.org/wiki/Gibbs_phenomenon
(which is not as incomprehensible as many wiki-math pages!)
http://mathworld.wolfram.com/GibbsPhenomenon.html has some more references at the end if you feel like following up!

 


Why is it exactly that lapace^2_V = 0 for many cases, and what does that exactly mean?

It's an amazingly prevalent PDE in physics! Not sure I have a super-simple explanation, though in E&M, it arises pretty directly:

The voltage and E-field are related by E = - gradient(V) (that's pretty much *exactly* the same equation as our familiar F = -gradient(U) that we've used all term)

But Coulomb's law says that the divergence of E is proportional to charge density. (That's a statement that "E fields are created by charges", or perhaps more accurately, the divergence of E fields is created by charges. It's one of the Maxwell equations)

Well, when you combine those, you get that the divergence of the gradient of V is proportional to charge density. (And mathematically, divergence of gradient is what we've been calling the Laplacian, or "Del^2") So if there is no SOURCE (no charges) around , then this says Del^2V = 0. And so in general, when there is physics that has "sources" that produce divergence (which is common), then in source-free regions, in steady state, you'll get del^2 (something)=0. Like in our upcoming example in class, if there's no source of heat, then del^2(temperature)=0 in steady state.

I don't know if that helps, though -it's kind of formal. It will be the subject of a big chunk of your next (E&M) course!

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Question while reviewing for exam 2:
I thought that the work done by a force field when a particle is moved from a to b would be the same no matter what path was
taken. Taylor's example 4.1 shows that the work can be different depending on what path is taken?

The work done by a *conservative* force field from a to b is the same no matter what path is taken.
But not all fields are conservative!

Taylor's example 4.1 is F = (y,2x). I claim that one is definitely not conservative (it can't be since the different paths give different results!) But d
Do you remember the quick mathematical trick to decide if a given F is conservative without doing any integrals, only taking derivatives?

Hope this helps, Steve

Hey Steve, here is the info about the guy who built a one-man jetpack and flew across the English channel with it.

Video: http://www.livevideo.com/video/BC47350638CC4018A469219DD963B578/yves-rossi-jet-packs-accross-e.aspx

Wikipedia: http://en.wikipedia.org/wiki/Yves_Rossy

Pretty cool if you ask me, very dangerous though (they talk about the many life-threatening problems he's had in some of the articles). Anyway, thought you might enjoy that and it might be useful.

Agreed: cool, dangerous, interesting! Connecting back to our "jetpack" homework problem, though, I'd just note that a rocket is quite different from a jet, (the latter takes advantage of the presence of air for lift of the vehicle!)

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Hey Steve,

I understand the effect of a driving frequency interacting with a natural frequency of say a bridge when the two are equal, but how exactly does a bridge have a natural frequency? It seems as though there has to be an outside source of perturbation on the bridge for there to be any sort natural frequency, and it seems that this must be the driving frequency itself that causes the displacement necessary for a natural frequency. Can you help me through this?

Anything that has an equilibrium configuration, and a restoring force when you LEAVE that equilibrium, will have a natural resonant frequency. This natural frequency really has nothing to do with outside forces, or outside drivers, it's a static property of the system. It comes from how "flat" or "curved" the potential energy function is for the system near the bottom.

Remember when I sketched a "generic" potential energy function on the board, and pointed out that if it has a minimum, that's a point where F=0, it's an equilibrium. And if you "push" the system (doesn't matter HOW you push it) away, it will feel a restoring force back towards equilibrium. So a bridge, e.g. sits there, in equilibrium. If you push on it, internal stresses immediately build up, trying to bring it back to the equilibirum shape. Those forces have to do with the material, the shape, the size... they are natural properties of the bridge. If you're a little away from equilibirum the restoring force will be -kx, where k is a "force constant". k could be measured statically (twist the bridge a little and try to hold it. How hard to you have to hold to keep it like that?) So k is a NATURAL constant associated with the system, it has nothing to do with "resonance" per se, it's just how strongly the system wants to return to equilibrium. But that restoring force automatically means there is ALSO a natural resonance frequency, Sqrt[k/m], because if you do perturb it, it will oscillate around the minimum. If you *drive* the system at a frequency equal to this natural frequency, you'll have resonance.

This helping, or am I missing the essence of your question?

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Preflight #11:
Note that these are *preflights*, meant to have you read and think about the book before class, so that we don't have to spend as much class time talking about the straightforward stuff, but can focus on the deeper concepts, and subtler/harder points! If you are confused about these points, no worries - wait till we've had the lectures on them... at which point, though, you should understand this stuff.

My answers
Q1) We switch from cos(wt) to exp[i w t] because the ODE is easier to solve with the exponential (you can "guess the solution" for the particular solution) And in the end, you can easily take real (or imaginary) part to get back to what you want. The main issue is with the FIRST derivative terms (the drag term) because that flips cos to sin, but doesn't do anything to exp[i w t] except bring down a constant.

Q2) What's different about the plots? One is as a function of w (the driving frequency) and the other is a function of w0 (the resonant, or natural frequency). You would be interested in plotting vs w if THAT is what you vary experimentally. This is most common, you have a system you're investigating (e.g. a molecule) that has a natural frequency. Nothing much you can do manipulate THAT, but you are free to drive it at whatever frequency you like, and it will resonate when you hit the natural frequency. On rare occasions, you can TUNE the natural frequency but can't do much to change the driver, then you want the other plot. This would be the case e.g. when you tune your FM radio to "find" a station, the driver is the EM wave coming in from the broadcaster, you can't change that, but you CAN alter the capacitance or inductance in your radio to change your natural vibration frequency

Q3) The curve 5.17 is a function of w, this one peaks just LEFT of w0 (at the value w2 described in the text). The smaller the damping, the subtler this effect.

And now, some of your comments and our responses:

I didn't know soldiers break step while marching over bridges to avoid breaking the bridge.

I just went to Wikipedia, and here's what they say about it:

http://en.wikipedia.org/wiki/Mechanical_resonance

Resonance may cause violent swaying motions in improperly constructed structures, such as bridges and buildings. The London Millennium Footbridge (nicknamed the Wobbly Bridge) exhibited this problem. A faulty bridge can even be destroyed by its resonance (see "Angers Bridge"; that is why soldiers are trained not to march in lockstep across a bridge, although it is suspected to be a myth, see e.g., MythBusters' 'Breakstep Bridge'.

So, I'm not sure if it's TRUE or a myth, but it does make a nice story. (And it's quite possible that solders break step out of a tradition that began with a concern for breaking the bridge, even if that concern was not well physically grounded!)

Thanks for participating in the preflight, as always

Cheers,
Steve P.



Why is the peak in Fig 5.17 left of the natural frequency? Is it because of the damping?

Yes indeed. You'll get to work this out on the next homework, but it's entirely because the damping term's contribution to the amplitude is ALSO dependent on the driving frequency, so they sort of "couple together". It's not very FAR left of the natural frequency, and for weak damping (which is the only case that "resonance" is physically very interesting) the difference gets really tiny, so one tends to think of it as happening AT the natural frequency, it's an excellent approximation -Steve

Technically speaking are there any real world examples (i.e. not deep space in some extremely rare and almost purely hypothetical situation) where there are no damping terms or are all these examples simply where the damping term is negligible? There seem to be very few occurrences of truly undamped harmonic motion.

I think in "real life" that you would be hard pressed to find a physical system that has NO damping. It would be a system that, once set ringing, would ring *forever*. Doesn't seem very likely. I suppose there are theoretical cases in quantum field theory where you consider systems of undamped oscillators as a sort of "baseline" from which you start, but I guess that's what you called "hypothetical".

It's all a question of degree - there are LOTS of systems where the damping is negligible for the physics purposes you are considering. (Like, the crystal oscillator in your watch, which might as well be ideal during the period over which you are measuring it) For atomic and molecular systems, often the "damping" is again negligible, and thus the "resonance line" is fantastically sharp - think of the extremely well defined frequencies of light emitted from hydrogen case. Those lines DO have a finite width (which might be of interest in SOME situations), but they are so narrow that for many other purposes the damping may as well be ignored.

In any case, I guess we study the undamped case first in part because it's mathematically simplest, and because it's a *starting point* from which to consider the smaller effects that sit on top. -Steve

Why is the amplitude most important in classical mechanics but the phase shift is most important in quantum mechanics?

