## Rabu, 26 Januari 2011

### PUSTAKA FISIKA (PF)

1. Sebuah Visi Pengumpulan 100.000 Buah Buku yang terkait dg Fisika
2. Pengumpulan Data-Data Kefisikaan sebesar 1 Terra byte

Tempat Pengumpulan dan Pendataan Buku-buku fisika via Internet

Fisika Komputasi

1. Numerical Methods and Modeling fro Chemical Engineers
2. Excel for Scientists and Engineers
3. Nonlinear Finite Element Methods
4. Introductory Computational Physics
5. Natural Language Processing with Python
6. Numerical Methods Using Mathlab
7. Numerical Methods in Engineering with Mathlab
8. Computational Electrodynamics: FDTD domain Method 2nd edition
9. Computational Physics Fortran Version
10. The Finite Difference Time Domain Method for Electromagnetics

Sumber:
FISIKA FOREVERMORE
Media Saling Berbagi Ilmu dan Informasi

### Fisika untuk Universitas

Fisika untuk Universitas

Ditujukan untuk meningkatkan kualitas proses dan hasil perkuliahan Fisika di tingkat Universitas

# 13: Equation of Motion for Simple Harmonic Oscillators

All right...

long weekend ahead of us.

One more lecture to go.

If I have an object, mass m, in gravitational field, gravitational force is in this direction, if this is my increasing value of y, then this force, vectorially written, equals minus mg y roof.

Since this is a one-dimensional problem, we will often simply write F equals minus mg.

This minus sign is important because that's the increasing value of y.

If this level here is y equals zero, then I could call this gravitational potential energy zero.

And this is y...

Then the gravitational potential energy here equals plus mg y.

This is u.

So if I make a plot of the gravitational potential energy as a function of y, then I would get a straight line.

This is zero.

So this equals u...

equals mg y, plus sign.

If I'm here at point A and I move that object to point B, I, Walter Lewin, move it, I have to do positive work.

Notice that the gravitational potential energy increases.

If I do positive work, the gravity is doing negative work.

If I go from A to some other point--

call it B prime--

then I do negative work.

Notice the gravitational potential energy goes down.

If I do negative work, then gravity is doing positive work.

I could have chosen my zero point of potential energy anywhere I please.

I could have chosen it right here and nothing would change other than that I offset the zero point of my potential energy.

But again, if I go from A to B, the gravitational potential energy increases by exactly the same amount--

I have to do exactly the same work.

So you are free to choose, when you are near Earth, where you choose your zero.

Now we take the situation whereby we are not so close to the Earth.

Here is the Earth itself.

Of course you can also replace that by the sun if you want to.

And this is increasing value of r.

The distance between here and this object m equals r.

I now know that there is a gravitational force on this object--

Newton's Universal Law of Gravity--

and that gravitational force equals minus m M-Earth G divided by this r squared, r roof so this is a vectorial notation.

Since it is really one-dimensional, we would...

Just like we did there, we would delete the arrow and we would delete the unit vector in the positive r direction and so we would simply write it this way.

The gravitational potential energy we derived last time equals minus m M-Earth G divided by r--

and notice, here is r and here is r squared--

and if you plot that, then the plot goes sort of like this.

This is r, this is increasing potential energy--

all these values here are negative--

and you get a curve which is sort of like this.

This is proportional to one over r.

Now, of course, if the Earth had a radius which is this big, then, of course, this curve does not exist, it stops right here.

If I move from point A to point B, with a mass m in my hand, notice that the gravitational potential energy increases.

I have to do positive work, there is no difference.

If I go from A to another point, B prime, which is closer to the Earth, notice that the gravitational potential energy decreases.

I do negative work.

If I do positive work, gravity is doing negative work.

If I do negative work, gravity is doing positive work.

Ucapan Terima Kasih Kepada:

1. Para Dosen MIT di Departemen Fisika

a. Prof. Walter Lewin, Ph.D.

b. Prof. Bernd Surrow, Ph.D.

2. Para Dosen Pendidikan Fisika, FPMIPA, Universitas Pendidikan Indonesia.

Terima Kasih Semoga Bermanfaat dan mohon Maaf apabila ada kesalahan.

## Senin, 24 Januari 2011

### Fisika Komputasi

"Dengan menggunakan komputer, kita bisa menghitung suatu sistem fisis yang mendekati kenyataan"

Fisika bukan hanya melulu berurusan dengan rumus. Bukan hanya dengan perhitungan yang rumit dan terkadang membuat anak-anak SMA merasa kesulitan mempelajari Fisika. Fisika semestinya dipandang sebagai suatu ide tentang suatu kejadian fisis sehari-hari yang kita alami setiap hari, dan bukan ribetnya rumus dan perhitungannya. Fisika memiliki suatu cabang keilmuan (bisa dikatakan demikian) yang memanfaatkan suatu tools yang dapat dimanfaatkan untuk membuat perhitungan menjadi lebih mudah dan cepat.

Tools itu adalah komputer dan cabang dari Fisika itu adalah Fisika Komputasi. Komputer dapat dipandang kini bukan hanya untuk mengolah data praktikum atau membuat dokumen ilmiah, namun bisa digunakan untuk menghitung suatu perhitungan yang rumit, yang sulit (bahkan mustahil) diselesaikan dengan tangan (secara analitik).

Komputer dapat melakukan perhitungan dengan lebih cepat dibandingkan manusia. Secepat-cepatnya manusia menghitung, komputer akan selalu lebih cepat. Dengan demikian, para fisikawan dapat lebih berkonsentrasi pada konsep dan ide yang lebih besar dan menyerahkan perhitungan kepada komputer.

Computational physics is the study and implementation of numerical algorithms to solve problems in physics for which a quantitative theory already exists. It is often regarded as a subdiscipline of theoretical physics but some consider it an intermediate branch between theoretical and experimental physics. It is a subset of computational science (or "scientific computing"), which covers all of science rather than just physics.

