Problem Solving Strategies

@JohnRennie Can you link the guide to install and use Mathjax for windows PC?

I'm not able to find it

If you use Chrome there's an addon for it that is much simpler ...

Oh, but I use edge, any idea how to enable it in edge?

A: Any chance of MathJax in chat?

6:31 AM

31

As a workaround while this request is pending, there exist several client-side workarounds that can be used to enable LaTeX rendering in chat, including: ChatJax, a set of bookmarklets by robjohn to enable dynamic MathJax support in chat. Commonly used in the Mathematics chat room. An altern...

That's what I used to use, but I started using Chrome and the addon instead.

Yeah, but I stopped using chrome after I started using a laptop. Edge is better for laptops and consumes less battery.

@JohnRennie Btw, there should be a square root here right?

Oops yes, $v = \sqrt{T/\rho}$

6:49 AM

@JohnRennie Hi

Sorry it took a while

Hi

I switched to mozilla for using chatjax

We can continue now

I've now forgotten where we got to ...

25 mins ago, by John Rennie

Now suppose halfway down the rope the linear density of the rope increases, so this means the wave velocity decreases. The point of this is it's like the rope has entered a higher refractive index region, since higher refractive index means the wave velocity decreases.

24 mins ago, by John Rennie

Since $v = f\lambda$ this means in the lower velocity region either the frequency, or the wavelength or both have to decrease.

OK so far?

@JohnRennie yes, okay.

But consider a point on the rope just before the density change, and a point on the rope just after the density change.

We can take these two points to be arbitrarily close to the point where the density changes.

These two points have to be moving at the same frequency because they're adjacent points on a rope.

And the rope is continuous.

Does this all make sense so far?

6:54 AM

Ah yes, makes sense

The same argument applies to the electric and magnetic fields at the boundary between the two media, where the refractive index changes. The electric and magnetic fields have to be continuous at the boundary.

Just like the rope.

yes, makes sense. Thank you!

:-)

7:29 AM

@JohnRennie hello sir

@user8718165 hi

@JohnRennie, Hi sir :)

Sir, I have a doubt in this answer - "...however this force is felt after the weight gets cancelled out by the normal contact force." How can this be possible sir. Could you please explain if possible?

@Intellex hi :-)

@JohnRennie Before that doubt, should I want to include nose in :) ?

like this - :-) ?

@JohnRennie Sir please help! I'm frequently getting stuck as I proceed :( :(

7:40 AM

@Intellex I think this is what the question means - suppose the ball is just sitting stationary on the floor. It experiences an upwards force of $mg$ from the floor due to its weight. Yes?

@Intellex that way, you'll have to leave the shift key for a while and that's a bit of work when you want to write fast. Otherwise go with it :-)

@JohnRennie Yes sir. But when it's in contact with the floor, the floor always exerts a force on the ball and the ball experiences it all the time right sir?

But acc. to this statement "...however this force is felt after the weight gets cancelled out by the normal contact force." the force is felt only after some time after contact.

The point is that the constant upwards force exerted by the floor isn't changing the momentum of the ball.

@JohnRennie First of all, I think the force exerted by the ground is not constant sir.

@Intellex I'm specifically considering the case where the ball is stationary on the floor.

My point is that if the total upwards force is $F$ then the momentum change is not proportional to $F$ but to $F - mg$:

$$ \frac{dp}{dt} = F - mg $$

7:45 AM

@JohnRennie Ok sir. Understood till this point sir.

The answer is just defining $F_1 = F - mg$ so that $\frac{dp}{dt} = F_1$

You can think of $F_1$ as the excess force due to the motion of the ball.

@JohnRennie Thank you sir, here is $F$ a constant or variable? Usually I see a somewhat downward opening parabola for impulsive forces.

Q: Does juggling balls reduce the total weight of the juggler and balls?

$F$ is variable. When the ball is not in contact with th floor $F=0$ and when the ball is in contact with the floor $F = \frac{dp}{dt} + mg$

The point of the question is to find the average value of $F$.

It's actually the same question as this one that I have answered:

87

A friend offered me a brain teaser to which the solution involves a $195$ pound man juggling two $3$-pound balls to traverse a bridge having a maximum capacity of only $200$ pounds. He explained that since the man only ever holds one $3$-pound object at a time, the maximum combined weight at any ...

newtonian-mechanics forces newtonian-gravity

@JohnRennie I understand that sir. I'm just considering the duration of time when the ball is in contact with the floor. Will it have a continuous change from 0 to $F$ or discontinuously jump?

