In At The Deep End - 2017-08-09

John Rennie

7:05 AM

room mode changed to Gallery: anyone may enter, but only approved users can talk

John Rennie

10:41 AM

→ 3 messages moved to trash

Hi (again)

I thought the seniors weren't allowed to talk to you. I trust it wasn't some form of ragging ...

user228700

We were afraid that we might get ragged etc.

user228700

But no, it was only to brief us about the rules of the hostel. Rules, as it happens, that they have created.

user228700

Well, that and to urge us to participate in the marathon coming Sunday.

John Rennie

Ah yes. You were going to run but don't have your running shoes with you?

user228700

But we came back only an hour later and by that time, I had completely forgotten about the conversation that we had been having for I was too engrossed in all of the things they had said.

user228700

10:44 AM

@JohnRennie No, I don't, but I'll go anyway :-P Lots of us are going so I will too.

user228700

I'm so sorry! :-(

John Rennie

I recall going to pretty much every social event there was in my early days at college whether I was interested or not.

It was a good way to meet lots of people - a few of whom became close friends :-)

user228700

^ I plan to do the same within my own capacity for social interaction :-P

user228700

@JohnRennie What were you going to say?

John Rennie

@Kaumudi.H you're not going to like this, because it involves sending you something, but that something is genuinely stuff I have lying around.

user228700

10:48 AM

I'm getting suspicious...

John Rennie

:-)

user228700

Go on?

John Rennie

All I was going to suggest is that I send you some spare disks for your laptop.

Because ...

user228700

Ah. And what will I do with them?

John Rennie

I think it would be good for you to have Windows 10 installed. I had a play with the Windows 10 unix shell last night and it works very well.

user228700

10:50 AM

Oh, wow, I see.

John Rennie

But given that the laptop is pretty important it's a bit of a risk to either try and upgrade or wipe the existing disk and install W10.

user228700

Right, right, I see.

John Rennie

What I do in these circumstances is take out the hard disk and bung in a spare disk and do the install on that. If it all goes well then that's great. I now have W10 installed.

If there's a problem I take out the spare disk, put in the original disk and everything is back to where it was before I started messing around.

user228700

Ah, I see. That sounds quite neat, actually.

John Rennie

So I was going to suggest I send you a spare disk for you to install W10 on

All you have to do to swap the disks is undo 2 screws on the base of the laptop then you can slide out the disk.

user228700

10:54 AM

Hmm, I see. How much does one cost, usually?

John Rennie

If you want to buy a brand new disk then ... actually I don't know because I get all my disks from the bust laptops I get.

I can't remember the last time I actually bought a disk.

user228700

Ah, I Googled it and found that it will cost around Rs. 4000!

John Rennie

That seems a lot. What size disk is that?

user228700

Ah, never mind, that one's 1 TB.

user228700

My bad; it's an external one.

John Rennie

10:56 AM

Ah :-) I was going to suggest a 256GB disk.

That should be plenty big enough unless you want to store hundreds of films

user228700

Huh, I can't find anything on Amazon...

John Rennie

Let me look ...

amazon.co.uk/320gb-2-5-internal-Drive-Laptop/dp/B00JLCWJ7K

user228700

Sorry about that.

user228700

Item doesn't ship to India :-/

user228700

OK, I will look around some more and see if it's possible for me to find a cheap one. May I let you know in two days or so?

John Rennie

11:03 AM

I wasn't suggesting you buy it. In any case you need the drive caddy to make it fit in your laptop.

I was going to suggest I can send you one of my large pile of spare drives.

user228700

Ah, right. And what will that cost you? ::Stares suspiciously at screen(you)::

John Rennie

If you did want to buy one yourself I guess you'd need to look on Amazon.in

It will just be the postage, and ... I don't know what the postage will be.

But these disks are tiny little things so it will go in the normal letter post.

user228700

Ah. Hmm...

John Rennie

It's not like having to post a laptop that is large and heavy.

user228700

Right, right.

user228700

11:05 AM

Can we sort this out later tonight, then?