It's a good question. I suspect the answer is that phase shift is often harder to *measure* classically, it often has very little physical effect, it's just something you sort of measure or notice after the fact. When an oscillator is oscillating, you SEE its motion, phase is an arbitrary feature, determined by when you set time t=0. The relative phase between the oscillator and the driver is not arbitrary, but it just doesn't MATTER much in a lot of applications).

In Quantum Mechanics, wave functions are complex, and systems can interfere. In interference, a phase difference has direct physical consequences - if you have a phase difference of 180 degrees, you can "cancel out" the presence of the object - think of water waves or light waves interfering: constructively in some "bright" spots, but destructively resulting in "dark" spots. Interference is key in lots of quantum situations, so the phase shift can be very important there. (But, of course, there are also plenty of places where it matters in classical mechanics too, in particular if you extend your definition of "mechanics" ever so slightly to include optics, waves, and electromagnetism.

Cheers, Steve

 

I was wondering if you could clarify the line 5.59 saying "Multiply (5.58) by i and add it to 5.57." I see what is being done, but i don't quite understand how. Also, I was wondering where 5.63 came from. It has been used before, but I don't really know where it came from, or if it is just a guess we should know (like 5.61).

Probably easiest to sit down together in office hours and walk through this, it's pretty much just algebra. I would say the "multiply by i and add..." business is just a way of pointing out that if you can solve the ODE with e^[i w t] on the right side, then you can ALSO solve the ODE with cos[wt] and/or sin[wt] on the right side instead. The latter are *physically* interesting (because we tend to "drive" systems sinusoidally), but the former is *mathematically* preferred... because it's easy to solve! So we end up solving this fictitious math problem with complex right hand sides, (because it's easier) knowing full well that at the end of the day, we can take that solution and quickly use it to find the answer to the real problem we started with!

As for 5.63, because of Euler's theorem, it is ALWAYS the case that any complex number at all can always be written in one of two equivalent ways:
either C = Re(C) + i Im(c)
or, C = A e^[i delta], where A is the ampitude, and delta is the phase. See Taylor's figure 2.14, I think it should help! But, if not, let me know, because we'll be using this idea often...

Thanks for asking, Steve

I don't fully understand the "Phase at resonance" section of the text.

Well, I personally think that's the toughest part of this chapter, so you located the right place to focus on :-) We'll certainly talk about it in class (including a Tutorial today, and a homework problem coming up next week). But, if you feel confused after the week is out, let us know, we can talk it through! -S

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Preflight #10:
Note that these are *preflights*, meant to have you read and think about the book before class, so that we don't have to spend as much class time talking about the straightforward stuff, but can focus on the deeper concepts, and subtler/harder points! If you are confused about these points, no worries - wait till we've had the lectures on them... at which point, though, you should understand this stuff.

My answers

Q1-4: I coded these in MMA (the code is now availabe on our lecture notes link). You can try it for yourself. In all 4 cases, I set omega_0=1, and simply varied beta. In ALL cases I set x0=1, and v0=0 (you can see that in the graphs, they all start off with zero slope at the origin)
In the first case beta was small, about 0.3, and you could clearly see "ringing", a couple of zero crossings. That's underdamped.
In the second graph, I set beta = .7. That's still underdamped, but getting CLOSE to critical. It's hard to spot, but if you look closely you can see it overshoots x=0 once. If you start with zero velocity, and still overshoot the origin, I guarantee you are underdamped, there is NO "ringing" that goes on in overdamped cases!

In the third graph, I set beta=1. That's critical damping, and the curve goes to zero rapidly without ever crossing

In the fourth graph, I set beta = 3 or so. That's overdamped, and the curve slowly (but surely) does go to zero.

If I had started with a nonzero, NEGATIVE speed (towards the origin), you might have seen one zero crossing in the overdamped and critical cases. Try it out for yourself, explore the MMA code a little

Q5: We'll talk about this (mostly after spring break). The critical thing to realize is that when you drive an oscillator, you get the "homogeneous" solution (which dies away with that exponential e^[-beta t] term!) and a "particular" solution which is totally fixed/determined by the driver, NOT by your initial conditions. So after a long time, the motion is fully determined, all the dependence on initial conditions is "damped away". We'll talk more about this, keep thinking about it, it's a key point when thinking about driven systems!

Q6: Following up on this, the frequency of the "long term" response will always match the driving frequency. Exactly! The amplitude may not be big (it won't be if your driver is far from the natural frequency), but the frequency matches precisely with the driver.

And now, some of your comments and our responses:

Ok so my question has to do with over damped systems if the answer to this question is covered in the book then I'm sorry. I want to say that overly damped systems approach there equilibrium locations but don't ever reach it. Thanks for your time.

It's a good question, and involves the subtle idea of limits. Technically, if you plot x(t) for an overdamped system, you are correct, x(t) never formally reaches zero. Yet from experience in life, you feel that it certainly reaches zero, and quite quickly. The resolution is that the exponential function, e^[- beta t] never formally goes to zero, but it gets VERY CLOSE, and indeed, relatively fast. In time t=3/beta, for instance, 95% of the original motion is gone. In twice that time, 99.8% of the motion is gone. Would you notice a residual displacement of just 0.2% of the initial? Not likely. And this is a relatively short time! If you start at x=1 m, then after the still modest time of t=20/beta, you are within one atomic distance of x=0! So, in *practice*, the damped oscillator "reaches zero" quite quickly, even if the math says, no, you still have .000000001% of the way to go. This help?

I want to know what exactly what is on the test.

I don't have a question, but can you post a simple review sheet with the basic equations or core concepts you think we should review, as well as the chapters those equations and concepts are from.

I really am just worried about this exam right now.

Have you noticed the "exams" link (followed by "specific exam details") now has a pretty extensive description of what's new on this test. But I'm sure you can accurately guess, we're just following Taylor (and, in the case of the Gravity section, my lecture notes). As far as equations - have you noticed that Taylor does such a summary for you, at the end of each chapter? And, I did the same thing in my lecture notes for the gravity section, (my notes, page 10)

Good luck! Steve

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Preflight #9:

Note that these are *preflights*, meant to have you read and think about the book before class, so that we don't have to spend as much class time talking about the straightforward stuff, but can focus on the deeper concepts, and subtler/harder points! If you are confused about these points, no worries - wait till we've had the lectures on them... at which point, though, you should understand this stuff.

My answers

Q1: The relations we need, mathematically, are that B1=Acos(delta) and B2=Asin(delta). That triangle is the one which exhibits these relations

Q2: Taylor explains it all (it's the buoyant force that results in the SHM), I just wanted you to think it through rather than "accepting" the final result.

Q3: All parameters are the same for those two curves except for phase, they are "phase shifted".
cos(w t + phi) lags a little BEHIND cos(w t), (Think about it, convince yourself)
and vice versa, cos(w t - phi) is a little AHEAD of cos(w t)
In the figure, oscillator two is a little ahead, so phi(2) is a little SMALLER than phi(i)

And now, some of your comments and our responses:

For equation 5.20 on page 171 I don't understand why he says we can dispose of the phase in the x direction but not in the y direction.

What in the world does "redefining the origin of time" mean and why does it allow you to throw away the x component of the phase shift?

In equation 5.20 and 5.23 we are able to get rid of the phi term but I'm not sure why.

in equation 5.20 why cant we dispose of the phase constant for y as we did for x

All of you had a similar question - why can we get rid of phase in one expression, but not the other?

First of all, consider 1-D oscillations, where x(t)=Acos(w t - delta) is the most general solution. In theory, you need two constants (Amplitude A and phase delta) to describe the two initial conditions (say, x and v at t=0). But, IF you grant me the freedom to "start my clock" whenever I want, then I can always get rid of delta, just set it to 0. That amounts simply to a shift of time, i.e starting t=0 at a moment when x is maximized. Since this is periodic motion, it seems pretty reasonable to choose "t=0" at a time wen that's convenient. For one oscillator, there IS always a time (actually many of them, separated by full periods!) when the position will be extreme, so just pick any one of those as your t=0!

This is fine if you have just one oscillator... maybe. There might be situations where you cannot do that, some OTHER reason determines when you want to call t=0. (Maybe somebody else is holding the stopwatch) But if you have nothing but one oscillator in 1D, you could just pick a coordinate system where delta=0. That seem ok?