@JohnRennie Thank you sir. Let me read that answer and ask if I have any doubts.

For a real ball the change is continuous because when the surface of the ball first touches the floor it is easily compressed. Imaging dropping a spring onto the floor. The force in the spring is $F=kx$ where $x$ is the compression.

When the spring first touches the floor the compression is zero so the force is zero.

Then as the spring compresses the force increases to a maximum, then decreases again as the spring rebounds.

7:54 AM

@JohnRennie Fine sir. But, in elastic collision, at least in my book, it's explained in terms of springs between two colliding blocks. So, why is it a discontinuous change here but a continuous one for a real ball?

I think both real and ideal elastic collisions are continuous changes.

In real life the change is never discontinuous.

@JohnRennie But in ideal cases as above, it is discontinuous, am I right sir?

We often use simplified models where the change is discontinuous to make the calculation easier, and in many cases the change is fast enough that to describe it as discontinuous is not a bad approximation.

@JohnRennie Ok sir. Thank you. If that's the average force asked in the question then as per the OP, the force stays $0$ most of the time and hence must add to zero as time tends to infinity, am I right sir?

@user8718165 Thanks!

Since the motion is periodic we don't need t average it for an infinite time. We can just average over one period. The periodicity means the average over all the other periods will be the same. Yes?

7:58 AM

@JohnRennie Yes sir.

The force is zero for part, but not all of the period, and it is non-zero for the remainder of the period. So when we average over the whole period we get a non-zero average.

@JohnRennie The impulsive force acts on the ball only for a short duration of time. Most of the time the force is zero. I agree the avg. force must be non zero. But how can it have such a large value equal to $mg$ sir?

Look at it this way: what is the net momentum change of the ball over one period?

@JohnRennie $2mv$ where $v$ is the velocity of impact or rebound sir.

The motion is periodic, so the velocity of the ball has to be the same at the beginning and end of the period. Yes?

8:02 AM

@JohnRennie Yes

For example we could take the period to start and end when the ball is at the apex of its trajectory.

@JohnRennie Yes sir.

At this point the velocity of the ball is zero. So the velocity at the beginning and end of the period is zero.

@JohnRennie Understood till this sir.

Or we could take the period to start at the instant the ball contacts the floor. In that case the velocity would be given by something like $v^2 = 2gh$, where $h$ is the height of the bounce.

But once again the velocity is the same at the start and end of the period.

So what is the net momentum change over one period?

8:05 AM

$0$?

Correct! :-)

So what is the net force on the ball averaged over one period?

@JohnRennie Again $0$ sir as $F=dP/dt$

Correct again!

And there is a constant downwards force of $mg$ due to gravity, so the average downwards force is $mg$. Can you see where I'm going with this?

@JohnRennie That is the answer! But why have you neglected the force exerted by the ground sir?

I haven't neglected the force exerted by the ground.

We are agreed there is an average downwards force of $mg$.

And we are agreed the average net force is zero.

Yes?

8:11 AM

@JohnRennie No sir

@JohnRennie Yes sir.

@Intellex no?

@JohnRennie No for "We are agreed there is an average downwards force of $mg$" alone sir.

@JohnRennie I'm sorry because I'm confused

There are only two forces acting:
1. the downwards gravitational force $mg$
2. the upwards force due to the floor

Agree sir.

And the downwards force is a constant $mg$, so the average value of the downwards force is just $mg$. Yes?

8:15 AM

@JohnRennie Average value of downward force is $mg$ - I agree, but how to determine average value of net force sir which is non zero?

The average value of the net force has to be zero, otherwise there would be a net momentum change over one period.

@JohnRennie sir, I want to ask a question

@JohnRennie Ok sir. Now I understood. [Average value of downward force is $mg$. Since average net force is $0$, the average upward force is $mg$ and this solved my doubt. Am I right sir?]

BOOM! :-)

@JohnRennie Thank you very much sir :)

8:18 AM

@user8718165 good timing, we've just finished with bouncing balls.

@JohnRennie Ha ha :) That really bounced a lot causing a big trouble in my brain!

@JohnRennie okay sir...I badly got stuck...

@user8718165 OK ... ?

@JohnRennie sir, why does heat flow across finite heat difference generate entropy?

It's a non-equilibrium process so it not reversible. That means $dS > dq/T$

8:20 AM

@JohnRennie Its given that because its irreversible process that's why...but where is the more entropy being generated?