Physicists often have a very precise mathematical theory describing how a system will behave. Unfortunately, it is often the case that solving the theory's equations ab initio in order to produce a useful prediction is not practical. This is especially true with quantum mechanics, where only a handful of simple models admit closed-form, analytic solutions. In cases where the equations can only be solved approximately, computational methods are often used.

## Applications of computational physics

Computation now represents an essential component of modern research in accelerator physics, astrophysics, fluid mechanics (computational fluid dynamics), lattice field theory/lattice gauge theory (especially lattice quantum chromodynamics), plasma physics (see plasma modeling), solid state physics and soft condensed matter physics. Computational solid state physics, for example, uses density functional theory to calculate properties of solids, a method similar to that used by chemists to study molecules. Molecular dynamics, which simulates the motion of interacting atoms and molecules, is an important technique in materials science, chemical physics and the modeling of biomolecules.

## Methods and algorithms

As these topics are explored, many more general numerical and mathematical problems are encountered in the process of calculating physical properties of the modeled systems. These include, but are not limited to
Computational physics also encompasses the tuning of the software or hardware structure to solve problems. Approaches to solving the problems are often very demanding in terms of processing power and/or memory requests.

Sumber:

1. Wikipedia
2. Christian Fredy Naa, M.Sc.

## Sabtu, 22 Januari 2011

### Fisika untuk Universitas

Fisika untuk Universitas

Ditujukan untuk meningkatkan kualitas proses dan hasil perkuliahan Fisika di tingkat Universitas

Kelistrikan dan Kemagnetan

Topics covered:

Inductance
RL Circuits
Magnetic Field Energy

Instructor/speaker: Prof. Walter Lewin

### Video

• iTunes U (MP4 - 109MB)
• Internet Archive (MP4 - 213MB)

Today, I will quantify the ability of a circuit to fight a magnetic flux that is produced by the circuits themselves.

If you have a circuit and you run a current through the circuit, then you create some magnetic fields, and if the currents are changing then the magnetic fields are changing.

And so there will be an induced EMF in that circuit that fights the change, and we express that in terms of a self-inductance: L, self-inductance, and the word self speaks for itself.

It's doing it to itself.

Magnetic flux that is produced by a circuit is always proportional to the current.

You double the current, the magnetic flux doubles.

And so it is the proportionality constant that we call L, that is the self-inductance, and so therefore the induced EMF equals minus d phi/dt, that is Faraday's Law.

And so that becomes minus L dI/dt.

L is only a matter of geometry.

L is not a function of the current itself.

I will calculate for you a very simple case of the self-inductance of a solenoid.

Let this be a solenoid and this is a closed circuit, and we run a current I through the solenoid, and the radius of these windings is little r.

Let's say there're N windings and the length of the solenoid is little l.

Perhaps you'll remember that we earlier derived, using Ampere's Law, that the magnetic field inside the solenoid is mu 0 times I times capital N divided by l.

This is the number of windings per meter.

If we attach an open surface to this closed loop, very difficult to imagine what that open surface looks like -- we discussed it many times -- inside this solenoid you have sort of a staircase-like of surface.

That magnetic field penetrates that surface N times because you have N loops.

And so the magnetic flux, phi of B, is simply the area by little r squared, which is the surface area of one loop, because I assume that the magnetic field is constant inside the solenoid, and I assume that it is 0 outside, which is a very good approximation.

So we get pi little r squared surface area of one loop, but we have N loops and then we have to multiply that by that constant magnetic field.

So we get an N squared because we have an N here, mu 0 I divided by L.

And this we call L times I.

That's our definition for self-inductance.

And so the self-inductance L is purely geometry.

It's pi little R squared, capital N squared divided by L times mu 0.

Let me check this.

Pi little r squared, I have a capital N squared, mu 0, that's correct, divided by little l.

And so we can calculate, for instance, what this self-inductance is for a solenoid that we have used in class several times.

We had one whereby we had 2800 windings.

R I think was something like 5 centimeters -- you have to work SI of course, be careful -- and we had a length, was 0.6 meters.

We had it several times out here, and if you substitute those numbers in there, you will find that the self-inductance of that solenoid is 0.1 in SI units and we call the SI units Henry, capital H.

It would be the same as volt-seconds per ampere, but no one would ever use that.

We call that Henry.

Every circuit has a finite value for the self-inductance, however small that may be.

Sometimes it's so small that we ignore it, but if you take a simple loop, a simple current, just one wire that goes around -- whether it is a rectangle or whether it is a circle it doesn't make any difference -- it always produces a magnetic field.

It always produces a magnetic flux through the surface, and so it always has a finite self-inductance.

Maybe only 9- nano Henrys, maybe only micro Henrys, but it's never 0.

And so now what I want to do is to show you the remarkable consequences of the presence of a self-inductor in a circuit, and I start very simple.

I have here a battery which has EMF V.

I have here a switch, and here are the self-inductor.

We always draw a self-inductor in a circuit with these coils, and we also have in series a resistor, which we always indicate with this, these teeth.

And I close this switch when there is no current running.

In other words, at time T equals 0 when I close the switch, there is no current.

When I close this switch the current wants to increase, but the self-inductance says uh-huh, uh-huh, take it easy, Lenz law, I don't like the change of such a current.

So the self-inductance is fighting the current that wants to go through it.

There comes a time that the self-inductance loses the fight, if you wait long enough, and then of course the current has reached a maximum volume, which you can find with Ohm's Law, because the self-inductance itself has no resistor.

Think of the self-inductance as made of super-conducting material.

There's no resistance.

And so without knowing much about physics, you can make a plot about the current that is going to flow as a function of time.

You start out with 0 and then ultimately, if you wait long enough, you reach a maximum current which is given by Ohm's Law, which is simply V divided by R.

And you slowly approach that value.

And how slowly depends on the value of the self-inductance.

If the self-inductance is very high, it might climb up like this, so this is a high value for L.

If the self-inductance is very low, that is a low value for L.

