The h Bar

Yashas Samaga

6:00 PM

You have your wave equation

y=$6sin(\pi x/10)cos(100\pi t)$

the first term gives the amplitude at a position $x$

user123733

@dmckee in this total energy is constant

FrancescoS

Hi guys, does anyone know anything about Conformal Field Theories?

Yashas Samaga

the second part tells that the particle at position $x$ is a harmonic oscilator

user123733

@YashasSamaga yes

dmckee

@user123733 Yes. Just like the SHO.

Yashas Samaga

6:01 PM

@user123733 Do you understand the partial differentiation notation?

user123733

@dmckee but in wave energy is not constanr

@YashasSamaga not much

Yashas Samaga

It is easy to understand.

user123733

I have just studied for high school level

Yashas Samaga

Differentiating a function partially = differentiate the function as if there was only one variable even though there are more than one variable

user123733

Okay

Yashas Samaga

6:02 PM

you understand that dy/dt gives the particle's velocity?

John Rennie

@user123733 the kinetic energy of the piece of the string between $x$ and $x+dx$ is: $$dE = \tfrac{1}{2}mv^2 = \tfrac{1}{2}\rho dx \left(\frac{dy}{dt}\right)^2 $$ where $\rho$ is the linear density. Does that make sense?

user123733

@YashasSamaga yeah

dmckee

@user123733 In a traveling wave the energy at a particular point in space isn't constant, but the energy of the whole is constant.

Actually, I'm not sure that is true. (The part about energy at a point).

Yashas Samaga

@user123733 if you differentiate your wave equation by keeping $6sin(\pi x/10)$ as constant, you get the velocity of the particle at the position $x$

dmckee

But certainly in a standing wave each point acts like a SHO so the energy is constant at every point.

user123733

6:05 PM

@JohnRennie yes

@YashasSamaga I think I have done that above

John Rennie

@user123733 Good. You've already calculated $$\frac{dy}{dt} = -600\sin\frac{\pi x}{10}\sin 100\pi t $$

Yashas Samaga

@user123733 consider the mass to be dm and write the kinetic energy equation as 1/2 dm v^2

user123733

@dmckee is energy is constant at every point then how energy is flowing

Yashas Samaga

then follow what John Rennie said

Energy isn't flowing in a standing wave.

John Rennie

@user123733 and the maximum velocity is when $\sin 100 \pi t = 1$ in which case: $$\frac{dy}{dt}_{max} = -600\sin\frac{\pi x}{10} $$

Yashas Samaga

6:07 PM

You have a standing wave because the other end is clipped. There is no displacement at the other end, hence no work.

user123733

@JohnRennie yes

John Rennie

@user123733 so just feed that into my expression for $dE$ and integrate from $x = 0$ to $x = L$ where $L$ is the length of the string.

user123733

@YashasSamaga energy is constant in that segment

Yashas Samaga

as a general tip, use letter to represent numbers while deriving things. In your equation, replace the 6 with A or something.

user123733

@JohnRennie 600$\pi sin(\pi x/10)$ from 0 to 100

John Rennie

6:10 PM

$$ E = \tfrac{1}{2}\rho \int_0^L dx \left( -600\sin\frac{\pi x}{10} \right)^2 $$

user123733

Which is zero

@YashasSamaga okay

dmckee

@user123733 In the standing wave it simply flips between potential and kinetic with no net transfer.

John Rennie

@user123733 you're integrating $\sin^2$ not $\sin$

user123733

@dmckee oh i see

@JohnRennie sorry

John Rennie

Cool. Does it make sense now?

user123733

6:13 PM

Yeah

John Rennie

There is another way to arrive at the same result if you're interested ...

Yashas Samaga

@user123733 prntscr.com/e9l84l

user123733

@JohnRennie I got the answer , thanks

@JohnRennie sure

Yashas Samaga

@user123733 after substituting for $dm$ in the K.E equation, you'll have to integrate with respect to x from 0 to L

John Rennie

Suppose you split up your string into chunks of length $dx$, so they have a mass $\rho dx$, then you can treat each chunk as a seperate harmonic oscillator with amplitude $A(x) = 6\sin\frac{\pi x}{10}$

dmckee

6:17 PM

In the standing wave, something has to continue to drive the wave from one end or another if you are to have the unending passage of peaks. The energy comes from that driver.

John Rennie

And the energy of a harmonic oscilaltor is $\tfrac{1}{2}kA^2$ where $k$ is the force constant.

user123733

@YashasSamaga thanks a lot

@JohnRennie yes

takashi

mr@JohnRennie can you advise me calculus book

John Rennie

@user123733 So the total energy is just the sum of the energies of all these elemenmts: $$E = \int_0^L \tfrac{1}{2}kA^2$$

user123733

@dmckee but in normal wave eqation energy is not constant at a point . Am I correct

takashi

6:21 PM

@JohnRennie and also a source where i can practice my skills in physics

John Rennie

And you get $k$ from it's relation to the angular frequency $\omega = \sqrt(k/m)$

Feed all that into your equation and it will turn into the same integral that we got from considering the kinetic energy.

user123733

@JohnRennie in this what we are integrating

Oh sorry I got it

John Rennie

@takashi To be honest I'm out of touch with modern physics books. Have you seen our book recommendations question‌?

takashi

let me have look

John Rennie

@user123733 Cool :-) I think integrating the KE is simpler, but either approach gives the same result.

user123733

6:24 PM

Thanks

Then what is this which is written in my book dK/dx =(1/2)$\mu A^2 \omega ^2 cos^2(\omega t-kx)$

Yashas Samaga

$(\omega t-kx)$ How did you end up with this?