Now we add a second oscillation (here, in the y direction). You can't "redefine time" AGAIN, if you do, you mess it up for the x-oscillation equation. So we can "start our clock" whenever we want, and Taylor (in Eq 5.20) has decided to do it when the x motion is at an extreme. I.e. he defined t=0 so that delta_x=0. But if x and y motion are independent, there's no guarantee the physical system will ALSO be at a y-extreme at that same time. And thus, we're stuck with one non-zero phase, in Taylor's case, the phase for the y-motion. (We could have set delta_y=0 if we wanted, and then delta_x would be the one we were stuck with. Or, we could have BOTH be non-zero, but why not make life as simple as possible by conveniently choosing when t=0?)

In a sense, what this tells us is that if you have independent oscillations (e.g. in x and y), what really matters is the DIFFERENCE of the two phases. There's just one number, the *relative* phase that is physically meaningful.
This help? Cheers, Steve P.

This extends a little beyond next class's scope, but it is bothering me that I can't resolve why a driven oscillator, physically (mathematically it is sensible), should behave purely like the driving frequency. When considering a linear damping term such as linear air drag, why should this air drag die off and disappear, leaving the oscillator unhindered?

We will see in more detail later what I am going to tell you. The point is that in general when you have driven system the total solution consists of two parts: One which is the solution of the homogeneous equation without a driving term and the other is a particular solution which directly reflects the driving term. In general the homogeneous solution decays with time and for long
times the steady state is determined by the driving term. If you drive your system periodically the system will just oscillate. AMR

I am a fan of simple harmonic motion; the potential physical application is really exciting. For the record, I want to make it known that Prof. DeGrand's class was very helpful in the grand scheme of understanding this concept. I hope that the SHM component continues in 2170 to benefit future semesters. More on topic (or maybe not), I am curious what the necessary features for a restoring force are to result in simple harmonic motion? We know that forces proportional to the distance of separation from equilibrium work to achieve SHM, but do forces that just change direction work to accomplish SHM, or maybe forces that are cubically related to the separation from equilibrium? I ask because I am curious how the use of a Magnetic trap varies from an optical trap (aka Magnetic and Optical tweezers). Does the fact that a linear restoring force is applied in the optical counterpart change it effectiveness in measuring a force induced, as opposed to the constant force of the magnetic well? This has been on my mind for a while and I can't quite figure it out. No problem if it is too specific to be applicable to the class.

We are glad to hear that Prof. DeGrand's class was very helpful foryou. We will certainly will try to keep the SHM component in 2170 .

Regarding your question: the harmonic oscillator motion is restricted only to linear restoring forces, i.e quadratic potentials. Since the quadratic term is the first non-zero term when you expand any potential close to equilibrium, SHO motion is the leading behavior of stable systems close to equilibrium. However if you drive your system far from equilibrium the motion is not longer SHO and the cubic, quartic corrections become important.In the later case you can still have periodic motion but it is not asimple sine or cosine. We will learn more about that when we study Fourier series.

Now related to optical magnetic trapping : one or the other could be more efficient depending on the type of particle you want to trap. The optical trapping relies on the dipole moment of the atom and the magnetic trapping on the magnetic moment. For example if you have alkali earth atoms (second column of the period table) you can not use magnetic trapping because the ground state has zero net magnetic moment. People need to use optical trapping methods do trap these atoms. AMR

 

How good of an approximation is it to say that the resistive forces are proportional to v? Is it more often the case for springs than it was for air resistance?

Actually the validity criteria is the one you just studied before. Linear applies to microscopic objects moving at small velocities.

However we deal with the linear term only because this gives rise to a linear ODE which is easy to solve. If you want to treat the quadratic case in the presence of an additional quadratic potential then the treatment becomes very complicated.

The linear drag is good for many microscopic situations. For example is an excellent approximation if we want to understand the behavior of a bound electron in a molecule which is exposed to radiation. In this case the damping term mainly comes from radiation damping. AMR

Preflight #8:

Note that these are *preflights*, meant to have you read and think about the book before class, so that we don't have to spend as much class time talking about the straightforward stuff, but can focus on the deeper concepts, and subtler/harder points! If you are confused about these points, no worries - wait till we've had the lectures on them... at which point, though, you should understand this stuff.

First, my answers, in brief:

Q1: |2-i| = Sqrt[2^2+1^2] = Sqrt[5]. Watch the signs here!
      (2-i)* = (2+i)

Q2: 2e^[I Pi/4]=2*(Cos[Pi/4]+i Sin[Pi/4]) = Sqrt[2] + Sqrt[2]*I

Q3: (1-i)/(1+i) = Sqrt[2]e^[-I Pi/4] / Sqrt[2]e^[+I Pi/4] = e^[-I Pi/2] (which is also -i)
Alternatively, multiply top and bottom by (1-i), to get (-2i)/(2) = -i.

Q4: Ae^(i omega t) is mathematically simpler than Cos[omega t] (in terms of, e.g. taking derivatives, or multiplying solutions, or in general just about any kind of manipulations) It IS complex, but the real part is just what we want, and for linear ODEs, if you have a complex solution, then the real part of that solution is ALSO, by itself, a solution. So we do the math with the complex form, (which, ironically, is *simpler* :-) and then at the very end we can always take the real part.

Alternatively, by including Be^[-i omega t] solutions, and choosing A and B appropriately, we can also get back to a real expression which satisfies all our initial conditions.

Q5: Resonance - read the lecture notes on this topic, or Taylor's chapter. It refers to damped, *driven* harmonic oscillators, and if you drive at (or near) the natural (undriven) frequency, the system responds very strongly. That's resonance.

And now, some of your comments and our responses:
Perhaps not directly connected to the reading but in all my experience with SHO's and solving ODE's there is an omnipresent need for us to plot the phase space diagrams of the solutions (f vs. f') and I have forever failed to see the usefulness of this practice. Maybe it's just me but I cannot extract any information from a plot of f vs. f', now f vs. t and f' vs t. are very useful but I am not so sure about f vs. f'

This is a good question! I agree, its use is not very apparent at first glance. I would say, first, that it's a compact way or representing the "state" of a system on a plot, even just as a single dot! It thus contains MORE information than a single dot of "f vs t", because for the latter, you also need to know the SLOPE (as well as the point) in order to then predict the *future* state. Being a new "representation" of a system, it gives us the chance to practice switching between graphical representations, something that will happen with more frequency in future classes, (and something that takes practice) it's just another way of "looking" at our system's motion in time.

The real value of this representation will come out in Statistical Mechanics, where you have ensembles of systems, each of which can be characterized at any time as a point in phase space, which evolves with time. There is a theorem (Liouville's theorem) about that evolution of states in phase space which proves to be very informative, and so this is a bit of a setup for that. If you have practice thinking about phase space for simple systems (like an oscillator), it will be easier to think about what it means/tells you when you have rather more complicated, many-body systems.

But for the moment, I would just consider it as an alternative way of thinking about/visualizing time evolution, that shows certain characteristics (like damping, or initial conditions) in a graphically different way than the usual graphs, and thus is useful as just another way of picturing what's going on. Cheers,Steve

When solving ODEs the idea of superposition principal can be applied when you have different solutions that independently solve the ODE. My question is how many solutions do you need to put to together to sufficiently solve the ODE?

For linear ODE it will depend on the order of your ODE.
If It is first order you need one, second order you need two linear independent solutions, third order you need three linear independent solutions etc..

How do simple harmonic oscillators relate to what we've learned about gravity?

In some cases (for example the problem you are solving in the Hw of the tunnel across the moon) the gravitational potential generated by a massive object can have stable equilibrium points. The motion of a particle close to these points will be a simple harmonic motion. AMR

It will be interesting to analyze different oscillatory motions and to determine what dimension they belong. Can the dimension of an oscillator be simplified by choosing one good single parameter similar to section 4.7?

Yes. That is correct. For example if you have a 2D harmonic oscillator in which the two coordinates are decoupled you can excite oscillations only along one direction keeping the other coordinate fixed. This gives rise to an effective 1D motion for this case.
It will depend on the initial conditions.
If the two coordinates are couple in general you can go to another basis and decouple then and similar considerations apply in the new basis. AMR

I grasp the math behind it and I can't argue with the solution, but I'm still finding it hard to intuit how a larger damping force (over damping) results in the oscillation dying out more slowly than the 'critically damped' system.