John Rennie

There's no hurry. I was going to suggest I send it to your parents for you to collect when you're back for the holiday. And that's a month or so yet I think.

user228700

Yep, that is! :-)

John Rennie

So it's not as if we have to decide right now.

user228700

Wokay! :-)

John Rennie

I'll leave it for you to think about.

user228700

11:07 AM

Thanks so much for offering! :-) Really.

user228700

Will you be around later tonight?

John Rennie

Yes. What time?

user228700

Oh, God, I almost fell over in my chair!

John Rennie

i.e. what time in Kochi? (I can subtract 4.5 from it :-)

user228700

@JohnRennie Same time as usual?

John Rennie

11:08 AM

20:30 your time?

user228700

How about 21:00?

user228700

Or 19:00?

user228700

(Apparently, dinner is at 20:30 and at 20:30 alone :-/)

John Rennie

Ah. For some reason I'd thought dinner was at 20:00

user228700

I had too but experienced differently yesterday :-/ To be safe, I've decided to go only at 20:30 everyday and I'll be done in 15 minutes or so; eating, washing, walking all the way back.

John Rennie

11:11 AM

19:00 is 14:30 my time and I'll most likely be out.

user228700

Ah, OK.

user228700

21:00, then? Will you have finished lunch by then?

John Rennie

Yes, I've normally finished by 4 p.m. - 20:30 your time - so 21:00 would be fine.

user228700

:-) Wokay! So I'll see you then?

John Rennie

Good. You're welcome to ping me earlier. If I'm not around the ping does no harm.

user228700

11:13 AM

Hehe, OK :-) Have a nice one!

John Rennie

Bye

user228700

3:34 PM

@JohnR: Hi! :-) I'm sorry that I'm 5 minutes late.

John Rennie

Hi :-)

Good dinner?

user228700

You will be happy to know that I have prepared questions today!

John Rennie

Uh oh! :-)

user228700

@JohnRennie Oh, great dinner! I'll tell you about it after :-)

user228700

I practiced Linux commands today as well and have a few more questions...

John Rennie

3:36 PM

OK ...

user228700

Here, look at this:

user228700

user228700

What is up with that?

John Rennie

Did you mean: cat >'Attempting creation'

i.e. create the file called "Attempting creation" with a space in the name?

user228700

Yep, precisely.

John Rennie

3:40 PM

Because you omitted the quotes the CLI interpreted Attempted creation as two separate arguments.

I think it probably tried to cat a file called creation and send the output to a new file called Attempting

user228700

Oh, was I really meant to put it in quotes? :-o My teacher didn't tell me about that!

user228700

John Rennie

If a file name has a space in it you have to surround with quotes so the CLI knows where the file name starts and ends.

user228700

> cat a file

user228700

What is that?

user228700

3:43 PM

@JohnRennie See, my teacher didn't tell me this. OK!

John Rennie

The command cat reads text from some source and outputs it to some destination

user228700

Ah. Hang on, I'll try something.

John Rennie

If you just type cat the cat program reads from stdin and writes to stdout

If you type cat somefile then the cat program reads from the file somefile and writes to stdout.

user228700

AH.

user228700

Right, right. I understand.

John Rennie

3:46 PM

The > character is the redirection symbol. It redirects anything written to stdout to the file name that follows >

So cat >somefile reads from stdin (i.e. your typing) and writes to the file somefile

user228700

I understand!

user228700

RIGHT.

John Rennie

Or cat afile >bfile copies the contents of the file afile to the file bfile

Or you could do cat <afile >bfile to do the same thing.

user228700

Oh, wow, nice.

John Rennie

The < character redirects standard input

user228700

3:49 PM

Right, right.

John Rennie

So cat <afile runs the cat program and takes its input from the file afile

user228700

OK, one more!

user228700

@JohnRennie And the >bfile redirects that input to bfile, correct?

John Rennie

>bfile redirects the standard output stream to the file bfile

user228700

What do you mean by "Standard output stream"?

John Rennie

3:51 PM

All unix programs have a standard input stream and a standard output stream. The term stream just means something that transports data.

user228700

And what does the term standard indicate?

John Rennie

By default the standard input stream is the console i.e. what you type.

user228700

Ah.

John Rennie

@Kaumudi.H good question, and I don't know. I think it's just a name and doesn't mean anything. Standard input kind of means default input i.e. the input you get if you don't mess about.

user228700

Ah, right, I see.