If the self-inductance were 0, it would come up instantaneously, but I just convinced you that there I no such thing as 0 self-inductance.

There's always something finite, no matter how small.

And so this is qualitatively what you would expect if you use your stomach and if you don't use your brains yet.

There's nothing wrong with using your stomach occasionally, but now I want to do this in a move civilized way, and I want to use my brains, and when I use my brains I have to set up an equation for this circuit.

And if you read your book, you will find that Mr. Giancoli tells you to use Kirchhoff's Loop Rule.

But Mr. Giancoli doesn't understand Faraday's Law, and he's not the only one.

Almost every college book that you read on physics do this wrong.

They advise you to use Kirchhoff's Loop Rule, which says that the closed loop integral around the circuit is 0.

That, of course, is utter nonsense.

How can it be 0?

Because there is a change in magnetic flux, and so it can only be minus d phi/dt.

I advise you to go to the 8.02 website and download a lecture supplement that you will find in which I address this issue and hit it very hard.

So the closed loop integral of E dot dL, if you go around the circuit, is not 0, is minus d phi/dt -- Faraday's Law -- so it's minus L dI/dt.

So we have to go around to circuit and we have to apply Faraday's law and not Kirchhoff's Loop Rule, which doesn't apply here.

This is the plus side of the battery and this is the minus side, so the electric field in the battery is in this direction.

The electric field in the self-inductance is 0 because the self-inductance has no resistance, it's super-conducting material, and so the electric field in the resistor -- if the current is in this direction, which it will be -- then the electric field in the resistor will be in this direction.

So now I am equipped to write down the closed loop integral of E dot dL.

I start here and I always go in the direction of the current, and I advise you to do the same.

I don't care that you guess the wrong direction for the current.

That's fine.

Later, minus signs will correct that, they will tell you that you really guessed the wrong direction, but always go around the loop in the direction of the current, because then the EMF is always minus L dI/dt.

If you go in the direction opposed to the current, then it is plus L dI/dt and that could become confusing.

So I always go in the direction of the current, and so I first go through the self-inductance.

There is no electric field in the self-inductance, so the integral E dot dL -- in going from here to here -- is 0.

This is where the books are wrong.

It is 0.

Now, I go through the resistor, and so now I get plus IR, E and dL are in the same direction.

Ohm's Law tells me it's IR.

In the battery, I go against the electric field, and so I get minus V.

That now equals minus L dI/dt, and this is the only thing and the only correct way to apply Faraday's Law in this circuit.

You can write it a little differently, which may give you some insight.

For instance, you could write that I can bring V and the L, and L, to one side -- so I can write down that V minus L dI/dt equals IR.

It's the same equation when you look, and the nice thing about writing it this way is that since dI/dt is positive here -- it's growing in time -- the induced EMF, which is this value -- notice it's always opposing the voltage of my battery -- and that's what Lenz's Law is all about.

It's not until dI/dt has become 0 that V equals IR, and that happens, of course, if you wait long enough.

And so we have to solve that differential equation, and what is often done that you bring all the terms to the left side and that you get an, a 0 on the right side.

And so what you often see is that L dI/dt plus IR minus V equals 0.

And because we have a 0 here, some physicist thinks that this is an application of Kirchhoff's Rule.

This is nonsense.

You can always make it 0 here by bringing all the terms to this side.

The closed loop integral of E dot dL is not 0, the closed loop integral of E dot dL is minus L dI/dt, but when I shift minus L dI/dt to this side I get 0 here.

And of course the people who write these books know that this is the right answer and so they manipulate it so they get this equation and they call that Kirchhoff's Rule.

Sad, and also embarrassing.

So, this is the equation that you have to solve.

Some of you may have solved this equation in 8.01 already.

Surely you didn't have an I here.

You may have had an X here for the position, but you probably solved it when you had friction.

Maybe you didn't.

I will give you the solution to that differential equation.

It's a very easy solution.

The current as a function of time is a maximum value times 1-e to the minus R divided by L times T, and I max -- that is the maximum current -- is V divided by R.

And let's look at this in a little bit more detail.

First, notice that when t equals 0 that indeed you find I equals 0.

Substitute in here t equals 0, you get 1-1.

So you find, indeed, that I equals 0.

Substituting there t goes to infinity, then you find that I indeed becomes V divided by R, which is exactly what you expect.

If t becomes infinity, then clearly the self-inductance has lost all its power, so to speak, and the current is simply V divided by R, the maximum current that you can have.

And so that's a must, that's a requirement.

If you wait L over R seconds -- and believe it or not, if you have some time, convince yourself that L over R indeed, as a unit, is seconds -- then the current I is about 63% of I max, because if t is L over R, then you get 1-1 divided by E, and that is about 0.63.

And if you wait double this time, then you have about 86% of the maximum current.

In other words, right here -- if I wait L over R seconds -- this value here is about 0.63 times the maximum value possible, and it's very, it's climbing up and asymptotically approaches then, ultimately, the maximum current which is V divided by R.

Make sure you download that lecture supplement that you'll find on the Web.

Now, what I'm going to do is all of a sudden I'm going to make this voltage 0.

The way I could do that is by simply shorting it out.

Of course on the blackboard, I can simply remove it.

So it's not there.

The current is still running and all of a sudden at a new time -- t 0, I define the time t 0 again -- the voltage is 0.

And now comes the question what is now going to happen?

Well, the self-inductance doesn't like the fact that the current is going down, so it's going to fight that change, and so you expect that the current is not going to die off right away, but you expect that the current is going to go down sort of like so.

And you want, when you wait long enough, you want that, at t goes to infinity, where t equals 0 the current is still maximum, so it's still V over R, but when you go to infinity -- if you wait long enough -- then, of course, the current has to become 0.

And as the current dies out, heat is being produced in that resistor at a rate of I square R joules per second, and then there comes a time that the current becomes almost 0 and then the whole show is over.

And so we can also calculate the exact time behavior by going back to our differential equation and make V 0.