We'll talk about this in class, but it might help to think of an extreme case. If you have really high damping, like thick molasses, and you displace an object away from equilibrium, then the system takes a REALLY long time to get back to equilibrium, right? (It has a super slow terminal velocity) I think it's nothing much more than that - once you're away from equilibrium, LESS damping will get you back faster. (If you're underdamped, however, then you start swinging PAST equilibrium, you overshoot. So, critical damping is really optimum for "return to the start".) That's presumably what car shocks, or swinging door pneumatics, are designed to try to do.

This help? Steve

I'm honestly not sure why the gravity equations are negative and the E&M equations are positive. I always thought gravity was negative because that was just the way the coordinate system was set up.

Gravitational forces are always attractive regardless of the coordinate system you use. Two massive objects tend to approach each other due to the gravitational force. If \vec{r} is the vector pointing from mass 1 to mass 2 the gravitational force is in the direction -\vec{r}. On the other hand electromagnetic forces can be both attractive or repulsive. Two charges with the same sign want to repeal each other whereas charges with opposite sign attract. - -> <- + (attractive ) , <- + +- > (repulsive). AMR

Help with complex numbers!

Check out my lecture notes (pp osc.3-6), or Boas Ch 2... A large majority of the class did ok on this preflight, but not everyone, so I will be reviewing (but not extensively!) The most essential thing is that you remind yourself about Euler's relation - check out Taylor 2.6, see if that helps!

Feel free to stop our office hours if you're feeling stuck on any of it. Steve

whats up with the complex numbers and their relation to SHM?

It's a beautiful thing, really :-) It all comes from Euler's relation, E^[I theta] = cos(theta) + I sin(theta). (See Taylor pp. 68 and 69 for a review of that) What that tells us is that complex numbers are intimately tied to sin and cos, which in turn are the solutions to SHM problems. So, complex numbers give us an elegant, efficient way to describe the *math* of SHM.

When I think of a complex number, I picture it as a point in the compex plane, rather like a vector. And, like any vector in the plane, you can think of it as
described by "Cartesian coordinates " (a,b) or equivalently in polar coordinates, as (r, theta). The connection is Euler's theorem, a+bI = r e^[I theta]

If you picture a "point in the complex plane", z, rotating with steady angular speed omega, i.e. z = r * e^[I omega t], then Euler's theorem tells you that the "Real part" of that is just r cos[omega t], the equation for SHM. So in other words, there is this quick, neat connection between oscillations (SHM) and rotations (with steady speed). Rotations are so simple, just something moving in a circle, so mathematically we tend to prefer thinking in that language.

We'll be talking about this coming up soon, hopefully all will become clear(er)! Let me know as we go along if it's all making sense. Steve

I was just wondering if we would ever use the sinh and cosh form of e^t and e^-t ever in the solution to Fxx = F. I have seen it before and I was wondering if we ever use this in practice in physics.

By "Fxx=F", I wasn't sure, but presume you mean an ODE of the form F''(t) = + omega^2 F (rather than a minus sign, which gives SHM and thus regular sin's and cos's) The plus sign there would indeed yield sinh and cosh instead of sin and cos. Physically, we're perhaps less likely to see such solutions in simple mechanics problems,... , this would be some sort of "unstable equilibrium" rather than stable equilibrium, but I'm sure it crops up from time to time. (We've already encountered hyperbolic functions in our solution to damped motion with gravity!)

Another place that your ODE (and thus the sinh and cosh solutions) will definitely re-appear is in E&M, when you are solving for voltage in regions of empty space. The boundary conditions there sometimes yield sin's and cos's, but they can ALSO perfectly well yield sinh's and cosh's, it just depends on the signs that appear when you separate variables. We likely will get to this later this term - we certainly haven't seen the last of sinh and cosh! - Steve

Can we at least touch on gravitational anomalies? I don't really understand how g can be different by very much just due to something like iron deposits in the earth near you.

If we have time, we'll touch on this in class today. g is just a measure of the local gravitational field. g is only going to be perfectly uniform if the earth is a perfect sphere. But it's not! Any "locally unusual extra mass" nearby is going to add to g, changing it from the simple, uniform result.

Just think of the superposition principle, suppose the earth was a perfect uniform sphere, with ONE iron deposit of very different density from the rest located underneath Boulder. Then, when you're in Boulder, the local field would be the superposition of the simple, uniform g, PLUS an extra contribution from the additional g-field arising from the above-average density of the iron. So, local "g" would be slightly higher here than everywhere else, that would be an "anomaly". ( And, if you were off to the edge of that field, it would have a slight sideways component too: radial from the uniform earth, plus a small extra gravitational force towards this extra chunk of mass).

Of course, this iron deposit affects g everywhere on the globe, but the force goes as 1/r^2, so its effect will dominate nearby.

Anomalous g can arise from anything that differs from the uniform solution. So, local high (or low) density ore deposits, or extra mass due to a mountain range, or plate structures deep underground resulting in non-uniformity of the mass distribution below us...

Turns out that although G is hard to measure accurately, g is not, it can be measured at way better than the parts per million level. That's pretty sensitive, there are companies that do this for a living, you can use it to look for underground deposits (so, mining companies are very interested!)

This making sense? - SJP

I think that complex numbers are really fun. I'm not sure why, but I'm excited to be using them again.I suppose the question I have is the same question I had last time complex-valued functions showed up: How is it that a mathematical abstraction such as the square root of a negative number wind up describing the "real" physical world? I mean, shouldn't the "real" world be completely described by "real" numbers? I know classically it can be, but in quantum mechanics it seems that complex values are an essential part of the field.

I agree, there's something really magical about e^(I Pi) = -1! I'm not sure I have a great answer to your question, I think the best I can come up with is that Euler's relation relates the MATH of sin's and cos's to the MATH of r E^{i theta}. So, the *physics* of simple harmonic motion is connected to the *math* of rotations in the complex plane. In this course, at least, we largely introduce complex numbers because they simplify the solution of equations. We could avoid them, and deal always and only with real quantities, it's just so much easier to handle the complex exponentials.

The real world IS always described by real numbers - in the end, we take the real part of our solutions to figure out motion (or voltage, or whatever). Even in quantum mechanics, where (I agree) complex numbers seem even more deeply entangled with the physics, still in the end physical observables are always real. You don't *observe* the complex phase of a wave function, not directly, it's still (I claim) just a mathematical tool that helps us describe what's going on. Indeed, the "realness" of observables serves as a constraint in Quantum mechanics (that's why physical observables are always represented by Hermitian operators, with real eigenvalues. But, that may be just jargon that might not make sense until you've had junior quantum and linear algebra)

Steve
________________________

Preflight #7:

Note that these are *preflights*, meant to have you read and think about the book before class, so that we don't have to spend as much class time talking about the straightforward stuff, but can focus on the deeper concepts, and subtler/harder points! If you are confused about these points, no worries - wait till we've had the lectures on them... at which point, though, you should understand this stuff.

First, my answers, in brief:

Q1: (This was whether you had figured out the geometry "trick" in Taylor's example for yourself. I hope so - for me, it just involved drawing my own picture, staring at it, and thinking for a little bit!)

Q2: Is the "weeble" 1D? The answer is a qualified yes. If you tip it by theta and let it go, as implied, then ONLY theta is needed to specify the state of the object. That's what we mean by 1D motion, it is fully parametrized by a single variable. (However, if you tip it by theta and then "kick" it perpendicular, then it has motion in both theta and phi directions - it's then 2D. And that changes things quite a bit. For instance, you'll get new, different "stable equilibria", e.g. if it rotates around the centerline in the phi direction, it can tilt outwards stably by "centrifugal" forces in its non-inertial reference frame. But this isn't what we're focusing on here, and I won't be asking you about the 2D problem on the upcoming homework!)

Q3: Where is speed greatest? At point D, where (E-U) is maximum, where the energy is farthest above the PE graph.
(Because E-U = T, simple as that!)

And now, some of your comments and our responses:

I was surprised that gravity has yet to be measured precisely, and that Newton was tortured over the constant itself.

I don't think Newton was tortured over G per se - he had no way to measure it! It wasn't till Cavendish, over 100 years later, that the first measurement of G was made!

For my own curiosity, was vector calculus used during Newton's time?
Well, it was Newton who *invented* calculus. So calculus was used (in the sense that he made it up as he went along :-) (Of course, Leibnitz might have disagreed about who invented calculus, but that's another story) However, Wikipedia says that what we now call "vector calculus" wasn't developed until the end of the 1800's, (this would be things like div, grad and curl)

On the main page, regarding the question about the gravity anomalies: Do the positive anomalies have to do with the fact that the Andes are higher, and therefore closer, to other bodies in the solar system (and vice versa for the negative anomalies)?