Exactly where the excess entropy is coming from will depend on the microscopic details of the system.

@JohnRennie yeah sir...its written...but where actually is the disorder increases?

@JohnRennie okay sir...can you please tell ?

@JohnRennie hello sir :(

Offhand I don't know. It's not something I have ever thought about.

@JohnRennie sir please...I mean can you give just a very rough example? Nobody else will answer :(

In general relating the macroscopic quantities used in thermodynamics to microscopic properties is hard.

8:37 AM

@JohnRennie okay sir...another qn

8:58 AM

@JohnRennie sir I want to ask you another question

@user8718165 hi

Sorry, I was doing something at work.

10:22 AM

This question has been bothering me for quite some time; What is entropy?

@BaiduryaMathaddict A measure of randomness

@BaiduryaMathaddict entropy has a precise definition, but that precise definition is a complicated one given by statistical mechanics.

@JohnRennie Sir, is that not about randomness then?

Boltzmann's entropy formula

It's given Boltzmann's equation:

In statistical mechanics, Boltzmann's equation is a probability equation relating the entropy S of an ideal gas to the quantity W, the number of real microstates corresponding to the gas' macrostate: where kB is the Boltzmann constant (also written as simply k) and equal to 1.38065 × 10−23 J/K. In short, the Boltzmann formula shows the relationship between entropy and the number of ways the atoms or molecules of a thermodynamic system can be arranged. == History == The equation was originally formulated by Ludwig Boltzmann between 1872 and 1875, but later put into its current form by Max Planck...

@BaiduryaMathaddict OK so far?

10:37 AM

@JohnRennie, Sir, may I type our today's discussion on bouncing balls :) as an answer there? I think it will be nice to have a non-mathematical answer there.

Yes, go ahead.

Thank you sir.

10:58 AM

@JohnRennie hello sir sorry I didn't respond to you

@user8718165 hi

@JohnRennie, I successfully wrote my first answer! If time permits, could you please see my answer?

@Intellex looks good :-)

One typo: in the last sentence it's should be its.

@JohnRennie Thank you very much sir :)

@JohnRennie Fixed it. Thank you.

:-)

11:06 AM

@JohnRennie sir, entropy increases in an isothermal reversible process..right?

In any reversible process the total entropy change is zero.

If we consider a reversible isothermal process then the entropy of the gas will change and the entropy of the surroundings will change by an equal and opposite amount. This makes the total change zero.

@JohnRennie oh sorry sir...in a non reversible process...the reservoir loses S while the working fluid gains S..is it sir?

@JohnRennie yeah sir...got it ...sorry :)

In a non-reversible process the entropy changes of the system and its surroundings are not equal and opposite so now the total entropy change is greater than zero.

@JohnRennie okay sir...got it

@JohnRennie could you kindly explain what entropy is, in simpler terms?

I thought I understood a bit of Wikipedia articles nowadays... And then you provided me the link to this statistical-mechanics article

11:12 AM

@BaiduryaMathaddict the problem is that the only accurate description is the one provided by Boltzmann's equation, and that is formulated using statistical mechanics which is hard.

@JohnRennie sir, I have another question

@JohnRennie well, what is statistical mechanics

@BaiduryaMathaddict So any other description of what entropy is can only be a rough guide and not a strictly correct description.

People often say that entropy is a measure of disorder, but it isn't really. It's true that disordered states generally have a higher entropy than ordered states, but this is not a definition of entropy.

@JohnRennie I understand. I do agree that comprehending a precise definition is the most useful thing to do, but considering my little knowledge of precise term, I think a tough intuition at first would do a lot of good!

So if you're asking for an explanation of entropy that is (1) accurate and (2) simple, then I have to say no such explanation of entropy exists.

11:16 AM

@JohnRennie I asked my teacher this exact thing - why is dq(rev)/dT=S? And she replied in two parts

1. It's the way nature is

2. That's higher knowledge, revealed through experiments and stuff...

@BaiduryaMathaddict that's called the Clausius inequality:

Clausius theorem

The Clausius theorem (1855) states that a system exchanging heat with external reservoirs and undergoing a cyclic process, is one that ultimately returns a system to its original state, ∮ ⁡ δ Q T s u r r ≤ 0 ,...

Could you suggest anything I should rather to gather a preliminary knowhow of this stuff?

The Wikipedia article does include a proof but it's not a trivial one.