Where is that differential equation?

Is it hiding?

Oh, there it is.

So I solved this differential equation, but this is now 0.

And the solution to that differential equation is that I as a function of time is I max times e to the minus R over L times t, and that exactly has all the quantities that you want it to have, because notice that at t equals 0 -- when you put in t equals 0 -- the current is indeed maximum, and that's what you require.

That's the moment that you make that capital V 0, the current was still running.

But notice that when t goes to infinity -- if you wait long enough -- that indeed the current goes to 0.

And if you wait L over R seconds, then you are down to about 37% of your maximum current.

So if you now go to I, I re-define t equals 0 here, so if now I wait L over R seconds then this value here is about 37% of that value.

So you've lost 63%.

And so you see, this is the consequence of the fact that the circuit is capable of fighting its own magnetic flux that it is creating.

When the current was running happily here, with the battery in place, the current was, let's say, all the time I max -- at least very close to that value -- V over R.

And so all the time, there was heat produced in the resistor.

I square R joules per second.

Who was providing that, uh, energy?

Well, of course, the battery.

But now, when I take the battery out, there is still current running, and that means while the current is dying there is still heat produced in that resistor, and that heat slowly comes out until the current ultimately becomes 0.

Now where does that energy come from?

Well that energy must come from the magnetic field that is present in the solenoid, and this idea -- that we have energy that comes out in the form of heat, which really was there ear-, earlier in the form of a magnetic field -- allows us to evaluate what we call the magnetic energy field density.

Let me first calculate how much heat is produced as the current goes from a maximum value down to 0.

Hmmmm, I'll have to erase something.

I'll erase this part here.

So at any moment in time, the current is producing heat in the resistor, and so if I, if my voltage becomes 0 at time t equals 0, then this is the amount of heat, uh-uh, no square here.

I square R dt, integrated from 0 to infinity, is the total heat that is produced as the current dies out, but I know what this current was -- I just erased it, if I still remember it -- so I can bring I max outside and I can bring the resistance outside and then I get the integral from 0 to infinity of e to the minus R over L times t dt.

And this is a trivial integral.

This integral is, uhm, L divided by 2 R.

Oh, by the way, it is I squared, so I have a 2 here.

It's very important.

Pengembangan Perkuliahan

1. Buatlah sebuah Esai mengenai materi perkuliahan ini

2. Buatlah sebuah kelompok berjumlah 5 orang untuk menganalisis materi perkuliahan ini

3. Lakukan Penelitian Sederhana dengan kelompok tersebut

4. Hasilkan sebuah produk yang dapat digunakan oleh masyarakat

5. Kembangkan produk tersebut dengan senantiasa meningkatkan kualitasnya

Ucapan Terima Kasih Kepada:

1. Para Dosen MIT di Departemen Fisika

a. Prof. Walter Lewin, Ph.D.

b. Prof. Bernd Surrow, Ph.D.

#### Staff

Visualizations:
Prof. John Belcher

Instructors:
Dr. Peter Dourmashkin
Prof. Bruce Knuteson
Prof. Gunther Roland
Prof. Bolek Wyslouch
Dr. Brian Wecht
Prof. Eric Katsavounidis
Prof. Robert Simcoe
Prof. Joseph Formaggio

Dr. Peter Dourmashkin
Prof. Robert Redwine

Technical Instructors:
Andy Neely
Matthew Strafuss

Course Material:
Dr. Peter Dourmashkin
Prof. Eric Hudson
Dr. Sen-Ben Liao

#### Acknowledgements

The TEAL project is supported by The Alex and Brit d'Arbeloff Fund for Excellence in MIT Education, MIT iCampus, the Davis Educational Foundation, the National Science Foundation, the Class of 1960 Endowment for Innovation in Education, the Class of 1951 Fund for Excellence in Education, the Class of 1955 Fund for Excellence in Teaching, and the Helena Foundation. Many people have contributed to the development of the course materials. (PDF)

2. Para Dosen Pendidikan Fisika, FPMIPA, Universitas Pendidikan Indonesia.

Terima Kasih Semoga Bermanfaat dan mohon Maaf apabila ada kesalahan.

## Selasa, 18 Januari 2011

### Fisika untuk Universitas

Fisika untuk Universitas

Ditujukan untuk meningkatkan kualitas proses dan hasil perkuliahan Fisika di tingkat Universitas

# 12: Resistive Forces

We're going to discuss today resistive forces and drag forces.

When you move an object through a medium, whether it's a gas or whether it's a liquid, it experiences a drag force.

This drag force depends on the shape of the object, the size of the object, the medium through which you move it and the speed of the object.

The medium is immediately obvious.

If it's air and you move through air, you feel the wind through your hair--

that's a drag force.

If you swim in water, you feel this drag force.

In oil, the drag force would be even larger.

This drag force, this resistive force is very, very different from the friction that we have discussed earlier when two surfaces move relative to each other.

There, the kinetic friction coefficient remains constant independent of the speed.

With the drag forces and the resistive forces, they are not at all independent of the speed.

In very general terms, the resistive force can be written as k1 times the velocity plus k2 times the velocity squared and always in the opposite direction of the velocity vector.

This v here is the speed, so all these signs--

k1, v and k2, and obviously v squared--

they all are positive values.

And the k values depend on the shape and the size of the object and on the kind of medium that I have.

Today I will restrict myself exclusively to spheres.

And when we deal with spheres, we're going to get that the force, the magnitude of the force--

so that's this part--

equals C1 times r times the speed plus C2 times r squared times v squared.

And again, it's always opposing the velocity vector.

C1 in our unit is kilograms per meters per second and C2 has the dimension of density kilogram per cubic meters.

We call this the viscous term, and we call this the pressure term.

The viscous term has to do with the stickiness of the medium.

If you take, for instance, liquids--

water and oil and tar--

there is a huge difference in stickiness.

Physicists also refer to that as viscosity.

If you have a high viscosity, it's very sticky, then this number, C1, will be very high.