Here's what Prof Wahr (who is involved with the GRACE project) told me when I asked him a closely related question:

Steve: These are maps of g as would be measured at a common elevation
(specifically, at "sea level"; so, at an elevation of 0) all around
the earth - NOT as measured at the surface of the Earth. They show g
after removing a reference value of g. That reference value is the
g-value you would measure at sea level, if the Earth were simply an
ellipsoid: i.e., if it had the shape of a sphere, but squashed down at
the poles and elongated at the equator because of its rotation.
Yes, the structure in the maps give you information about the mass
of the underlying topography plus in the underlying earth. You don't
usually see broad-scale topography very clearly, because high
topography tends to be compensated by low density roots in the
underlying earth - it's like icebergs floating on the ocean. In some
places, like the Andes, where the topography is there because of
collisions between plates, that topography is not floating but is
supported by stresses caused by the plate collision, so you see
un-compensated topographic mass, and the net gravity anomaly is
positive. In Canada, there's some debate as to whether the negative
anomaly is caused by thermally induced density anomalies, or to
post-glacial-rebound: the ongoing viscoelasatic response of the Earth
to the removal of the ice load at the end of the last ice age. That
response isn't completed yet; Hudson Bay is a remnant of that load,
and is still filling in. So, there's a mass deficiency there, which
is maybe part of what shows up in the gravity field maps. Probably,
what you see over Canada is a combination of the rebound and
thermally-induced density structure
.

What are the implications that come with the term "one-dimensional" in
the systems we are examining?

The implications is that those systems can be treated as effectively
1D systems and therefore offer a much simpler treatment.
You can visualize clearly the motion of a 1D object but a 3D motion
can be quite complicated. Also mathematical it's much simpler to deal
with a single coordinate and not three.
Always if you have a 3D problem first take a look and think if the
motion can be reduced to 1D. This will make your life simpler.

Ana M

 

When we say "the equilibrium is stable", does that mean the system at
equilibrium won't even move or whereas it moves it will oscillate
until it stops at the equilibrium point again?

It means if you displace the system out of equilibrium it is going
just to perform oscillation centered around its equilibrium point.
Actually it will only stop at the equilibrium point again if there is
some type of damping. Notice that conservation of mechanical energy
does not allow it to stop. In principle it will keep oscillating back
and forth.
The initial potential energy that it gets when you displace it from
the equilibrium must be dissipated in other form of energy if the
object stops moving. Ana M

___________________________

HW 6 question #2: On problem 2 on HW 6, are we including gravitational and/or drag forces?

Definitely no gravity or drag here - we should have been clearer, but Q2 is meant to be a straightforward application of the simplest "rocket equation" derived in Taylor (and the notes) (gravity is included Q3, and drag is Q4!)

Steve
________________________

Preflight #6:

Note that these are *preflights*, meant to have you read and think about the book before class, so that we don't have to spend as much class time talking about the straightforward stuff, but can focus on the deeper concepts, and subtler/harder points! If you are confused about these points, no worries - wait till we've had the lectures on them... at which point, though, you should understand this stuff.

First, my answers, in brief:

Q1) Force is greatest where the *gradient* of potential energy is largest, or in simpler words, where the "equipotential lines" are most closely spaced as you head away in a perpendicular direction. From the figure, B has the highest force, A is smaller, and C is smallest (lines are farthest apart at C)
(Note that the reason is NOT "because it looks like there might be a charge off to the left, and B is closest", because these equipotentials are not circles, so the source of this potential field is definitely NOT a single charge!)

Q2) Force is perpendicular to equipotentials, so the direction is up (or down) the page, slightly tilted, so it's perpendicular to the local dashed line. Without knowing more (I need to know which equipotentials are larger) I can't say which way - but if A is at higher potential, then a +q at B will head DOWN, towards lower PE. (And vice versa if A is at lower potential)


Q3) Figs 1 and 2 have no curl - any line integral "loop" (any one at all!) will aways be zero. But Fig 3 definitely does have a curl, it's not hard to spot some loops where the line integral going out one leg is BIGGER than the one going back the other way, and thus they don't cancel out. (The way I think about this graphically is, if you imagine putting a tiny little "paddlewheel" in that field, it would turn, and that means there's a curl!)

Q4) Conservative forces means curl=0, so Fig 1 and 2 could be conservative, but Fig 3 could not!

And now, some of your comments and our responses:
It would be nice to regain some intuition on why conservancy implies path-independance.

Conservative forces *means* that they have an associated potential energy (and thus, that KE+PE is defined and conserved. Hence the name!) If you think about it, potential energy is associated with a *place in space*, U(r), and the only way that you could have that place associated with a particular, definite potential energy is if the amount of work required to bring the object there is independent of HOW you brought it. That make sense? We'll talk more about this, but that's the main idea - if (and when) work is path-independent, then the amount of work can be "connected" to the endpoint, and not the path.

Thanks for participating!

 

The hardest thing for me to understand right now is the physical significance of curl and divergence. I remember the formulas from calc 3, but given that the Gradient vector has no physical representation, its hard to understand what, exactly curl and divergence are.

[Related q from another student: I suppose a question I have would be how to determine the curl of a field without being given the equation. I don't know that we will need to use it outside of this preflight, but its the only thing that I'm having any trouble with from the things that we have to do for tuesday.

We'll be working on that this week (and for many semesters to come :-) I would argue that the gradient of a scalar function (at least) does *indeed* have a pretty direct physical representation, and we'll do a Tutorial on that today in class. I think of the gradient of a scalar function as giving you a vector which points in the direction of most rapid change of the function. And, the length of the vector tells you how rapidly you're changing in that direction. It's the closest thing in 3-D to simply "the derivative" of a function! It contains the information of the rate of change of your function in all three directions, x, y, and z.
Curl and div will also have physical interpretations.
Curl is maybe hardest to compute, but possibly easiest to picture: curl is a good name, it means the field has some "rotation" or "curl" associated with it - small line integrals around loops give non-zero results. If you imagine putting a small "paddlewheel" into a vector field, and if it would start to *rotate*, then the field has a curl there.
Divergence is also a decent name: it tells you if the function is "diverging", that is, if there are MORE field lines exiting a point than entering... It tells you if the point in question is a "source" or "sink" of field lines.

Even after following Taylor's examples, I am struggling to understand where the minus sign comes from in F=-(Grad)U. Why is work equal to minus the change in PE? How does this make sense physically?

Good question, this is a deep one! I have two rather different ways I think about this minus sign.

Imagine a particle sitting in a force field. Just let the particle go - which way does the force push it? Downhill, right? Always! If "potential energy" is to sensibly represent energy, then things should want to go towards LOWER potential energy if you let them ((right? To conserve energy, if it's going naturally, it's speeding up - it's gaining kinetic, but losing potential energy) OK, so for "energy conservation" to make sense, we're saying that forces naturally tend to push things downhill, i.e. DOWN in potential energy. Ah, that's exactly the minus sign! The force is in the NEGATIVE gradient direction, downhill. (That's because "Gradient of a function" is simply telling you the direction of maximum increase, by its very definition. Think of derivatives in 1D, they tell you the slope, the direction of INCREASE of a function)

Grad(U) is the vector that mathematically tells you which direction U is increasing most rapidly, and thus -Grad(U) is a vector that tells you which direction U is *decreasing* most rapidly - and that's the natural direction of the force!

But here's another way to think of it - going back to your original question, "why is work=minus the change in PE". Imagine adding in a SECOND force to the problem, YOUR force. So you pick an object up that's sitting in a force field, and you move it around. If you move it at constant speed (so KE doesn't change or enter the story in any interesting way), and you push it OPPOSITE the direction of the force field, YOU are doing positive work (right? Think what it feels like to lift a book against the gravity field). But if energy is conserved, and YOU are adding positive energy to lift the book, where is the energy going? You're not speeding up, it's not stored as KE, it must be stored as PE. So if YOU push the object "up the hill", OPPOSITE the direction of the force field, you are *raising* its PE. Work done BY YOU is equal to the change in PE, but work done BY THE FIELD is opposite your work done, so NEGATIVE the change in PE.

Keep thinking about it! We'll prove it in notes/class, but it's more useful to make sense of it yourself!