John Rennie

3:53 PM

Standard output is normally the console i.e. anything written to standard output appears in your console.

So cat afile just types the contents of the file afile on your console

user228700

Right. Understood!

user228700

I've got one more:

John Rennie

We usually say stdin and stdout for short

user228700

@JohnRennie Noted!

user228700

3:54 PM

My teacher told me that the "man" command is meant to list the descriptions of a particular command. Clearly, that is not the case...

John Rennie

That's a bit odd. I guess the online shell you're using doesn't have all the man pages. What you typed should work.

user228700

Ah, so my teacher was right(?)

John Rennie

I've just tried `man mkdir' on the Windows 10 unix shell and it works

user228700

Right, so it will work. OK! On to Physics, then(?)

John Rennie

user228700

3:57 PM

@JohnRennie Oh, wow, nice.

John Rennie

This is why I think it would be good for you to be able to install Win10 or even Linux (if I sent you two spare disks).

user228700

Hmm, right, right. Tell you what; you tell me how much it will cost and I will let you know :-P

John Rennie

I've no idea. I'd have to look up the postage to India. But I'd guess 20-40 rupees

user228700

What? That's all?

user228700

Well, sure, then!

John Rennie

4:00 PM

Certainly less than 100 rupees.

It's only a small parcel. Little bigger than a letter.

user228700

Ah, and you were ready to buy me those Portal games so that's not too much, I guess...

user228700

I'll take 'em! :-D

John Rennie

OK let me rummage through my pile of disks and look for a reasonable fast one.

user228700

Wokay! :-) No hurry at all though; I will go back home only next month!

John Rennie

I assume your parents address hasn't changed?

user228700

4:02 PM

No, it hasn't :-)

John Rennie

Do you use Office on the laptop? If so I'll need to include the installation files for Office.

user228700

I...no, not very often but it would be nice to have them if it's not too much trouble.

user228700

I don't think I've got PPT at the moment.

John Rennie

No problem. I'll include them as well.

You have got Powerpoint.

user228700

Ah, yes, I do. Sorry :-P

John Rennie

4:04 PM

:-)

user228700

Do you have to go now or do you have time enough for some Physics (at last!)?

John Rennie

Yes, I have no urgent engagements elsewhere. I usually hang around the chat until 18:30 (23:00 your time)

user228700

Ah, so we've got some time!

user228700

Wokay, I was attempting to understand that derivation whose picture I showed you the other day, do you remember?

John Rennie

The damped SHO?

user228700

4:06 PM

The very same.

John Rennie

Can I give John's quick summary of the DSHO?

user228700

YES, please.

John Rennie

If you have an undamped SHO the equation of motion is: $$\frac{d^2x}{dt^2} = -\omega^2 x$$

user228700

@JohnRennie Yes, right.

John Rennie

And the solution is: $$x(t) = e^{i\omega t}$$ give or take a few constants. So far so good?

user228700

4:08 PM

Sort of, yes.

John Rennie

Write this as: $$x(t) = e^{C t}$$ where $c$ is an imaginary number. I've bolded imaginary for reasons that will become clear ina moment.

user228700

OK...

John Rennie

Now suppose we consider damped linear motion i.e. some object moving along ina straight line with a drag force proportional to its speed. The EOM is:

$$ \frac{d^2x}{dt^2} = -kv = -k\frac{dx}{dt} $$

user228700

@JohnRennie I have a question related to this statement but OK...

John Rennie

And the solution is: $$ v(t) = e^{-kt}$$ again with a few constants around.

user228700

4:12 PM

@JohnRennie Wait, where did the term to do with the restorative force disappear to?

John Rennie

This is undamped linear motion not simple harmonic motion

There is no restoring force, just the drag force

I do have a point to this - trust me :-)

user228700

@JohnRennie Huh? It's not an oscillatory motion?

John Rennie

56 secs ago, by John Rennie

I do have a point to this - trust me :-)

user228700

I'm very confused but OK...

John Rennie

My point is that for linear damped motion $x(t)$ is going to contain terms in $e^{-kt}$ where $k$ is real.

So for oscillatory motion we have an imaginary exponent and or non-oscillatory damped motion we have a real exponent.