So this we call the viscous term, and this we call the pressure term.

The C1 is a strong function of temperature.

We all know that if you take tar and you heat it that the viscosity goes down.

It is way more sticky when it is cold.

C2 is not very dependent on the temperature.

It's not so easy to see why this pressure term here has a v square.

Later in the course when we deal with transfer of momentum, we will understand why there is a v-square term there.

But the r square is very easy to see, because if you have a sphere and there is some fluid--

gas or liquid--

streaming onto it, then this has a cross-sectional area which is proportional to r squared, and so it's easy to see that the force that this object experiences--

we call it the pressure term--

is proportional to r square, so that's easy to see.

Two liquids with the very same density would have...

they could have very different values for C1.

They could differ by ten...

not ten, by four or five orders of magnitude.

But if they have the same density, the liquids, then the C2 is very much the same.

C2 is almost the density rho of the liquid--

not quite but almost--

but there is a very strong correlation between the C2 and the density.

If I drop an object and I just let it go, I take an object and I let it fall--

we're only dealing with spheres today--

then what you will see, I have a mass m, and so there is a force mg--

that is gravity.

Ucapan Terima Kasih Kepada:

1. Para Dosen MIT di Departemen Fisika

a. Prof. Walter Lewin, Ph.D.

b. Prof. Bernd Surrow, Ph.D.

2. Para Dosen Pendidikan Fisika, FPMIPA, Universitas Pendidikan Indonesia.

Terima Kasih Semoga Bermanfaat dan mohon Maaf apabila ada kesalahan.

## Minggu, 16 Januari 2011

### PUSTAKA FISIKA (PF)

1. Sebuah Visi Pengumpulan 100.000 Buah Buku yang terkait dengan Fisika

2. Pengumpulan Data-Data Kefisikaan sebesar 1 Terra byte Tempat Pengumpulan dan Pendataan Buku-buku fisika via Internet

Fisika Komputasi

• An Introduction to Parallel Programming
• Computational Physics
• Transport Phenomena In Porous Media III
• Fuzzy Logic with Engineering Applications
• Engineering with Mathcad
• Applied Numerical Methods Using Mathlab
• An Introduction to Programming and Numerical Methods in Mathlab
• Monte Carlo Methods for Applied Scientists
• Applied Numerical Methods with Mathlab
• Computer Simulation Methods in Theoritical Physics

• Sumber:
FISIKA FOREVERMORE
Media Saling Berbagi Ilmu dan Informasi

## Sabtu, 15 Januari 2011

### Fisika untuk Universitas

Fisika untuk Universitas

Ditujukan untuk meningkatkan kualitas proses dan hasil perkuliahan Fisika di tingkat Universitas

Kelistrikan dan Kemagnetan

Topics covered:

How do Magicians levitate women? (with demo)
Electric Shock Treatment (no demo)
Electrocardiogram (with demo)
Pacemakers
Superconductivity (with demo)
Levitating Bullet Trains
Aurora Borealis

Instructor/speaker: Prof. Walter Lewin

### Video

• iTunes U (MP4 - 104MB)
• Internet Archive (MP4 - 203MB)

You have ten days left for your motor, so that's a nice project for Spring Break.

I'll give you some hints.

Keep the friction of your rotor as low as you can.

You can't use any oil, of course; that's not allowed.

Balance your rotor to the best you can.

And try to avoid that the rotor begins to bounce, begins to vibrate, because when it vibrates it loses contact with the current when it needs it so there's no torque.

How will we test your motors?

We do it with a stroboscope, and I've decided to demonstrate to you how we're going to do that.

That's probably the best thing to do.

We here have a disk, and we're going to rotate the disk at 1000 RPM.

Let's assume that is your motor.

And we're going to strobe it with a strobe light until it stands still.

In this case, I have set the strobe so that it will stand still, roughly, and the strobe is now going at 500 RPM, and the motor is going at 1000 RPM.

So this clearly is not the rotation rate of your motor.

In fact, your motor goes twice around between the blinks.

And we'd have no way of knowing that, so we double the frequency.

I'm trying to double it now, double the frequency of the blinking of the strobe light.

And now it stands still again.

So now we may think that your motor is going 1000 RPM, but we don't know yet.

Maybe it's going 3000 R- 2000 RPM.

Maybe 3000 RPM.

So what are we going to do now, we're going to double the frequency.

And so we go now with the strobe light to 2000 RPM.

And what we see now is we see a double image.

So 2000 RPM is out, and any multiple of 2000 RPM is out.

So 4000 RPM is out, 6000, and 8000 is out.

But what is not yet out is 3000 and 5000 and 7000.

So we would have to test for that.

On the other hand, I told you already that this motor is going 1000 RPM, so there's no sense us testing that now.

But during the actual contest, of course, we will continue all the way until we are convinced that we have the right RPM for your motor.

And so that's the way we will do it.

We will put a little bit of white paint on one side of your rotor, so that's the way it will be done.

Of course, if your motor is highly unstable in terms of rotation rate, it will not be easy to get a right correct number.

I want to talk with you about the heart.

The heart, our heart has four chambers.

Looks sort of like this.

The left atrium and right atrium.

Maybe this is why it's -- this is why it's called the heart.

And here is the left and the ri- and the right ventricle.

And here is the aorta.

The sole purpose of the heart is to pump blood.

About 5 quarts per minute, which is 75 gallons per hour, which is 70 barrels per day, which is about 2 million barrels in 75 years.

And it pumps about 70 times per minute.

If the blood to your brain stops for about 5 seconds, you lose consciousness.

So it's five skips of the heartbeat, and you're down on the floor.

And four minutes later, permanent brain damage.

The way the heart works is absolutely mind-boggling.

Extremely complicated.

Nature had one billion years to design it, but nevertheless it's impressive.

Each heart cell is a mini chemical battery, and it pumps ions in or out as it pleases.

In the normal state, each heart cell is minus 80 millivolts on the inside relative to the outside.