Ah! It is the same!

They used a trig identity to combine the two terms.

If you move the dx to the other side, you will get the same equation

wait

you wanted to calculate the maximum K.E, right?

and it was for a standing wave?

Your equation doesn't seem to be that of a standing wave.

user123733

Yeah , But I got that . Now I wanted to know what tis equation

Yashas Samaga

Is this a new question?

user123733

@YashasSamaga i think it is for general wave

Yashas Samaga

$\omega t - kx$ appears for travelling waves

yea

That had nothing to do with what you just derived.

user123733

6:30 PM

Ohk

@YashasSamaga If instead of stationary wave we have travelling wave example y=Asin($\omega t-kx$)

Then how can we find KE

Yashas Samaga

It is the same process.

user123733

Or PE or total energy

Yashas Samaga

dy/dt gives the kinetic energy of a particle

take x as constant

user123733

Then (1/2)dm (dy/dt)$^2$

Integrating it

What about PE

Yashas Samaga

P.E in travelling wave. Nice question :P

Does your textbook ask for it? I don't see any use of it. We usually calculate power transmitted by a travelling wave but not potential energy associated with it.

The potential energy is localized to the part of the string which is currently being stretched. Elastic potential energy.

user123733

6:37 PM

@YashasSamaga no

0

Q: In wave motion of a string both kinetic energy and potential energy are minimum at $y=y_\text{max}$ then why does the string comes down again?

In wave motion of a string both kinetic energy and potential energy are minimum at $y=y_\text{max}$ then why does the string comes down again? As everything in tries to attain lowest energy possible what brings that string element back to its original position?

Thanks a lot @YashasSamaga

@JohnRennie @dmckee

FrancescoS

1

Q: States on spheres in radial quantization

Following the explanation of radial quantization of Slava Rychkov in link the states are defined on the spheres with which we foliate the space-time and the generator that moves the states from one surface to another is the dilations generator $D$ which plays the role of the Hamiltonian. He says...

conformal-field-theory

Anyone?? :)

Ismasou

7:02 PM

is someone here workin on qcd ?

FrancescoS

@Ismasou It depends... what do you need to know?

DanielSank

7:32 PM

@heather Yo there!

skullpetrol

Yo

anonymous

8:10 PM

-2

A: February 20th Ask Me Anything with heather: Question Pool

I note your interest in quantum computing. I have a deep interest in physics, but I'm a IT guy by profession, and I've taken careful note of the way digital electronic or "ordinary" computing has advanced in leaps and bounds over recent decades. Advances in computing has changed our lives, for th...

"In that time I've seen a lot of articles about quantum computing. See for example physicsworld. It's been hailed as something wondrous for year after year. See the timeline of quantum computing on Wikipedia. It's been kicking around for about thirty-five years now. And yet it has delivered nothing. Apart from jobs for researchers who promise jam tomorrow. In fact, I'd go so far as to say it's been promoted and peddled and pimped for decades, even though it's delivered diddly squat. "

Arrgh! I feel so bad when someone says that Quantum Computing has achieved nothing !

skullpetrol

No need to feel "bad" about that source.

AccidentalFourierTransform

quantum computing has achieved nothing.

Josh Pilipovsky

Yo guys can someone clarify something for me

skullpetrol

Askaway

Josh Pilipovsky

Silly question but I was doing the energy levels for An electron in a magnetic field and I found plus minus hbar/2 but I'm not so sure what negative energy means in this case. I know for a potential it would mean bound state but this is different. Can u clarify for me

David Z

8:21 PM

@JoshPilipovsky not a silly question, but we could probably use a bit more information. Quoting the problem statement would help. This would be a good question to ask on the main site using the homework-and-exercises tag (along with other appropriate tags)

AccidentalFourierTransform

^

skullpetrol

^^

Josh Pilipovsky

lol I was going to but I didn't think it was necessary. It's just an electron in a magnetic field and the Hamiltonian is H= omega Sz

skullpetrol

Give it a try, it can't hurt.

Josh Pilipovsky

Ok

skullpetrol

8:24 PM

Good luck.

David Z

@JoshPilipovsky A lot of people say that, but actually it's a good question. "Just a [whatever]" isn't a problem! This is not an all-star standout question, sure, but it is exactly the "bread and butter" of the site - it's just the kind of thing we want. It's definitely much better than all the "how do I solve this problem please help" junk we get; you're not asking for an answer, you're asking for physical insight.

Josh Pilipovsky

😊

2