And (drum roll because this is the big reveal) ...

user228700

4:17 PM

I understand that, but what is non-oscillatory damped motion?

John Rennie

If we combine the two to get damped oscillatory motion we have solutions: $$x(t) = e^{Ct}$$ where $C$ is a complex number i.e. $C$ has real and imaginary parts

user228700

I...hmm, I do follow but only vaguely.

John Rennie

When solving differential equations we usually guess what the solution looks like and feed it into the differential equation to see if it works.

user228700

OK...

John Rennie

The differential equation for DSHO looks like: $$\frac{d^2x}{dt^2} = -k\frac{dx}{dt} -\omega^2 x$$ Yes?

Silence ...

user228700

4:21 PM

Yep!

user228700

@JohnRennie I'm sorry :-/ My internet can be a little annoying.

John Rennie

So what I'm saying is take, as your guess: $$x(t) = e^{Ct}$$ where $C$ is a complex number. Feed it into your differential equation and you'll get a couple of equations you can solve to get the complex number $C$.

The real part of the complex number $C$ gives you the damping term and the imaginary part gives you the oscillatory term.

user228700

My understanding remains quite vague.

user228700

...is there some text I can read to understand this better? I tried reading the prescribed textbook but that didn't go well.

John Rennie

We'd have to go through the working to convince you and I'm not sure that's profitable just now.

user228700

4:25 PM

The working of what?

John Rennie

@Kaumudi.H feed my proposed solutin into the differential equation and see what happens.

user228700

Ah.

user228700

I think it's worth adding that this is the first time I'm dealing with this stuff.

John Rennie

But you'd find that some of the apparently odd equations your lecturer wrote down fall out of the working.

But, let's go back to you.

user228700

@JohnRennie I think I lost the physical significance of things right about here.

John Rennie

4:28 PM

If I fire a bullet that's linear motion (ignoring gravity for the moment)

user228700

Hmm, I think it would be better if I gave this a few more reads and come back tomorrow with a clearer picture (and head).

user228700

(Like I mentioned yesterday, I haven't slept well in over 3 days and it's starting to take its toll on me...)

John Rennie

Maybe just forget my analogy for now and we can come back to it if I think it helps. Obviously it hasn't helped so far.

What did you want to ask about DSHO?

user228700

No, no, it was interesting and quite useful too, the place you were taking things but I didn't understand because I don't have a firm background in this yet.

user228700

6 mins ago, by Kaumudi. H

...is there some text I can read to understand this better? I tried reading the prescribed textbook but that didn't go well.

John Rennie

4:32 PM

I'm not really the person to ask about textbooks. I haven't read a mechanics textbook for thirty years

user228700

:-/ Ah, hmm.

John Rennie

But if you find you've got lost following the lecture notes maybe I could help if you explain where you are lost.

user228700

OK.

user228700

How is $\sqrt{C/m}= {\omega}_o$?

user228700

I'm sure that you really are the correct person to ask about this because you still remember!

John Rennie

4:35 PM

If we have an undamped SHO then its equation is basically just $F=ma$

The force is proportional to the displament so: $$F = -Cx$$ for some constant $C$.

user228700

Right, right.

John Rennie

So we have $-Cx = ma$ or with a quick rearrangement $$a = \frac{d^2x}{dt^2} = -\frac{C}{m}x$$ OK so far

user228700

Yep!

John Rennie

And we guess that the solution is $x(t) = e^{i\omega t}$

user228700

What's the $\omega$ doing over there?

John Rennie

4:39 PM

If we differentiate $x(t)$ once we get: $$\frac{dx}{dt} = i\omega e^{i\omega t}$$

If we differentiate it again we get: $$\frac{d^2x}{dt^2} = -\omega^2 e^{i\omega t}$$

OK so far?

user228700

Yes...

John Rennie

So now we take our equation: $$\frac{d^2x}{dt^2} = -\frac{C}{m}x$$ and we substitute for $d^2x/dt^2$ on the left side using the second derivative I just calculated. And on the right side we subsitute for $x$.

user228700

Right.

John Rennie

Ans what we get is: $$-\omega^2 e^{i\omega t} = -\frac{C}{m}e^{i\omega t} $$

Is this clear, because this is the crucial step?

user228700

Absolutely.

