« first day (3395 days earlier)      last day (1521 days later) » 
05:00 - 20:0020:00 - 00:00

8:01 PM
c) Draw the graph of natural logarithm of the amplitudes as a function of time and find the relaxiation time ($\tau$).
d) Find the damping coefficient $b$.
That's the question. Let me give you the data.
Graph 7.3 Period as a function of mass
Mass (gr) Period (s)
50 - 0,72
100 - 0,85
150 - 0,96
200 - 1,06
250 - 1,16
300 - 1,23
Graph 7.4 Vibration amplitude as a function of time for a 150gr mass.
 
I don't think you need to post all the data
 
Time (s) - Amplitude (cm)
0 - 4.5
30 - 4.0
And so on.
Now, I draw a graph of $T^2(m)-m$ graph.
Because the question 4.a tells it.
Let me show you the graph.
 
@Knight Sorry for the delayed response -- busy day at work today. Your clarification makes it clear that perhaps you didn't mean the exact way it came across, but you still intended to insult ACM. We can't allow people to direct insults at anybody or use ad hominem attacks on our site, which includes all chat rooms. That's true when discussing mods, as well as regular users.
So some general advice for better ways to handle this moving forward -- if you feel somebody has been rude or something towards you, it's okay to try and address it with them directly. There's a lot of great ways to express that you're upset with somebody's actions, without attacking the person. If you want some help on how to do that, I think any of the mods would be happy to give examples of good and bad ways to have that conversation
 
If that doesn't work, it's okay to walk away and/or flag the interaction for the mods to look at (or the community management team if needed). But try to take the high road and don't resort to insulting people or being rude to them, even if you feel they did it first.
 
8:10 PM
@ICCQBE Also looks fine to me (also, if you're not aware, people can see your real name at that link, or at least the one you use for your Google account)
 
It's fine, that's for informing :P
 
If you're not using the mobile site, you can use the "upload" button to the right of the chat text box to upload images directly to SE chat
Just saying, enough people have let it slip by accident :P
 
I wish I could use the desktop website, but unfortunately I'm a mobile user :(
Yeah, you are right :D
 
Ah, bummer
 
@ICCQBE You can use your mobile browser and tell it to request the full/desktop site
 
8:13 PM
I don't know, if we think the question is about a natural/non-damped harmonic oscillator, we can use $T^2=\frac{4M\pi^2}{C}$
 
It will give you the desktop chat interface, complete with insanely tiny buttons and stuff. But with zooming in, you can hit the image upload button
 
@tpg2114 Yeah, but my phone is not really good. It can't easily handle the desktop view and also it decreases my user experience :(
 
@ICCQBE Yes, but you need to be mindful of how the $M$ in that equation relates to the masses you're given
 
No worries -- just a suggestion. I will switch into that just to upload an image, then switch back. If the image was worth the effort to do all that :)
 
@tpg2114 Hmm, really nice solution. I'll try this next time :P
@ACuriousMind I don't understand what you mean :/
 
8:15 PM
@ICCQBE The $M$ in that equation is not the same as the 50 g, 100g, etc... in the data you have
 
Huh, why?
 
That's why the exercise says that There mass values given on the table don't include the mass of the spring.
 
So, they include only the mass of the attached object, right?
 
@ICCQBE Because the $M$ is the full mass of the oscillating system, i.e. the mass of the weight + the mass of the spring.
 
It's the only mass of the object. Because, it doesn't include the mass of the spring.
Isn't that correct?
 
8:19 PM
The mass in the table is only the object, correct.
But the mass $M$ in the formula you're using is the mass of the object and the spring
 
Hmmmmm...
Now I get it.
 
The exercise is asking you to figure out how to use that information to determine the mass of the spring from the data
 
So, we can use the formula as $T^2=\frac{4(M+M_{s})\pi^2}{C}$.
 
Yep, that's the way :)
 
But we will have 2 unknown terms here, $M_{s}, C$.
Hmm...
 
8:22 PM
Yeah, you will have to also determine the $C$ along the way
 
I thinknI can't find $C$ from the Table 7.3, I should use the Table 7.4
 
No, you can find both unknowns from 7.3
 
Amplitutes as a function of time and time graph.
Hmm, please let me think. I want figure it out.
 