 

I'm not so clear on why a field is conservative if the curl is zero; that fact is just kind of thrown at us in both books. I'd also like to take just a bit of time to discuss what divergence and curl actually mean since my calc 3 professor never really told us, he just made us calculate them.

The best "proof" I have relies on Stoke's theorem. If you don't have an intuitive sense for that one, it's a little tough. Stokes theorem says that if the curl is zero, then ANY line integral around ANY loop will always give you zero. If we're talking about "line integral of F dot dr", then this says "if the curl of F is zero, then the work done moving around ANY closed loop is always zero. And that's basically conservation of energy - if you take an object, move it around any way you like, but end up where you started - no work is done, there's no change in energy. Energy is conserved!

So, if you buy Stokes' theorem, then the "field is conservative <=> curl is zero" connection is pretty easy. If you don't remember (or have intuition) for Stoke's theorem - well, I can show you a proof that's only a few lines long... but, it really needs a blackboard or piece of paper :-) Come to office hours some time! We won't be really using Stokes' theorem much this term - it'll come back big time in E&M I and II, though.

As for div and curl - check out the questions above this one, lots of people are asking about that!

why are we studying point charges in classical mechanics?

IMHO: Because E&M is part of classical physics! And because charges are one of the great *examples* of "point-like particles" that help us make sense of Newtonian physics. And because the mathematics of point charges involves div, grad, and curl in ways that are about the easiest cases to visualize and make sense of, letting us build intuitions for more complicated situations (like plasmas, or fluid fields, or quantum fields) that will come in future physics courses.

________________________

Preflight #5:

Note that these are *preflights*, meant to have you read and think about the book before class, so that we don't have to spend as much class time talking about the straightforward stuff, but can focus on the deeper concepts, and subtler/harder points! If you are confused about these points, no worries - wait till we've had the lectures on them... at which point, though, you should understand this stuff.

First, my answers, in brief:

#1) Explain in your own words how you draw the conclusion that planetary orbits stay in 2D, from "rxF=0". First, torque = rxF is clearly zero in this case, because r (sun to planet) and force (planet towards sun) are "antiparallel". 0 torque means L is constant. But, L is a vector. A constant vector means ALL COMPONENTS are constant. The L vector points in some particular direction in space, and this doesn't change with time. But L is perpendicular to "r", so the radius vector pointing to the planet is always perpendicular to a fixed vector. "Perpendicular to a fixed vector" defines a plane, so r must lie in the plane for all time. That's why the planets orbit is planar, and stays that way. (Note that an EXTERNAL torque could change this!)

 

#2) What does it mean to expand in a series around a point? I would say, most fundamentally, this is an approximation method by which functions can be approximated NEAR a point, by knowing about the function and its derivatives AT that point. The idea would be to approximate the function - first as a straight line near the point, then as a parabola near the point, then as a quartic near the point... The farther from the point you go, the more "terms" you need (more derivatives at that point, higher order...). If the function is analytic, then (remarkably) there is no limit as to how far away from the point you can go, and still have the series (ultimately) converge.

 

#3) What are you shakiest on? Whatever your answer was - keep working on it! We'll have regular office hours this week, so you're welcome to come talk to us about it.

 

And now, some of your comments and our responses:
First: some exam questions:

None of our homeworks had a problem on rockets and thrust, so I guess my only difficulty is that I don't have enough practice working with it. Maybe I'll make that the problem I pick for this weeks homework..

Just FYI, according to our "exam info page", section 3.2 is not on the first midterm (meaning, we're not going to explicitly have a rocket/thrust problem) Not to worry, you'll get several problems on this for next week's homework!

I guess I'm more worried about relating the topics together. I think I understand them individually, but if we were to have a problem that combines them I don't think I'd perform very well. Can you give me an example of a problem that will be on the test? Will you be combining topics or separating them?

It's a good comment/question - I think as we move further in this class, I'm going to try to make homeworks (and to some extent, exam questions) that combine topics more. But, we're trying to build up, so I think it's safe to say that for the most part, first midterm questions will pretty clearly "look like" they came from a particular topic that we've studied so far. (I guess you'll have to tell me afterwards if you agree :-)

But yeah, ideally, we're always trying to craft problems that are not too computationally nasty, but get at multiple (central) ideas... So it's certainly good to be thinking about how all the different elements of the course fit together, what's the "big picture"?

And the rest, on upcoming topics:

I should probably know this, but it's been a while. If I'm taking the line integral along a function, do I need to parametrize it as Taylor shows, or can I use the function as bounds in some way?

We'll definitely review the line integral - lots of people will need to "relearn" that. There are an infinite number of ways to "parametrize" a path, there's no firm rule about it. You look at the path, think about a SINGLE variable that will "march you along" the path, and that's how you parameterize. Taylor shows a couple of ways, Boas Ch 6.8 has others, my lecture notes also show a string of examples. It's usually pretty obvious from the context of a given problem (and indeed, "dx" or "dy" is often the simplest choice... but not always!) Cheers,Steve

I'm a little confused about how the line integral depends on the path to get to the point, because I was under the impression that Work does not depend on the path.

We'll talk about this soon in class. Work is independent of the path *if the force is conservative*. But, not all forces are conservative! Drag, like we've been considering, is for instance most definitely not conservative, and the work done when something experiences drag is strongly dependent on how far (and how fast) you move it)

Cheers,
Steve



I tried problem 4.2 (KE/work problem involving 3 different paths) and kept getting different answers....
After reading section 4.2, my doubts have only risen, as one of the requirements for a force to be conservative is that F should only depend on the particle's position; it seems to me that I should be getting the same answer for each path.

It's true that all conservative forces have an F that depends only on position, but the converse is NOT true: just because F depends only on position does not guarantee F is conservative! The force in Taylor 4.2 is NOT conservative, so the answer does depend on path. Next week we will discover the simple mathematical trick to decide immediately if a force is conservative (take its curl. If it vanishes, THEN it's conservative! Otherwise, it is not!) -SJP

Are all forces technically conservative so long as all energy transfers (like the heat energy from friction) are tracked or are they really only conservative if the transferred energy an easily be converted back to kinetic energy? Does this really come down to defining a system and as long as the energy is still in the system then the force is conservative?

Non-conservative forces are NOT merely an issue of "defining a system" carefully, or carefully keeping tabs of the energy. The 2nd law of thermodynamics is at the heart of non-conservative forces - these are forces which transfer energy into the form of (random) thermal energy, and then there is (statistically) NO WAY in principle to get it all back into, say, kinetic energy. This distinguishes non-conservative forces (and thermal energy) from conservative forces (and potential energy).

So I'll say no - it's not just "defining a system", and expanding your system so the energy is still "in it" doesn't make the forces conservative. It's interesting that all fundamental forces of nature, acting on *pointlike consituents* are conservative, yet when they act on/between macroscopic ensembles of particles, these forces effectively produce real non-conservative forces. There's no getting around it, friction is not conservative in any bookkeeping scheme!

This gets beyond Phys 2210 though - get stoked for Phys 4230 :-)

If you expand a system to include the universe there aren't any nonconservative forces are there?

There ARE definitely nonconservative forces. If you talk about *fundamental* forces acting on pointlike constituents, there are no nonconservative forces. Nonconservative forces are "effective" forces (like friction) acting on collections of objects, and capable of moving/transforming energy into the form of thermal energy (thus, not "recoverable" in the way conservative forces' work is, by the laws of thermodynamics) Expanding your scale to include "the universe" won't help you get away from the 2nd law of Thermodynamics, friction is STILL nonconservative even if you consider all objects everywhere! -Steve

________________________

Preflight #4:

Note that these are *preflights*, meant to have you read and think about the book before class, so that we don't have to spend as much class time talking about the straightforward stuff, but can focus on the deeper concepts, and subtler/harder points! If you are confused about these points, no worries - wait till we've had the lectures on them... at which point, though, you should understand this stuff.

First, my answers, in brief:

#1) The double sum is F12+F13+F21+F23+F31+F32. (6 terms = N(N-1) with N=3, as it should be)
(Note that F12 and F21 cancel, as do F13 and F31, and F23 and F32, this whole thing is zero by N-III!)

The single sum is Fext,1+Fext,2+Fext,3, that's Fnet, ext.

