Problem Solving Strategies

4:09 AM

@jeea You may ask in the HBar, the main physics chat. User Kyle Kanos is into these. You may ask him and I'm sure he'll help you.

yuvraj singh

4:30 AM

@jeea try this ProjectileMotionApp.java

frame.setSize(900, 600);
frame.setVisible(true);

timer = new java.util.Timer();

timer.schedule(new TimerTask() {

@Override
public void run() {

// has moving objects
if(gui.getScene().inFlux()) {
gui.getScene().updateBalls();
gui.getScene().repaint();
}
}
}, 0, DELAY);
}
}

That, s what we made.

@jeea can you specify what kind of help you want.

yuvraj singh

5:32 AM

@JohnRennie hi.

Morning %%)

John Rennie

@yuvrajsingh hi :-)

yuvraj singh

@JohnRennie have you seen jeea qsn.

John Rennie

Yes

yuvraj singh

So, have you seen my comment.

M. Guru Vishnu

@JohnRennie, Hi sir :-)

John Rennie

5:39 AM

@M.GuruVishnu morning :-)

M. Guru Vishnu

@JohnRennie May I ask a doubt after yuvraj's?

John Rennie

@yuvrajsingh I can't comment because I've never used the box2d engine and I don't know it's capabilities. I would generally favour using an existing engine over trying to write your own.

yuvraj singh

Then it's OK.

John Rennie

@M.GuruVishnu You can ask now.

M. Guru Vishnu

@JohnRennie, Ok sir. Question: What colour does a beam of red light appear to an observer under water? My Approach: We know that $\lambda=\lambda_0 /mu$ where $lambda_0$ is the wavelength in vacuum and $\mu$ is the refractive index of the medium (here it's water). $\lambda_0\approx720$ nm and so $\lambda=540$nm which is green. So it must appear green to the observer under water.

But in one website I saw, colour depends only on frequency and frequency remains same in refraction so it appear red to the observer underwater.

John Rennie

5:47 AM

The colour is determined by the energy of the photons. When light enters your eye it excites electrons in chemicals called rhodopsins in your retina.

There are different types of rhodopsin. Basically there are three types sensitive to red, green and blue photons, and you brain judges the colour based on which rhodopsin molecules interact with the light.

And the energy of the photon determines which type of rhodopsin it interacts with.

And the energy of a photon is $E = h\nu$, which as you say depends only on the frequency.

M. Guru Vishnu

Ok sir. So energy is Planck's constant X frequency so it appear red, am I right sir?

@JohnRennie Thank you sir. Now understood.

I am sorry, I don't know how my formatting went wrong:

5 mins ago, by M. Guru Vishnu

@JohnRennie, Ok sir. Question: What colour does a beam of red light appear to an observer under water? My Approach: We know that $\lambda=\lambda_0 /mu$ where $lambda_0$ is the wavelength in vacuum and $\mu$ is the refractive index of the medium (here it's water). $\lambda_0\approx720$ nm and so $\lambda=540$nm which is green. So it must appear green to the observer under water.

John Rennie

@M.GuruVishnu you missed a backslash :-)

yuvraj singh

@JohnRennie morning sir, Alesha here.

John Rennie

@yuvrajsingh morning :-)

yuvraj singh

Can you check my answer.

John Rennie

5:54 AM

@yuvrajsingh Can you post a link to it?

yuvraj singh

So question is two mirror are inclined at an angle theta, a light enters parallel to one of the mirrors. What should be the value of theta so that it retrace, it, s path.

M. Guru Vishnu

@JohnRennie Yes sir. But for me everything is doubled like this -$\lambda=540\lambda=540$. This happened after I edited it the second time.

yuvraj singh

Is my diagram is correct. i.stack.imgur.com/OvSDv.jpg @JohnRennie,,,,,

John Rennie

@M.GuruVishnu it looks like this here:

@yuvrajsingh that looks fine to me. So $\theta = 45°$

yuvraj singh

But sir, book. Says 30°

Let me post the book diagram.

John Rennie

5:58 AM

@yuvrajsingh let me draw a quick diagram ...

M. Guru Vishnu

@JohnRennie Ok sir. Then it must be an error on my side:

yuvraj singh

@JohnRennie that, s what book diagram look like.