(Maybe it helps you to think geometrically about what "properties" of the line you've drawn $C$ and $M_s$ control. Maybe not. People are different. )
 
@tpg2114 I think addressing the actions not the person is something that is hard to overstate. And in that vein, I think it's helpful to take a step back when you feel insulted and see if the comments are being directed at you personally, or if you are taking comments about actions personally when that isn't the intention.
 
8:26 PM
It's tangent value gives me $\frac{T^2}{M}$ values...
 
By "tangent value" you mean slope?
 
Yeah, exactly. Sorry for my English :P
 
Don't worry about it
 
@JMac Great points -- and it's also useful to step away and collect your thoughts before trying to have a conversation if one goes that route. Drafting and re-drafting the issues to raise makes it much easier to have a productive conversation than winging it and mis-speaking.
I have to do that frequently and it sometimes takes me awhile to hash out how to address the problem in a constructive way
 
Ah! The unit of the slope value is $s^2/kg$.
Spring coefficient's unit is $kg/s^2$.
The slope value may equal to $\frac{1}{C}$.
 
8:32 PM
@tpg2114 I debated with myself for a solid hour before I even responded to that "Good People" chat room a month or so ago. I didn't really want to stir the pot, so I was reluctant to say anything; but I decided in that case not addressing the perceived rudeness wouldn't help the situation either.
 
Did I find it @ACuriousMind? :P
 
@ICCQBE Well, proportional to - there's a constant factor missing there. Look at your equation for $T^2(M)$ - can you read the slope off it?
 
@JMac I wonder why I wasn't a part of a "good people" chat room
 
@AaronStevens To be fair, I think you were indirectly part of it...
 
@ACuriousMind $\frac{T^2}{(M+M_{s})}=\frac{4\pi^2}{C}$.
 
8:35 PM
@JMac Uh oh. I was not even aware
 
But the slope value gives $\frac{T^2}{M}$.
Because the mass values don't include the mass of the spring.
 
Do you know that if you write a line as $y(x) = ax + b$, then $a$ is the slope and $b$ is the y-intercept?
 
Yeah, I do.
 
Okay, I'm not quite sure what's confusing you. The slope is indeed $\frac{4\pi^2}{C}$.
 
@AaronStevens Yeah, I don't want to get too into it, because I was pretty pissed at the time; but it was about 'hypothetical' high rep users and their poor behaviour. This didn't name anyone; but they did post 3 reputation numbers that happened to match you, emilio's, and my reputation at the time.
 
8:39 PM
How is the slope $\frac{4\pi^2}{C}$? :/
 
@ICCQBE You have the equation for your line as $T^2(M) = \frac{4\pi^2}{C}M + \frac{4\pi^2}{C}M_s $, right?
(all I've done is re-arrange the terms so it matches the $ax + b$ pattern)
 
@JMac Ah, interesting. Odd that no one has actually just talked to me about any issues they have. I would totally be fine with discussing it.
 
Yeah, abbsoultely. Let me see the re-arrangement, please.
Oh!
$ax=\frac{4\pi^2}{C}M$ and $b=\frac{4\pi^2}{C}M_{s}$ right?
 
Yep!
 
Well, then the slope $a$ equals to $\frac{4\pi^2}{C}$!
 
8:44 PM
@AaronStevens That's what really bugged me about the situation. I only became aware of it when I stumbled on that chat on my own. But when I called them out on it they denied that they were even referring to me, so it kinda killed any attempt at really working it out even after I was aware.
 
You're a genius! :D
Hmm! Let me calculate the slope please.
 
@JMac Odd... Oh well. I usually don't have problems with people here. If I do I forget about it in about a day
 
@AaronStevens Yeah I really shouldn't have even reminded myself of it. All it does is make me salty.
 
@JMac Yeah I can understand. I would probably be just as upset had I seen it
 
@JMac Don't be salty. Too much sodium is unhealthy.
 
8:54 PM
@EmilioPisanty I definitely remember a webcomic (smbc?) about "God's patch notes" that included an item like "Fixed: Human existence included unintentional difficulty spikes" but for the life of me I can't find it again
 
@tpg2114 Yeah I'm not a fan of being salty. It's a pretty crappy feeling. Thankfully it passes quickly almost always. I think my talent with procrastination pays off in that sense, since you can just procrastinate the saltyness indefinitely; given it's lack of deadlines.
 