#2) Using m1=2m2, m1*v1=(2m2,2m2), m2*v2=(m2,-2*m2). Adding these gives p(total) = (3m2,0).
This must be equal to M(total)*v(final), and M(total)=2m2+m2=3m2, which says that v(final) = (1,0)
We asked for the angle between this (1,0) and v1=(1,1). That's easy, draw a picture (or use the dot product divided by the magnitudes), to get 45 degrees.

#3) You have to look at the picture of the cone, and realize that the edge of the cone is a straight line, which starts from r=0, z=0, and rises to r=R, Z=h.
That's a straight line with slope h/R, so the formula for the line is z = (h/R) r. Inverting that gives what we want, r = Rz/h. It simply tells you that, given z, r can run "out to" that straight line. You always have to "parametrize" your shape in some way, there's no rule, you just stare at the picture and invent the solution!

Now, some of your comments and our responses:

It would be awesome if you would start printing out copies of the homework for those that need it! Just a thought. Thanks!

Done! They're in a folder taped to the wall just outside my office! (F1013) They typically go up there Fridays, which is also when we post them online.
Thanks for the suggestion - always happy to try to accomodate if we can! (We've been doing this for a couple of weeks now) Cheers,Steve P.

Why are we reading about inelastic collisions when collisions in the real world are elastic?

Collisions between ideal pointlike particles are always elastic, but real objects are compound. Even an ATOM is not pointlike, and it can have inelastic collisions. Inelastic just means that some of the kinetic energy in the collision is converted to some other form of energy. (E.g, the internal electrical potential energy of the atom). In the macro-world, essentially ALL collisions are inelastic (!) because some energy goes into sound energy, light, thermal (the big one, usually!), breaking of bonds ("broken glass/dented metal"), etc. Think of the collision of a ball of putty with something solid - they thwack, stick together, and we have a (very) inelastic collision! So it's funny - at the most fundamental level, all collisions are elastic. But in the effective world we live in, NO collisions are elastic! Cheers, Steve

In the book it talked about completely inelastic collision. Whats an example of a partially inelastic collision with two particles?
This can be an idea of a partially inelastic collision: Imagine a pullet that penetrates a block of wood. Inside the wood the projectile fragments and only half of it manage to get out the block. The other part remains inside. AMR

The concept of defining the center of mass as a vector is very difficult for me, since I always imagined it as just a point with coordinates, not going anywhere, as the vector implies. It makes sense when I think of the whole system moving, but when it's a stationary like the cone in Example 3.2, I have a hard time visualizing the vector.
I think the easier way to understand the CM concept is to first think about a system made of discrete masses. If you have a collection of masses distributed in space then the CM is just the average position of the system's mass. If you have a solid body you just need to imagine is made of very tiny blocks of mass dm and the CM is the average position of this tiny blocks. Ana Maria
(P.S. from SJP: Think of CM as "the position vector of a complex object". So yes, it is a point with coordinates, but it's best to think of that as an arrow FROM the origin TO that point. Just like "r" for a pointlike particle is just a point with coordinates, but we think of it as "the position vector". )

I wanted to know if as far as momentum is concerned we will go over elastic as well as inelastic collisions?
Yes, we will cover both. Hw5 will have for example a problem on elastic collisions. (AMR)

what do center-of-mass calculations look like in polar coordinates? It would seem useful to be able to do...
Yes, certainly to use polar coordinates will be useful in some cases. Imagine for example a cylinder with inhomogeous mass density \rho( r,\phi,z). In this case it is better to write the vector R=(x,y,z) in cartesian coordinates, because remember only in cartesian coordinates unit vectors are constant. You have to compute three integrals (1/M\int \rho x dV, 1/M\int \rho y dV,1/M\int \rho z dV). But then, each of the integrals can be calculated easily if we use cylindrical coordinates because they have the same symmetry than the object you are dealing with.

Why can you split a vector valued integral into three separate integrals, one for each component? I guess it's because of how Cartesian vectors add up with each other, it's component-wise. Oh yeah this reminds me, how do you add up vectors in polar or cylindrical coordinates? It's hard since the unit vectors are functions of position. Do you have to convert to Cartesian first?
Yes, you are correct. In general the easiest way to proceed is to use cartesian coordinates to split R. As you pointed out only cartesian unit vectors are constant. Each of the three integrals, which are now scalar quantities can be evaluated using cartesian, spherical or cylindrical coordinates depending on the geometry of the system you are dealing with. For a cylinder for example I would choose cylindrical coordinates. (AMR)

How are we going to incorporate this with drag?
We may have a homework problem combining Rocket motion with drag. Drag is always present in real life. For example check problem 3.14 (AMR)

________________________

Preflight #3:
Note that these are *preflights*, meant to have you read and think about the book before class, so that we don't have to spend as much class time talking about the straightforward stuff, but can focus on the deeper concepts, and subtler/harder points! If you are confused about these points, no worries - wait till we've had the lectures on them... at which point, though, you should understand this stuff.

First, my answers, in brief:

#1) What is tau? It is the natural time scale in problems with drag, a combination of constants of your system (with units of time) that will enter the solution as a "natural relaxation" time.

#2) The larger object has larger v_t. (!) If 2 spheres are of the same material, but one is bigger, the bigger one will have a mass which is bigger by the CUBE of the dimension (because mass grows proportional to volume). In the terminal velocity formulas, v_t grows with mass, and also drops explicitly like an inverse power of dimension, but in both linear and quadratic cases, this "dropoff" is not rapid enough to compensate for the growth with mass. So, bigger objects "fall faster" (the old Aristotelean idea!) Terminal velocity for an ant is so small that it can survive a fall. Terminal velocity for an elephant is much larger than for a human.

#3) With drag, the x motion of the falling object gets quickly damped out, so it does not travel as far forward as it would have. So, you must release it later, to compensate.

Now, some of your comments and our responses:

I'm still a little unclear on why the quadratic drag never causes a terminal velocity.
Oh, it does. (Eq 2.53). Perhaps what you are thinking of is the fact that for quadratic drag in 1D (with ONLY drag), the object never reaches an asymptotic *position*, it keeps on moving and "reaches" infinity (albeit, very slowly) Or, did you have something else in mind?, Steve

Drag is really interesting, but how does it apply to the coordinatesystems we're learning in class? Drag seems to focus on the cartesian
coordinate systems.?

Yes. Newtons's equations are independent for the coordinate system you use. You only need to write them in a consistent way. We are dealing with cartesian coordinates because in general the ODEs in cartesian coordinates are much simpler for the projectile motion
problems we are solving. However there might be problems in which other coordinates systems might be useful.

For example consider a pendulum in a viscous medium. Probably in this case might be easier to go to spherical coordinates. ,Best Ana M

How could one model the changing effect of friction upon an object dropping through a medium with density changing as a function of altitude? How much of an effect does the changing density of the atmosphere have on the results computed when a uniform

This is a very interesting question What we need to do is to change the dragg coefficient, which so far we
have assumed constant by a time depending one. c ->c (t). Now you need to add also an ODE which tell is how c(t) is changing, i.e d c/dt= f(t).

The solutions can become complicated but in principle you could solve them in a computer.

With respect on how the change in atmosphere composition will modify the motion I think this will depend on where are you at. Close to the earth surface it can not be that much but I am sure this is something scientist must take care when modeling Rockets coming back to the earth. Best, Ana M

thought it was interesting that the terminal velocity of a very small diameter object increases as the object increases in size (up until it's no longer modeled linearly, obviously). I don't have a good intuitive understanding of why that makes sense.

If we are thinking about changing the diameter of an object the reason why the terminal velocities increases is that as you increase
the diameter is that you increase the mass. In that respect the fact that heavy objects reach faster terminal velocity is not that counter
intuitive. They just feel larger gravitational force and thus larger net acceleration.

The other case is that imagine you compare objects made of different materials and same total mass but different diameter. In this case the objects with larger diameter reach smaller terminal velocity since as the area or diameter increase their corresponding drag coefficient increases as well. Best, Ana M

That textbook really needs to arrive in the mail now.
Yikes! In the meantime, you do know about the "reserve" books in the library (right across the hall from our classroom), right? You can get either text (and others) for a couple hours, any time. Probably a Very Good Idea about now, if your book hasn't come yet! Steve

I would like to spend more time on the polar co-ordinate system. It seems to be a very powerful tool, good for solving all sorts of complex physics problems.
Oh yes, it will come back... (Turns out projectiles and rockets are typically more well suited for Cartesian, though, which is what we're focusing on for a bit, now ) Steve

Is there any way to solve example 2.6 without using a computer program?
Not that I know of! Separable equations are like precious gems in physics :-) Steve

If a projectile is given an initial velocity greater than its terminal velocity, like a baseball pitch or meteorite entering the earth's atmosphere will it slow down to its terminal velocity given enough time?
Check out Taylor Eq 2.30, and take the limit as t-> infinity. Yup, if you start off faster, or slower, than vterm, doesn't matter, you end up at vterm!