There are some cells which are called pacemaker cells.

They are located in a very small area, about 1 square millimeter, near the atrium, the right atrium, and they change their potential from minus 80 millivolts to plus 20 millivolts.

Now why they do that is a different story, which I will not address.

Once they go to plus 20 millivolts, the neighboring cells follow, and a wave propagates over the heart.

I'll make you a drawing shortly.

So the wave first moves over the atrial chambers and then over the ventricle chambers.

And when the cells are at plus 20 millivolts inside relative to the outside, they contract.

So they form a muscle.

The whole heart is one big muscle.

And after about 2/10 of a second, the cells return to minus 80 millivolts, and this wave goes from below to above.

And then the whole thing waits again for another message from the pacemaker cells.

Takes about one second, and then the whole process starts all over.

Now I want to be more precise.

Here is one heart cell.

So this is about 10 microns in size.

And this cell has 80 millivolts with respect to the outside.

So that means it has repelled positive ions, and so the inside is negative.

And there is no E field here outside, because if you put a Gaussian surface around here, there is no net charge inside.

But there is, of course, a electric field across the walls here, from plus to minus.

Now the depolarization, which is the change to the plus 20 millivolt state starts, and it starts from above.

And I will assume now that it is not plus 20 but 0 millivolts, and it's easier to see.

If we have this cell, and the wave is, say, halfway down, and this is now 0 millivolts, then there is no longer minus charge here and no longer plus charge here, because 0 millivolts relative to the outside world.

So there is no electric field across here anymore.

In other words, what the cell has done, it has moved positive ions back in.

But here the situation is still as it was before, so this is still at your minus 80 millivolts.

And if you look now, you have here a minus layer on top of a positive layer.

Positive here, minus on top.

And that creates an electric field, which has roughly the shape of a electric dipole.

It has this shape.

So as the wave goes through the cells, only then do they create a dipole.

And we call this the depolarization.

A little later in time, when this wave has passed, the whole thing is plus 20 millivolts.

I chose 0 here, but it really goes to plus 20.

This is just easier to explain.

So that means that now the inside is plus, so positive ions are now inside, negative ions are outside, and the E field here is again 0.

Now there is the repolarization wave, which comes from below, when it goes back to minus 80 millivolts.

And I will again do the same trick that I did before; I will just assume the wave is halfway, that it is not minus 80 but that it is 0 millivolts.

So there are no charges here, but the charges here are unchanged.

So what do you have now here?

You have a minus layer on top of a plus layer.

So you have exactly what you had before.

So again you get an electric field, which is an electric dipole field, which has again the same shape.

So what's going to happen is the depolarization wave is going to run down, leaves behind here the cells at plus 20 millivolts, when they are contracted, so this part of the heart has already pumped, and it moves down.

And only the cells where the depolarization occurs, that's only the ones on the ring, contribute to that electric dipole field.

If there is no wave, which is a sizeable fraction of the heart, of the cycle, there is no wave, then there is no electric dipole field.

And when the repolarization goes in the other direction, when the heart relaxes because the cells go back to minus 80 millivolts, then again there is an electric dipole field, but only from the cells through which the repolarization wave moves.

And you can very easily see that the electric dipole fields of all these cells here support each other.

So you get a dipole field from the heart.

And so if I make you look at your heart -- so this is you, this is your body, your legs, and this is your arms, and here is your heart, and there goes this wave.

And so here is your electric field that is generated while the wave is going, either depolarization down or repolarization up.

But if there is an electric field, there's going to be a potential difference between different parts of your body.

You look here at your belly button, and you follow this electric field line at your head, there is an E field.

The integral E dot dL gives you a potential difference.

And so now you see that there're going to be potential differences between the various parts of your body.

And that's the idea behind an electrocardiogram.

Typically there are 12 electrodes attached to arms, legs, head, and chest to get as much information about the heart as we can.

And the maximum potential difference between two electrodes, in general, is not more than about 2 to 3 millivolts.

I'd like to show you a healthy heart cardiogram, of a healthy person.

I have that here.

The time here is about 1 second, and from here to here is about 1 millivolt.

The P wave -- we call this the P wave -- that is observed when the atrium is being depolarized, so when the depolarization wave goes over the atrium.

A little later it goes over the ventricle, and you get a larger potential difference because there is more muscle in the ventricle.

So that's why this R wave is higher.

The T wave is the repolarization, when the wave goes back over the ventricles.

The dipole field is in the same direction, remember.

That's the T wave.

It's not known, at least it wasn't known recent- until recently, what causes the U wave.

I talked to a heart expert about this, Professor Cohen at MIT, and I was surprised to learn that it's not known what the U wave is about.

Not everyone's cardiogram looks as healthy as this one.

There is a terrible disease, which 4000 people die of per year in the United States, which is known as ventricular fibrillation, also known as sudden death.

The ventricles fire without any message from the pace wave, pacemaker wave, and there is random, non-synchronous depolarization.

So the heart doesn't pump anymore, in 5 seconds you lose consciousness, on the floor, and in four minutes you, um, have permanent brain damage.

In hospitals, heart patients are being monitored, and as soon as it's noticed that there is something wrong like this, so severe as the fibrillation, ventricular fibrillation, then they apply electric shock treatment.

So you have to be fast, you only have a few minutes before you get brain damage.

And 3000 volts is applied, 1 amperes, for about a tenth of a second.

Large plates are being used on each side of the chest.

And this, of course, is enough to kill the patient.

But it makes little difference, because the patient would have died anyhow.

Heart patients can also get synchronization problems and then they implant a pacemaker, that's a circuit.

And this pacemaker takes over the role from the pacemaker cells.

When the heartbeat rate falls below a certain rate, the artificial pacemaker takes over.

About 10 milliamperes for half a millisecond, and it does it 60 times per minute.

And so it triggers, then, the depolarization wave.

These pacemakers are susceptible to influences from the outside world, and one person's pacemaker stopped, for instance, every 10 seconds due to a radar sweep from a police car.