Sorry, I can not upload photos by using desktop view :/
The button is not reacting to my clicks :D
 
@JMac We can start a chat room of "formerly salty people"
Or should it be "future salty people"?
Depends on the reference frame I suppose
 
@ICCQBE Aside from rounding too early (no need to round the fraction to 3.92), looks fine to me
 
8:58 PM
@AaronStevens And if you believe in closed timelike curves.
 
"people with partially salty worldlines"
 
@JMac "people who could be formerly or soon to be salty depending on your belief of closed timelike curves" chat room
 
"Salty Worldline" is what I imagine a snack at a GR conference to be called, come to think of it
 
Has a nice ring to it
 
The bar serves "Closed Winelike Curves"
 
9:00 PM
I'm thinking "Desalination Chamber" has a nice ring to it.
 
@ACuriousMind That might be confusing for people who want to keep an open tab
Supermassive black bean burrito?
I am just trying to think of GR related foods now
 
Does caloric content change based on whether you eat it while moving?
 
@tpg2114 I guess as long as you are moving relative to the food
 
"Dark Matter Sandwich" (it's just two slices of bread)
 
@ACuriousMind The first time a scientific discovery was born out of a lawsuit
"I asked them to prove to me that there was dark matter in their sandwich, so they did"
Are you technically lying if you make claims no one can prove is true or false?
 
9:06 PM
Any time you start to ask the question "Is this technically a lie?" is a good sign to rethink the choices you've made :P
 
@ACuriousMind What if that question came up while rethinking the choices I have made?
 
re-re-think them!
 
You'll need an exit condition
Add a loop counter and exit to beer
 
@tpg2114 Why not just cut out the middle man and exit to beer regardless? Who needs the loop at all?
 
Because you can't exit if you are already there
 
9:09 PM
@JMac Nobody said the maximum loop counter had to be greater than the starting index
 
"This program is supposed to help me make decisions. Every time I run it, it just tells me to drink another beer. I've been very impressed with the results so far."
 
For(loop = 1, loop <10, loop++){ If loop >0 then beer and exit loop)}
 
I was just writing the Fortran version of that loop...
 
I don't think I picked any specific language
 
It's... C-esque
 
9:12 PM
That is probably the last time I used a For loop, so it makes sense
My first introduction to programming was C++, but really the class was just "Introduction to programming, and I guess we will use C++ to teach it"
 
I keep telling myself I need to screw around with coding again and I never do. Closest I've gotten is the Zachtronics video games in the past like 5 years.
 
Only afterwards did I realize that we didn't get into anything that was actually specifically (or mostly?) unique to C++
 
@AaronStevens Hey I took one of those too! And the "C++" I learned looks nothing like any other beginner C++ stuff I've looked at.
 
@JMac Haha yeah! I did have a Java class that was actually about Java though
That was a fun class
We spent a couple of week ins Java II learning how to program Android mobile apps
 
My intro to programming class in college was Matlab, with a short (like 3 classes) section on Java
 
9:15 PM
Sadly that was my only actual class focused on coding. My numerical methods class taught us some matlab; but it wasn't formal at all.
 
@ACuriousMind Sorry, my room mate called me. I'm back.
 
I feel like Matlab is just one of those things you pick up because you need to
I never needed to, so I never did :P Although I did dabble in it for part of a project I was working on
 
I have a question about the question. We have solved this question and found answers to matters a and b by using the $T^2=\frac{4M\pi^2}{C}$
 
It was a really cool tool to play around with, especially with some of the add-ons. Actually getting an accurate looking plot of the heat distribution in a 2D object was so satisfying, even though it was fairly trivial.
 
So at GT, Matlab was the engineer's intro programming class -- all non-engineering majors took Python.
But I think that changed and now everybody takes Python
MathWorks does a fantastic job of brainwashing people to learn how to program using Matlab, so when they get to the workplace their employer has to pay $20k+ for licenses
 
9:18 PM
My undergrad didn't have anything with Python. So since going to graduate school I have felt like a fraud
 
@ACuriousMind We solved the question by considering the system is non-damped/natural harmonic oscillator. But if we consider that thr system is a damped harmonic oscillator, the things would change, right?
 
@tpg2114 The joys of learning Cashlab
*Matlab
 
The simulink on matlab was pretty friggin cool for systems. I think at one point we even had it hooked up to data collection, simulated a control system with simulink, and the outputted the results to the physical system again. That was really entertaining.
 