John Rennie

4:43 PM

OK, and the equation gives us: $$\omega = \sqrt{\frac{C}{m}}$$

user228700

AH.

user228700

See, in my textbook, they started out by stating this and went on with the derivation with this assumption in mind.

John Rennie

So we guessed a form for the solution with an undetermined constant $\omega$, and discovered that YES our solution works provided $\omega$ has the value above.

user228700

@JohnRennie Please move to Kochi and take the place of my Physics teacher :'-( That is so incredibly clear!

John Rennie

Maybe now you can see better what I was getting at with all my babbling about the damped SHO?

user228700

4:46 PM

A little bit, yeah!

John Rennie

With the damped SHO the equation changes slightly because it includes the damping term. So it becomes: $$ \frac{d^2x}{dt^2} = -k \frac{dx}{dt} -\frac{C}{m}x$$

Do you understand where that extra term comes from i.e. what it means physically?

user228700

@JohnRennie I do, but I don't understand why it's proportional to velocity...

John Rennie

Have you heard of Stokes drag?

Stokes' law

In 1851, George Gabriel Stokes derived an expression, now known as Stokes' law, for the frictional force – also called drag force – exerted on spherical objects with very small Reynolds numbers in a viscous fluid. Stokes' law is derived by solving the Stokes flow limit for small Reynolds numbers of the Navier–Stokes equations. == Statement of the law == The force of viscosity on a small sphere moving through a viscous fluid is given by: F d = 6 π η R ...

user228700

Reading now!

John Rennie

Just read the introduction. You don't need all the detail.

I'm just going to get a glass of water while you're reading.

user228700

4:49 PM

Hmm.

user228700

@JohnRennie OK.

John Rennie

The point is that in many physical systems the drag force is proportional to velocity.

user228700

I read this:

user228700

10

Q: Explanation that air drag is proportional to speed or square speed?

A falling object with no initial velocity with mass $m$ is influenced by a gravitational force $g$ and the drag (air resistance) which is proportional to the object's speed. By Newton´s laws this can be written as: $mg-kv=ma$ (for low speeds) $mg-kv^2=ma$ (for high speeds). I assume that $k...

user228700

Understood.

user228700

4:52 PM

I probably shouldn't care about that this much.

John Rennie

If you take a pendulum that's undamped motion. If we now immerse the pendulm in water the water produces a drag on the pendulum bob that is proportional to the velocity of the bob.

user228700

@JohnRennie Other than this, yes, I do understand.

user228700

@JohnRennie Right. I get it.

John Rennie

The link you posted describes quadratic drag.

user228700

Yeah, but I get the idea. And also that I should drop it.

John Rennie

4:53 PM

As a general rule we get quadratic drag in turbulent systems i.e. motion at high speeds. We get linear (Stokes) drag in non-turbulent systems i.e. low speeds.

The damped SHO assumes speeds are low enough that the drag is linear.

user228700

Right, right. I understand.

John Rennie

Imagine a pendulum underwater.

user228700

...but not in maple syrup, no?

John Rennie

Maple syrup if you want. Any fluid really.

user228700

...hmm, ah, but that would be overdamped, no?

John Rennie

4:56 PM

It's the same physics. The same differential equation. All that changes is the ratio of the damping force to the restoring force.

user228700

Simply by observing the ways in which I am responding, I'm sure you are able to judge just how well I understand the subject, lol.

John Rennie

That's fine. It's easy to forget after 30 years that we all had a first time for learning this.

user228700

Hmm :-/ You know what, let's stop here today.

John Rennie

Let me write my equation again since it's scrolled away: $$ \frac{d^2x}{dt^2} = -\frac{k}{m} \frac{dx}{dt} -\frac{C}{m}x $$

A thin fluid would have a small value for $k$ while maple syrup would have a large value for $k$

That's the only difference.

user228700

Right, right. I understand.

user228700

4:59 PM

To save you this much extra trouble, I will come back tomorrow, having revised a lot better.

John Rennie

Let me go on for just a minute longer ...

user228700

Sure.

John Rennie

With the undamped SHO I guessed a form for $x(t)$ then by putting my guess in the equation I could show my guess was correct and work out what the angular frequency $\omega$ had to be.