John Rennie

OK, 30° works. But then so does 45°.

yuvraj singh

Do you mean both are correct. @JohnRennie

John Rennie

@yuvrajsingh let me double check 45°

Yes 45° definitely works as well.

yuvraj singh

6:09 AM

Yes the thing is that, the reflected ray from the first mirror, incident 90° with respect to second mirror. Right sir. @JohnRennie

John Rennie

I wonder if there are other solutions as well ...

@yuvrajsingh yes.

yuvraj singh

@JohnRennie, leave it, sir if two questions more.

John Rennie

I suspect there are an infinite number of solutions ...

yuvraj singh

Can I ask?.

@JohnRennie

John Rennie

@yuvrajsingh Yes

yuvraj singh

6:12 AM

So first one very easy, but I do not know why my logic was incorrect, sir before answering the question can you confirm that my way of approaching this question, was correct, or is it useful in exam like JEE.

So first one was, I need to MOI.

Of right angle triangle along the length of hypotense.

As the axis.

Each side has a length l. So what I tried I assume this to be an square.

Of side length l.

John Rennie

An isoceles right angle triangle?

yuvraj singh

And I know the MOI was ml^4/12.

So moment of inertia for half of the triangle is ml^2/6.

Yes.@JohnRennie

John Rennie

You are correct that the MOI of the triangle is half the MOI of the square. That's because MOIs add, and the square is made up from two equal triangles.

However half of $ml^4/12 = ml^4/24$ not $ml^4/6$

yuvraj singh

OK, sir that, s the easy question, but genuinely I found difficulty in deriving this from basics. @JohnRennie

Ah, OK sir.

So how do I do it from basics?

John Rennie

@yuvrajsingh let me draw a diagram ...

Consider the element I've drawn at a distance $x$ from the axis. The MOI of this element is $dI = x^2 dm$. Yes?

yuvraj singh

6:25 AM

OK.

John Rennie

If the side of the square is $a$ then the length of the diagonal is $a\sqrt{2}$ and the height of the triangle is $a/\sqrt{2}$.

yuvraj singh

Yes.

John Rennie

So the length of the element is $\ell = a\sqrt2 \times (1 - x/(a/\sqrt2))$

And $dm = \sigma \ell dx$, where $\sigma$ is the area density.

So:

$$ dI = \sigma a\sqrt2 \left(1 - \frac{x\sqrt2}{a} \right) x^2 dx $$

And the area density is $\sigma = 2m/a^2$

Just integrate that and you should get the result.

yuvraj singh

Got it. Sir.

M. Guru Vishnu

6:41 AM

@JohnRennie, Sir, are you free now? Can we discuss about Acoustic shadows?

John Rennie

@M.GuruVishnu yes, what did you want to ask?

M. Guru Vishnu

@JohnRennie, Are diffraction effects from sound waves larger than that of light waves?

I'm having this doubt as sound and light waves are of different nature - one longitudinal and the other transverse.

John Rennie

It's true that sound waves are longitudinal not transverse, but they diffract in exactly the same way as transverse waves.

Diffraction just depends on the relative phases of the two interacting waves. If the phases differ by 180° you get a minimum whether the waves are longitudinal or transverse.

M. Guru Vishnu

Ok sir. I know diffraction effects depend on the wavelength of the waves. So, if both sound and light waves have the same wavelengths the optical and acoustic shadows are the same region. Am I right sir?

John Rennie

@yuvrajsingh if you're interested, I've found that there are an infinite number of solutions for that mirror question. The angles are given by $90 - n\theta = \theta$ where $n = 1, 2, 3, ...$.

@M.GuruVishnu Yes.

yuvraj singh

6:50 AM

@JohnRennie, it means the question is wrong?

John Rennie

After $n$ reflections the angle to the mirror (not the angle to the normal) is $90 - n\theta$.

yuvraj singh

But if I think it correctly, that, can, t be possible because for every angle light ray retrace path.

M. Guru Vishnu

@JohnRennie Ok sir. In general, do sound waves have a larger wavelengths than light waves? If yes, is this why we are able to hear a sound source behind a carboard sheet but not able to see a light source behind the cardboard? (assuming the position of source, cardboard, and the observer are at the same positions)

yuvraj singh

But in the case of your diagram, yes it can be, but question haven't said that light ray fall like this only.