It's also possible that you get a built-in defibrillator, in other words, a system that gives you electric shocks when sudden death might otherwise occur.

So it senses that something is wrong, that the ventricle is going into fibrillation, and then it applies, all by itself, 650 volts, about 5.5 milliseconds, up to 5 to 10 amperes.

And that's not enough to kill the patient, and the whole idea is sort of a wake-up call to the heart to get it back into synchronization, to get this depolarization wave being synchronized again.

So clearly I would like to show now a heart cardiogram of a student, and I prefer to have a healthy one to avoid some difficulties.

You feel strong?

You a healthy person?

You don't mind volunteering?

Tight pants, we have to do something about.

OK, why don't you sit down.

[laughter].

Well, there's nothing I -- come in.

We'll, we'll, we will, we'll, we'll find a way.

All right, so we have to attach -- we don't have twelve electrodes, we only use three.

And the first one -- that's why I was worried about your tight pants.

Can you roll them up a little?

OK.

Oh, this one goes here.

Let's hope that it makes good contact.

Now the others go on your arm, and we need very good electrical contact, and therefore we put some conducting grease on there.

It will make it a little -- it will make it a bit of a mess, but we'll give you a chance later to clean up.

So let's first put this one -- you're relaxed, right?

Yes, of course.

So can you roll up your sleeve there?

Very good.

And mayb- oh, oh, and maybe you can put this over your arm, yeah, over your -- yeah, that's good.

High up.

Oh man, boy, you have muscles.

[laughter].

Pengembangan Perkuliahan

1. Buatlah sebuah Esai mengenai materi perkuliahan ini

2. Buatlah sebuah kelompok berjumlah 5 orang untuk menganalisis materi perkuliahan ini

3. Lakukan Penelitian Sederhana dengan kelompok tersebut

4. Hasilkan sebuah produk yang dapat digunakan oleh masyarakat

5. Kembangkan produk tersebut dengan senantiasa meningkatkan kualitasnya

Ucapan Terima Kasih Kepada:

1. Para Dosen MIT di Departemen Fisika

a. Prof. Walter Lewin, Ph.D.

b. Prof. Bernd Surrow, Ph.D.

#### Staff

Visualizations:
Prof. John Belcher

Instructors:
Dr. Peter Dourmashkin
Prof. Bruce Knuteson
Prof. Gunther Roland
Prof. Bolek Wyslouch
Dr. Brian Wecht
Prof. Eric Katsavounidis
Prof. Robert Simcoe
Prof. Joseph Formaggio

Dr. Peter Dourmashkin
Prof. Robert Redwine

Technical Instructors:
Andy Neely
Matthew Strafuss

Course Material:
Dr. Peter Dourmashkin
Prof. Eric Hudson
Dr. Sen-Ben Liao

#### Acknowledgements

The TEAL project is supported by The Alex and Brit d'Arbeloff Fund for Excellence in MIT Education, MIT iCampus, the Davis Educational Foundation, the National Science Foundation, the Class of 1960 Endowment for Innovation in Education, the Class of 1951 Fund for Excellence in Education, the Class of 1955 Fund for Excellence in Teaching, and the Helena Foundation. Many people have contributed to the development of the course materials. (PDF)

2. Para Dosen Pendidikan Fisika, FPMIPA, Universitas Pendidikan Indonesia.

Terima Kasih Semoga Bermanfaat dan mohon Maaf apabila ada kesalahan.

## Senin, 10 Januari 2011

### Fisika untuk Universitas

Fisika untuk Universitas

Ditujukan untuk meningkatkan kualitas proses dan hasil perkuliahan Fisika di tingkat Universitas

Kelistrikan dan Kemagnetan

Topics covered:

Displacement Current (Difficult Concept)
Synchronous Motors
Induction Motors
Secret Top, How does it work?

Instructor/speaker: Prof. Walter Lewin

### Video

• iTunes U (MP4 - 107MB)
• Internet Archive (MP4 - 210MB)

Today, I'm going to take a critical look at Ampere's Law.

I'm going to run a current through a wire, as we did before, but now I'm going to also put a capacitor in that line and so we are charging a capacitor.

Here is that capacitor.

And here is the wire.

We are running a current I.

And as we are running this current, clearly, we get a changing electric field inside the capacitor.

The electric field inside the capacitor, sigma free divided by kappa epsilon 0, which is also Q free divided by the area.

This is a circular plate capacitor.

Capital R, is the radius of this capacitor, so we get pi R squared kappa epsilon 0.

But since I run a current the Q free is building up all the time, and so the current per definition is dQ/dT, and so I now have ex- a changing electric field inside, dE/dt, which is the current I divided by pi R squared, kappa epsilon 0, because I simply take the derivative of this equation, I get dQ/dT, and dQ/dT is I.

And only if the current is 0 is there no changing electric field inside.

So how does this affect the magnetic field?

Well, if I take here a point P1 at a distance little r from the wire, if you're far away from this capacitor it's hard to believe that Ampere's Law would not give the right answer.

And we will apply that very shortly, Ampere's Law.

It's on the blackboard there.

Suppose you are at the same distance from this line here at point P2.

Well, yeah, you've got to admit there's an interruption of current now.

There is no current going through this space and so you expect that the magnetic field here would be a little lower perhaps than it is here.

But not very much.

So the question is, how can we now calculate the magnetic field here and there, now that we have this opening in the wire.

Well, Biot-Savart could handle it but I wouldn't know how to do it because if there's a current flowing like this there's also a current going up on these plates, and one like so, and I wouldn't know how to apply Biot-Savart.

In principle, yeah, but in practice, no.

How about Ampere's Law?

Well, let's give Ampere's Law a shot.

This is a cylindrical symmetric problem, so I choose a closed loop, which of course itself is a circle with radius R, and I apply -- I attach to this closed loop an open surface.

That's mandatory.

And I give myself an easy time, I make it a flat surface.