And we do exactly the same for the damped SHO.

user228700

@JohnRennie Yes, right. Well...

user228700

Did we assume that $\omega$ would be the angular freqeuncy going in?

John Rennie

5:02 PM

@Kaumudi.H my guess was $x(t) = e^{i\omega t}$ yes?

user228700

Yep.

John Rennie

And $e^{i\omega t} = \cos\omega t + i \sin\omega t$

So my guess is an oscillating function and in that function $\omega$ is the angular frequency.

user228700

Ah, right.

user228700

I think I understand.

John Rennie

One last desperate crawl towards the finishing line, then we'll stop for the day ...

user228700

5:05 PM

OK...

John Rennie

With the dampded SHO my guess for $x(t)$ is slightly different. This time it's: $$x(t) = e^{(a + ib)t}$$ So now I have two unknown constants instead of just the one constant $\omega$.

But the method we use is exactly the same.

user228700

Right, right, I see.

John Rennie

Start from my $x(t)$ I calculate $dx/dt$ and $d^2x/dt^2$, which is straightforward

Then I go back to my DSHO equation $$ \frac{d^2x}{dt^2} = -\frac{k}{m} \frac{dx}{dt} -\frac{C}{m}x $$

user228700

Yes...

John Rennie

and I just substitute for $x$, $dx/dt$ and $d^2/dt^2$

user228700

5:09 PM

Right.

John Rennie

And that gives me an equation that relates the constants in my guess, $a$ and $b$, to the constants in the equation $k$, $C$ and $m$, and from that I can work out what $a$ and $b$ need to be.

If you take my guess $$x(t) = e^{(a + ib)t}$$ we can write it as: $$ x(t) = e^{at} e^{ibt} $$ Yes?

user228700

RIGHT.

user228700

Yep.

John Rennie

And the $e^{ibt}$ is just the same oscillatory function we got from the undamped equation.

So we what we get is an oscillating function multiplied by an exponential.

That number $a$ is going to turn out to be negative, so it's actually $$x(t) = e^{-at} e^{ibt} $$ i.e. oscilaltions bt muliplied by exponential decrease with time.

user228700

@JohnRennie Wait, don't you mean the other term?

John Rennie

5:14 PM

With the undamped SHO we get $e^{i \omega t}$ and this time we get $e^{ibt}$.

In the limit of the damping going to zero obviously $b = \omega$

user228700

Ah, hmm, I think I see your point.

John Rennie

In the equation: $$x(t) = e^{-at} e^{ibt}$$

the $e^{ibt}$ term just oscillates continuously with time - for ever!

But the $e^{-at}$ term decays with time and tends exponentially towards zero.

user228700

Ah, right, right, I see...

John Rennie

Multiply the two together and you get oscillations but their amplitude decays exponentially with time i.e. they are damped

That's what a damped simple harmonic oscillator is.

user228700

Hmm, right...

John Rennie

5:18 PM

So you can see why my guess has a good chance of working

user228700

Yep, I think so...

John Rennie

Your lecturer is just solving the equation using this method, though his notation is different from mine.

user228700

Right :-/

John Rennie

And what he's getting out is the frequency of the oscillation and the decay constant for the damping - effectively my constants $b$ and $a$.

user228700

Right. right.

user228700

5:21 PM

Thanks so so so much. Really. I will revise this in the morning one more time to make sure I understand it a little better than I do right now.

John Rennie

OK it's about the time I usually settle into my armchair anyway :-)

user228700

Or, I'll come ask in the evening first.

user228700

:-) OK. Have a nice one. Thanks so much! I'll see you tomorrow, same time?

John Rennie

Like I say, I'm generally around so just ping me.

user228700

Thanks so much! My brain is a little fried so I'm sorry if I sucked today :-/ Goodnight :-)

user228700

5:23 PM

OK, will do :-)

John Rennie

See you tomorrow. Bye.

John Rennie

5:40 PM

@Kaumudi.H: for tomorrow: I worked through putting my guess equation into the DSHO equation (just to check I could actually do it :-) and actually it's only a few lines. If you want I can go though it with you on Thursday.

Transcript for

♦ In At The Deep End