@JohnRennie

John Rennie

And for this to be parallel to the other mirror you need te angle to be equal to $\theta$. Hence $90 - n\theta = \theta$.

@yuvrajsingh Note that I'm starting with a 90° reflection and tracing the light ray back. So the light ray I've drawn always reflects back along its own path.

@M.GuruVishnu Light waves have wavelengths from about 400nm to 700nm.

yuvraj singh

6:55 AM

What happened in the last of every light ray that it land up 90° with respect to other mirror to retrace path.

John Rennie

@M.GuruVishnu Sound waves gave wavelengths of around a metre. So there is a seven order of magnitude difference!!!

@yuvrajsingh you can trace the light ray in either direction because light propagation is symmetric. I know that to retrace its path the last reflection must be at 90°. So to draw my diagram I started with the 90° reflection and traced the light ray back.

yuvraj singh

Ah......

John Rennie

And on each subsequent reflection the angle decreases by $\theta$

M. Guru Vishnu

@JohnRennie Thank you sir :-)

yuvraj singh

@JohnRennie, hi sir is M. Guru vishu finish.

John Rennie

7:04 AM

@yuvrajsingh yes, I think so

M. Guru Vishnu

@yuvrajsingh A typo in my second name

yuvraj singh

Oops. @M.GuruVishnu.

@JohnRennie shall I ask my new question.

M. Guru Vishnu

@JohnRennie I'll ask after reading some more about this topic on the net sir. So yes.

yuvraj singh

@JohnRennie so can I.

John Rennie

@yuvrajsingh yes

yuvraj singh

7:07 AM

Let me post diagram.

@JohnRennie i.stack.imgur.com/lZ7he.jpg

I need to find 100x,if mass of container is M.

John Rennie

@yuvrajsingh I'm not sure I understand what the diagram shows ...

yuvraj singh

It is like hinged container at one end.

@JohnRennie

John Rennie

A square container hinged at the top edge?

yuvraj singh

Yes.

John Rennie

So what is making it tilt?

yuvraj singh

7:20 AM

Actually, both upper parts were hinged.

If I remove one nail from the top part.

@JohnRennie

John Rennie

In that case haven't you drawn it tilting the wrong way?

yuvraj singh

Sorry, can you make new onem

@JohnRennie

John Rennie

If it is hinged at the red dot then it will rotate like this.

yuvraj singh

Oh yes!

John Rennie

It rotates until the centre of mass lies on the dashed line.

So to find the angle you just need to find the centre of mass.

yuvraj singh

7:28 AM

If I understand you correctly it will rotate until com lines on dash line specify that then it will not further rotate because axis passes through com.@JohnRennie

John Rennie

@yuvrajsingh correct!

Because the torque is proportional to the distance of the COM from the dashed line.

yuvraj singh

OK sir, now I can do it.

@JohnRennie

John Rennie

@yuvrajsingh OK. I'll be around if you want to ask anything more.

@M.GuruVishnu what did you want to ask?

M. Guru Vishnu

@JohnRennie, Can we continue our Discussion on Acoustic shadows?

John Rennie

@M.GuruVishnu yes

yuvraj singh

7:34 AM

@JohnRennie droppers effects.

John Rennie

What are "droppers effects"

Doppler?

M. Guru Vishnu

@JohnRennie I think he meant doppler

John Rennie

Phone auto-correct strikes again! :-)

M. Guru Vishnu

Typing

John Rennie

@M.GuruVishnu OK ... ?

yuvraj singh

7:37 AM

@JohnRennie that, s was a big typo, I mean doppler effect.

M. Guru Vishnu

Wavelength of sound waves in (a) is more than that in (b). So, in (a) diffraction effects are more than that in the second case. So as we gradually increase frequency of sound waves, the acoustic shadow approaches the optical shadow. Am I right sir? (This topic is very interesting for me)

yuvraj singh

@M.GuruVishnu I am not he I am she. My name is Alesha.

I am uv sister.

John Rennie

@M.GuruVishnu do you know about diffraction at a straight edge?

M. Guru Vishnu

@JohnRennie I know only for a single slit sir

@yuvrajsingh UV=Ultra violet ?

yuvraj singh

Tha, s funny. Hahaha.