So now I apply Ampere's Law.

You see it there on the blackboard.

Anywhere on that closed loop, the magnetic field will have the same strength, for reasons of symmetry, and so we get B times 2 pi r equals mu 0 times I pen, and pen means the current that penetrates my open surface.

Well, that's I.

I goes right through that surface.

And so the magnetic field at that point, P1, mu 0 times I divided by 2 pi r.

We've seen this several times before.

Now I wonder about P2.

Can I apply Ampere's Law for point P2?

Well, yeah, you can try.

So now I attach a closed loop to this point.

Circle again, radius little r, and I use this flat surface and I apply Ampere's Law.

Well, I'm in for a shock, because B times 2 pi r is not changing but there is no current that penetrates that surface.

And so I is 0, and so I have to conclude that the magnetic field at point P2 is 0 which is absurd.

Couldn't be.

I can make the situation even worse.

I'm going to revisit point P1, and here is my capacitor, and here is my point P1.

My current is flowing like so.

Here's my closed loop.

According to Ampere's Law, int- closed loop integral B dot dL.

Why should I choose a flat surface?

I'm entitled to any surface! I like surfaces like this.

They are attached to a closed loop, so I will choose that kind of a surface.

The surface now goes like so.

[whistle] Right through the capacitor plates, and I apply Ampere's Law, it's open here.

B times 2 pi r, the radius is little r.

Mu 0 times I, but there is no I going through that surface.

Nowhere through this surface is a current poking, because there is no current going between the capacitor plates, so now I have to conclude that the magnetic field at P1, which we first concluded was this, is now also 0.

So something stinks.

So Ampere's Law is inadequate.

And so of course, Faraday and Ampere were both perfectly aware of this.

But yet it was Maxwell who zeroed in on this and he argued that any open surface that you attach to a closed loop should give you exactly the same result, same answer.

And so he suggested that we amend Ampere's Law, and so he asked himself the question, what is so special about in-between the capacitor plates?

Well, what is special there is in-between the capacitor plates there is a changing electric field.

And Maxwell reasoned, gee, Faraday's Law tells me that a changing magnetic flux gives rise to an electric field, so he says maybe a changing electric flux gives rise to a magnetic field.

And I want to remind you what an electric flux is.

Phi of E is the integral.

In this case it would an open surface of E dot dA.

That is an electric flux.

With Gauss's Law that you see on the blackboard there, we had a closed surface.

I'm talking now about an open surface.

That is an open surface.

This is an open surface, and this is an open surface.

And so Maxwell suggested that we have to add a term which contains the derivative of the electric flux.

And that's what I'm going to do there now, walking over to Ampere's Law.

I'm going to amend it in a way that Maxwell suggested.

He adds a term here, epsilon 0 kappa, d/dT, of the integral over an open surface attached to that closed loop of E dot dA.

This current, which is the one that penetrates, remember, through the surface is really a real current.

This term here, Maxwell called "displacement current.

I want to make sure that I have no slip of the pen, because I hate slips of the pen.

That is correct.

I have everything in place.

You may think now that we can start a party because all four Maxwell's equations are now in place.

Not quite.

We're going to make one small adjustment after spring break, and that adjustment is going to be made in this one, and then we'll have our party.

So now, I would like to use the new law and see whether we can clean up that mess.

So I'm going to revisit my point P1 and I'm going to apply the new law by first having a flat surface, that surface that we have here, and then trying this surface.

And I want to get the same answer.

Pengembangan Perkuliahan

1. Buatlah sebuah Esai mengenai materi perkuliahan ini

2. Buatlah sebuah kelompok berjumlah 5 orang untuk menganalisis materi perkuliahan ini

3. Lakukan Penelitian Sederhana dengan kelompok tersebut

4. Hasilkan sebuah produk yang dapat digunakan oleh masyarakat

5. Kembangkan produk tersebut dengan senantiasa meningkatkan kualitasnya

Ucapan Terima Kasih Kepada:

1. Para Dosen MIT di Departemen Fisika

a. Prof. Walter Lewin, Ph.D.

b. Prof. Bernd Surrow, Ph.D.

#### Staff

Visualizations:
Prof. John Belcher

Instructors:
Dr. Peter Dourmashkin
Prof. Bruce Knuteson
Prof. Gunther Roland
Prof. Bolek Wyslouch
Dr. Brian Wecht
Prof. Eric Katsavounidis
Prof. Robert Simcoe
Prof. Joseph Formaggio

Dr. Peter Dourmashkin
Prof. Robert Redwine

Technical Instructors:
Andy Neely
Matthew Strafuss

Course Material:
Dr. Peter Dourmashkin
Prof. Eric Hudson
Dr. Sen-Ben Liao

#### Acknowledgements

The TEAL project is supported by The Alex and Brit d'Arbeloff Fund for Excellence in MIT Education, MIT iCampus, the Davis Educational Foundation, the National Science Foundation, the Class of 1960 Endowment for Innovation in Education, the Class of 1951 Fund for Excellence in Education, the Class of 1955 Fund for Excellence in Teaching, and the Helena Foundation. Many people have contributed to the development of the course materials. (PDF)

2. Para Dosen Pendidikan Fisika, FPMIPA, Universitas Pendidikan Indonesia.

Terima Kasih Semoga Bermanfaat dan mohon Maaf apabila ada kesalahan.

## Kamis, 06 Januari 2011

### PUSTAKA FISIKA (PF)

1. Sebuah Visi Pengumpulan 100.000 Buah Buku yang terkait dengan Fisika

2. Pengumpulan Data-Data Kefisikaan sebesar 1 Terra byte

Tempat Pengumpulan dan Pendataan Buku-buku fisika via Internet

Instrumentasi Fisika

• Soldering in Electronics Assembly
• Radiation Dosimetry, Instrumentation and Methods
• Theory and Problems of Basic Electricity (Schaum Series)
• Electric Circuits (Schaum Series)
• All New Electronics Self Teaching Guide