Presumably that's what happens to parachutists - just before you pull the rip cord, you are going at vterm(for a human body), but right after you pull the cord, there's a big chute behind you (thus, you have smaller vterm now, but you're instantaneously going FASTER than your current vterm). So you slow down (there's an upward jerk when the chute opens!) ending up at the (new) terminal velocity. Steve P.

A more general question, for my curiosity: Why is it custom to take an introductory quantum mechanics course only to switch to classical mechanics immediately after?
I'd say it's because we want to span the full "sweep" of physics in the 1st three semesters (Classical, E&M, Quantum/Modern), and then start *revisiting* them one at a time. It's the "spiral staircase" approach to physics education :-) It's not the only way, but very common (and has a lot to say for it). Returning to Classical now, with you having already had some QM/modern under your belt, lets you do a little better "compare and contrast" compared to focusing solely on Classical at a more advanced level before ever seeing any QM. And, when you go back (again) to QM in your junior year, you can build on some of the Classical physics (Hamiltonians, e.g.) that will come next term.

Plus, Quantum is just too cool to keep away from you for more than 2 semesters!

________________________
Professor Pollock,

My name is Justin Baacke and I am in your Phys 2210 course. I was trying to put a study group together and was hoping you could send this link: http://doodle.com/8nb3754a8guu6kwc out to the class to see who would be interested, and what times they would like to meet. Thank you,

Justin G. Baacke

____________

Preflight #2:

Note that these are *preflights*, meant to have you read and think about the book before class, so that we don't have to spend as much class time talking about the straightforward stuff, but can focus on the deeper concepts, and subtler/harder points! If you are confused about these points, no worries - wait till we've had the lectures on them... at which point, though, you should understand this stuff.

First, my answers, in brief: 1) there is little difference between Taylors "half pipe" and my "pendulum" (except that Taylor's half pipe could in principle be some other shape, whereas my pendulum is really stuck being a circle. And of course, the physical origin of his radial force is "normal" (contact), whereas mine was "tension" from the string

2) Tension = mg cos(phi) + m R phidot^2. (From my notes, or from Taylor)

Interpretation: the first term is the component of weight (mg) that the string has to hold up! At the bottom, phi=0, the string holds it ALL up (and has mg added to the tension). Higher up, the bob is falling, so you don't have to hold it ALL up. (If it was freely falling, like at phi=90, you don't have to hold ANY of it up with the string - it's falling!)

The second term is m R*(angular velocity)^2, that's the centripetal force! The string must supply that, there's more tension if you swing around faster. (Note that this term is present even if g=0, an astronaut in the shuttle must supply this tension force if she swings a mass around in a circle on a string)

3) A volleyball (and any macro object) is dominated by quadratic drag. This comes straight from Taylor 2.7, as long as D is of order 1 m (which is "human scale things), and v is of order 1 m/s, (which is walking speed) , then f(quad)/f(lin) = 1600, so quadradic is ~1-2000 times more important! When might linear dominate? Well, when D, and/or v is small, really small, like mm distances, or mm/sec speeds. Sounds like the realm of biology to me! (There are many other examples - raindrops, oil drops in the famous Millikin experiment, etc)

Here are some questions and responses, you might find interesting.

I could not find your notes. (I don't know why, I looked, swear!)
Main web site, upper left corner, 3rd link down is "lecture notes".

In equation 2.8 the linear term was shown to be negligible because the ratio was 600. Is there are specific ratio at which we can say one of the terms is negligible? judgement call?
You bet - it really just depends on the accuracy you desire in your answer. If you need results with 10% accuracy, a 1 part in 600 error will be negligable. Same if you need it to 1%, but at 0.1%, then 1 part in 600 will matter. In this course, we will usually tend to go with the term that "dominates" and neglect the other entirely, to get the main physics effect.


I found that I'm still struggling a bit with the angular components of these problems in polar coordinates. I'm still not 100% sure what phidot and phidoubledot mean. If we could quickly touch on that again, it would be great (obviously, a lot of this lack of understanding falls back on me and my own need to work through more problems).
We will briefly touch on this in class today, but will then move on. So, this would be a great question to tackle in an office hour - it won't take long to clear it up for you, I think! (Also, if you haven't done so, check out my lecture notes, around pp. 10-12, I cover this stuff just a little differently than Taylor.)

The bike demonstration was a bit confusing and hard to see from the back, I'll have to try it at home. The explanation of the r unit vector was a little confusing because we didn't discuss the importance of phi much. The class seemed to want to include phi into the equation for the r vector. It would've helped to explain that the r vector isn't dependent on phi but then go on to talk about how phi is important in determining location in cylindrical coordinates.The orthogonality of phi to r was also confusing.
Yeah - try it on your bike, it'll probably make a lot more sense when you can *feel* what's happening.

I'm happy to chat about the phi/r/phihat/rhat thing, maybe stop by office hours this week? I would argue that r vector DOES depend on phi, phi is *implicitly* present in rvector (as well as rhat), because the radius vector "knows" which way it's pointing (even though the symbol {\vec r} on the page doesn't tell that information, the ARROW you draw really does!) It's the implicit dependence that is critical when taking the derivative to find velocity. Not just in cylindrical coordinates. Take a look at my lecture notes (pp. 10-12, roughly) and the text, and definitely happy to chat this out with you more if you like.

Thus far I am struggling a bit with the Newtonian laws in polar coordinates. I have attempted the derivation of velocity and acceleration twice since class and I have bundled through both times rather shamefully. Is there another medium I could approach with this inquiry, beyond Taylor's pictorial derivation?
Have you looked at my lecture notes? Pp. 10-12 I go through this, in more detail (I think) than Taylor. Try that out, let me know if it's still confusing. (We'll review, but not rederive, in class today).

It was really tough to find somthing that would give linear drag, since the book gave us some of the most used examples.
I really liked your "corn syrup" example, where the viscosity coefficient is high. But I would argue that the ratio depends on "D*v", irrespective of the medium details, so SMALL objects and SLOW objects tend to be better modeled with linear friction. (Small size, slow speeds - sounds like biology examples to me! Materials moving around in cells, small creatures crawling in water, that sort of thing)

I was confused by the fact that your equation in your notes said that F=m*R*phidot^2 when I thought it should've been (m*phidot^2)/R. But that was about it. The text was pretty straightforward and well-written.
Remember that phidot has units of "inverse time" (rad/sec, but radians are unitless). So this formula has the right units, F = kg * m /sec^2. In Phys 1110, we have two formulas for centripetal acceleration, the more common v^2/R, but also later in the course, omega^2*R. They are equivalent because v = omega*R. Does that help with the location of the "R" factor? (phidot here is the angular velocity, what we called omega in Phys 1110)

It would also be nice if we could go over the derivation of the coefficients beta and gamma in class.
You bet - preflights just aim at the basics, but we'll go over a lot of it again in class. (My lecture notes also cover this physics of beta and gamma too)

Not strictly related to the reading, but the use of ODE's made me wonder: would it be bad form for me to use the Laplace transformation to solve things if the differential equations get tough later on? I remember it being so much easier.
In general, if you can solve homework or exam problems using a different method, *and* can show/explain what you're doing, we're happy. But, there will also be times when we really want you to use a particular method (because it might have later generalized usefulness! And, though the methods you already know might work for this particular instance, they might not always work...) Really, our goal in this course is to broaden and deepen your math and physics toolbox. So, generally, practice/play with the methods we teach, and then in the end you can decide which method is best suited for specific problems you encounter. Make sense?

I found it very interesting that the time constant tau appeared in the equations for air resistance, as the last time i saw it was in the equations for capacitance.
Good point. The symbol "tau" is used quite generally to represent any "natural time constant". Note that the formula for tau in capacitance was totally different, but the USE of tau (telling you the exponential time decay of some physical quantity) was very similar. (The same differential equation in form, but completely different physics. I love it when that happens! And of course, it happens all the time throughout physics , precisely why it's useful to learn these general math methods :-)


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