*read

@JohnRennie actually I find the "Wikipedia language" as I call it, a bit hard

Any thermodynamics textbook will discuss the Clausius inequality.

Offhand I can't recommend a book as I'm out of touch with the modern literature.

11:21 AM

Thermodynamics textbook.. like? Well as for now all I'm reading is high school chemistry.

Do you have books in mind that might be old but helpful nevertheless?

If you're still at high school you just have to accept the equations without proof, as the proof requires degree level thermodynamics.

Umm, that's what I do really, accept!

The nice thing about continuing in physics is that you get to understand all the things that you had to take on faith before, but it takes a long time. Years! :-)

That's so very true. I've had a tad bit of realisations myself...

Mugging kinematics equations...

And then seeing them all come alive with time!

Well, I'll search for some degree level books online... And give them a shot maybe...? Trying isn't wrong I guess!

11:52 AM

@JohnRennie hello sir

@user8718165 hi

@JohnRennie sir why does entropy of the system remain constant in reversible adiabatic?

@user8718165 you should be able to answer this.

If it's reversible then $dS = dq/T$. Yes?

@JohnRennie by Clausius ineq sir?

Yes

11:58 AM

@JohnRennie Sir, suppose its adiabatic compression (rev) then initially its system volume will be more and its cold...finally, its volume is less while it gets hotter...

@JohnRennie but still entropy is 0

Yes. As a rough guide entropy increases with increasing volume and with increasing temperature.

So for an adiabatic compression the reduction in volume decreases the entropy and the increase in temperature increases it.

@JohnRennie yeah sir...so if we decrease volume and increase temp...what will happens sir?

@JohnRennie do the effects necessarily cancel sir?

@user8718165 for an adiabatic process the two effects necessarily cancel.

@JohnRennie okay sir...can you please tell me a proof?

The proof is simply that $dq=0$ and therefore $dS=0$.

You could manually calculate the entropy change during the process but offhand I don't know how to do it.

12:03 PM

@JohnRennie yeah sir...its written...but sir can we show that $\Delta S$ (due to increased temp)=$\Delta S$ (due to decreased volume)

I'm sure you can, but I don't know how to do that calculation and I'm insufficiently interested in thermodynamics to find out how to do it.

@JohnRennie okay sir...please sir, can you just give me a hint?

The derivation is here

But that derivation relies on the equality $dS = dq/T$ so it's just a long and complicated way of coming up with the same result.

Their final equation is:

@JohnRennie thank you very much sir

@JohnRennie okay sir

@JohnRennie I'll surely have a look

$$ \frac{\Delta S}{C_v} = \ln\left(\frac{T_2}{T_1}\right) + (\gamma -1 )\ln\left(\frac{V_2}{V_1}\right) $$

And you can use the adiabatic formula $TV^{\gamma-1} = constant$

12:14 PM

@JohnRennie got it sir those terms will cancel...

@JohnRennie yeah sir

@JohnRennie sir could we multiply both sides by $C_v$?

@user8718165 Yes

@JohnRennie $$ \Delta S= \left(\ln\left(\frac{T_2}{T_1}\right) + (\gamma -1 )\ln\left(\frac{V_2}{V_1}\right)\right)C_v $$

Yes.

@JohnRennie sir can it be used for irreversible adiabatic also? I think yes

The derivation relies on the Clausius equality $dS = dq/T$ and this is not true for an irreversble process.

So this equation will not apply to irreversible processes.

12:20 PM

@JohnRennie okay sir...sorry. Got it

@JohnRennie sir I have a qn

Yes?

@JohnRennie Sir, why is thermodynamics SO difficult? 😥

Thermodynamics isn't really that difficult, but it has a steep learning curve.

Statistical mechanics is far more difficult.

With a subject like mechanics you need only learn a few basic principles and you can immediately start solving problems.

@JohnRennie Yeah sir...you told me :(

With thermodynamics you need to learn a whole lot of stuff before you can start doing anything useful.

12:26 PM

@JohnRennie okay sir..

That makes thrmodynamics really hard at first because there are loads of quantities and equations that you don't know.

@JohnRennie okay sir...got it...another qn sir?

Is it quick? I need to go soon.

@JohnRennie yeah sir...

OK ...

12:29 PM

@JohnRennie sir that page you told me is from MIT...so is it that they teach these stuff which I don't know about (or atleast not told about) to their students in great detail?