@AaronStevens Meh -- that hard part to programming is learning how to translate something like "Go get a drink from the fridge" that seems like a single instruction to humans into the 10000 serial things a computer would need to do to execute it. Once you wrap your head around that, picking up new languages is pretty easy
@JMac I think Simulink is the only part of Matlab that is uniquely useful, but even then it's very narrowly focused on controls
Everything else can be found in much cheaper, faster, more general tools or languages
 
The expression for $T^2$ would change because $\omega^2=\omega_0^2-\beta^2=\frac{4\pi^2}{T^2}$. So, $T^2=\frac{4\pi^2}{(\omega_0^2-\beta^2)}$.
 
9:23 PM
@ICCQBE Yes, if you consider the damping, a factor of $\sqrt{1 - \zeta^2}$ where $\zeta$ is the damping coefficient enters.
 
@tpg2114 Yeah, I was about to say it does have other uses; but it's probably far from the best tool for the job in most cases. It wasn't too bad for me since one of my friends gave me a pirated copy for home use.
 
And here we have another unknown, $\beta$ the damping coefficient.
Hmm, can we solve this expression with only these data?
 
Even if you have to pay for it as a student, it was like $100 for the academic copy when I got it. I've been getting quotes for fluid solvers recently, it's astounding how expensive these things are
Like... highway robbery expensive.
Single-seats that can't run on more than 4 processors are like $60,000/year
 
@ICCQBE Well, everything works as before, you just have an undetermined $\zeta$ in your results until you've determined it from the decay of the amplitude in the second table of data.
 
@tpg2114 $100 is a lot to a student though :P But yeah I imagine something like a fluid solver is quite a bit of a different beast. Even just thinking about AutoCAD licenses and stuff, the prices for "professional" software like that are insane.
 
9:26 PM
Yeah, NX, SolidWorks, AutoCAD etc are nuts too. But they get the students hooked
 
So, the question is only about a damped harmonic oscillator instead of a natural one? :D
 
Wouldn't surprise me if the companies seeded some of the pirated copies that end up at universities
 
@tpg2114 A guy at my work got us in trouble with AutoCAD because he used a pirated copy on a laptop to try to open a model or something and they called us out on it like the very next day.
 
@ICCQBE Probably, yeah (also ideal, not natural - arguably the damped ones are the natural ones as you'll rarely find a completely undamped oscillator in nature ;) )
 
We even have a bunch of licenses; he just needed to open something in a hurry and didn't have access to one of those computers or something.
 
9:29 PM
I believe it. But if I were AutoDesk, I would let students steal copies so it ends up being the only tool they know how to use. So when they get to a company to work, I could charge them basically anything
 
@tpg2114 That's the genius of it. I don't think they give a crap about students with it. I think my friends had copies in university. But as soon as it touched a company CAD file they were on us right away.
Because they probably know they aren't going to get anything going after the students anyways.
 
@ACuriousMind Well, I'll mind this one :P Thanks. I'll try to solve the question as considering the system is a damped harmonic oscillator.
By the way, I forget that the equivalent of $\omega_0^2$. It equals to $\frac{C}{M}$ right?
 
I think so, yes
 
Alright, I think I have got the way of solving this question. I'll ask you again if I struggle :P
Thanks again!
 
you're welcome
 
9:50 PM
@ACuriousMind Should I take the natural logarithm of the time values in Table 7.4 along with amplitude values?
 
No, it's telling you to graph the logarithm of the amplitude against the time
 
Alright.
 
hmm, i gotta find some new way to wake up, coffee doesn't seem to work anymore
at least not in "healthy" quantities
 
10:13 PM
I drew the $\ln(A)-t$ graph.
It's a linear graphic.
As the question matter c, it want me to find the relaxiation time using this graph.
I know that if i can find the relaxiation time ($\tau$), I can find the $\beta$ the damping coefficient using $\beta=\frac{1}{2\tau}$.
Now, I have an idea to find $\tau$.
As the definition of $\tau$ (as I know) It's the moment when the object loses 63% of It's initial velocity.
Is it correct?
But, now I don't have any idea what is the physicsl description of $\ln(A)$ values and what is the relation of $\ln(A)$ and time.
Also, I only know one relation between amplitude and time; $x(t)=e^{-t\beta}x_{0}\sin(\omega t+\phi)$.
 
 
2 hours later…
11:56 PM
@EmilioPisanty ooo
@EmilioPisanty this sounds neat
 
05:00 - 20:0020:00 - 00:00

« first day (3395 days earlier)      last day (1521 days later) »