If you do a physics degree you will learn thermodynamics in great detail :-)

@JohnRennie These aren't written in the book :(

@JohnRennie Sir I'll take physics rather than computer in college :) Physics is very good :)

@JohnRennie I changed my mind :)

I must admit I never really liked thermodynamics and I did the minimum necessary to pass the exams.

That's why I now have to Google the answer whenever you lot ask me about thermodynamics :-)

@JohnRennie Correct sir...Its so complex :(

The mathematics involved isn't hard. It's just partial differential equations. It's just that there is so much detail you have to know.

12:34 PM

@JohnRennie yeah sir...just getting the derivations and the math sometimes don't help in understanding :)

@JohnRennie sir will you return in the evening?

I'll be around later, though my enthusiasm for answering complicated questions will be limited.

@JohnRennie You're soo good. Thanks a lot for helping me this much :-)

:-)

Now I'm off. Bye.

@JohnRennie Goodbye sir...have lunch nicely :)

4:36 PM

Hello sir :) @JohnRennie

@user8718165 hi :-)

@JohnRennie yeah :-)

Hello @AdvilSell 😊 How are you?

@user8718165 Hey @user8718165 I am Fine , how bout you ?

@AdvilSell Hello! Great to know you're doing fine. I'm fine too :) What branch are you pursuing? :)

@user8718165 Currently Biotechnology

I am hoping of getting into an IISER by next year

How's your prep going ?

4:41 PM

@AdvilSell That's pretty GREAT :) (Just casually asking) Any plans of changing branch?

@user8718165 Nah

I love Biotech

@AdvilSell I'm forgetting class 11 stuff :( So I have to practise more during free time :(

@AdvilSell GREAT :)

@AdvilSell Which college are you in now :)

@user8718165 Hm...pretty normal , my advice would be start solving papers as soon as possible

@user8718165 NIT Rourkela

@AdvilSell I tried math and chem...I was blown up...Didn't dare to touch Phy. I'm attempting them individually. Looking forward to solve more prev-years.

@AdvilSell That's a great NIT I (everyone) would say :)

@user8718165 indeed , But not a great place to learn anything new , Primarily the focus here are grades , though many people are doing great researches here but these are not accessible to firstt years , and there is no overall environment for science here

@user8718165 refrain yourself from doing this , solve all 3 together in one seating , This will improve your exam aptitude

4:54 PM

@AdvilSell actually that's the point. Most of the NITs and a few IITs are not good places for performing research. Rest of the IITs and IISERs are much better. I shouldn't really comment cuz as of now, I didn't even get into one of these.

@user8718165 yeah , probably

@AdvilSell Yes...I will should try them together. I tried math and chem at once many times. Most of the times...Chem takes the least time, relatively...however I could somehow solve at most 5 from math and time runs out... I will begin rigorous practice from Dec.

@user8718165 that's Great :-)

@AdvilSell As of now, my Phy is weak as compared to the other two. I should admit I didn't do much in class11. I still regret :(

@user8718165 That's fine you have time

5:01 PM

@AdvilSell Well...Thanks a lot for your advice :) You're in college...have fun ;)

@AdvilSell yeah I just have a month and a few days :(

@user8718165 I am sorry to break the reality but life's tough here also , and probably be tough for it's existence

Okay , Anyways bye I got to go now !!

@AdvilSell yeah...most of the people you'll come across are well...tough. Good people are hard to find :)

@AdvilSell yeah...cya :) Good Night

@user8718165 Night !

@JohnRennie Hello sir :) Did you have your lunch?

@user8718165 hi, yes, lunch is finished.

5:09 PM

@JohnRennie sir I have a very easy question about PE. It went unnoticed for all these days. I want to tell you sir. For how long will you be there?

For a few minutes more. What's the question?

@JohnRennie sir let's assume the body has no KE and we raise it to height $h$. At each point the net force is 0 so work done is also 0. But at the end there is PE stored...

@JohnRennie I saw David Z's answer. it was good but I have a few more confusions sir.

OK ... ?

@JohnRennie sir...isn't the net work 0 in the process? Because no net force was acting on the object sir?

If the net force is zero the acceleration is zero and you can't move the object.

This is kind of like the reversible process in thermodynamics.

5:16 PM

@JohnRennie yeah sir...but for raising a body without accln... can't it be uniform velocity sir?

Let's talk about this tomorrow.

@JohnRennie okay sir...you're very tired...I got it. See you tomorrow sir :) Good night