John Rennie

7:40 AM

Diffraction Effects

►

@M.GuruVishnu this seems like a nice explanation.

yuvraj singh

@JohnRennie have you done with vishnu.

M. Guru Vishnu

@JohnRennie Thank you for your suggestion sir. I like Bozeman's videos :-)

John Rennie

@M.GuruVishnu shall I get back to Alesha while you're watching that?

M. Guru Vishnu

@JohnRennie Sure sir.

John Rennie

@yuvrajsingh go ahead ...

yuvraj singh

7:44 AM

Actually, I am OK, with doppler effect questions.

But how we can derive or justify the it, s apparent frequency equation.

@JohnRennie

John Rennie

The derivation is straightforward.

yuvraj singh

OK.

John Rennie

Suppose a source is emitting waves with a frequency $f$ and velocity $v$ then the wavelength is $\lambda = v/f$ and the period is $T = 1/f$. OK so far?

yuvraj singh

Yes.

John Rennie

Now suppose the source is moving with velocity $u$. That means in the time $T$ it has moved a distance $d = uT$.

So when it emits the next wave the next wave is emitted a distance $\lambda - uT$ behind the first wave i.e. the wavelength has been decreased to $\lambda' = \lambda - uT$.

I can draw a diagram if it will help ...

yuvraj singh

7:51 AM

OK.

John Rennie

Is that OK you want me to draw a diagram?

yuvraj singh

If you can that will be good.

Sir my coffee is on gas. So you continue for answer. Let me..... Is it OK sir.?

@JohnRennie

John Rennie

@yuvrajsingh ping me when you want to continue

M. Guru Vishnu

@JohnRennie, Sir, I checked phet website for the simulation Paul used in that video. I didn't get that one. Do you know where to get that? I hope it gives a good insight on what I'm thinking on paper.

John Rennie

8:06 AM

@M.GuruVishnu I don't know I'm afraid. The diffraction at a straight edge is a fairly routine calculation but a complicated one as it requires an integral called the Fresnel integral that has to be done numerically.

But the bottom line is that the amount the wave diffracts around the straight edge increases as the wavelength increases.

So the reason the acoustic shadow decreases with increasing wavelength is that the increasing wavelength means the waves bend around the obstacle more so they move into the space behind it.

M. Guru Vishnu

Thank you very much sir :-)

yuvraj singh

8:42 AM

@JohnRennie are u there sorry, I was indulge in other things.

John Rennie

@yuvrajsingh hi :-)

yuvraj singh

Hi sir. @JohnRennie

John Rennie

Do you want to continue with the Doppler effect?

yuvraj singh

Yes

John Rennie

At the top we have a stationary source.

yuvraj singh

8:46 AM

Should n be there, because in the diagram, it look like a single wavelength of lambda completes the journey.

John Rennie

It emits a wave with a frequency $f$, so the wavelength is $\lambda = v/f$. The period is $T = 1/f$. OK so far?

@yuvrajsingh The image shows two wavefronts emitted by the source.

yuvraj singh

Ah, OK.

John Rennie

So the two wavefronts are separated by one wavelength.

OK so far?

yuvraj singh

Yes

John Rennie

Now in the lower diagram the source is moving to the right at velocity $u$. That means when it emits the second wavefront it has moved a distance $d = uT$ to the right.

So the spacing between the wavefronts is reduced by this distance $uT$.

I've just realised the text on the second diagram is wrong ...

yuvraj singh

8:51 AM

That should be lambda '

John Rennie

There ...

So we have $\lambda' = v/f - uT = v/f - u/f = (v-u)/f$

And $f' = v/\lambda'$

So:

$$ f' = \frac{v}{v-u} f $$

yuvraj singh

Got it! Thanks you sir.

@JohnRennie sir your correction diagram is still not displayed.

John Rennie

10:27 AM

Deleted

M. Guru Vishnu

11:27 AM

@JohnRennie, Hi sir. Are you there?

John Rennie

@M.GuruVishnu hi :-)

M. Guru Vishnu

@JohnRennie I have a doubt related to our previous discussion. Are you free now sir?

John Rennie

Yes, what's your question?

M. Guru Vishnu

Will the double slit experiment conducted for two waves where one is longitudinal and the other is transverse show any variation (small or big)? Or in other words, can we classify whether a wave is longitudinal or transverse by seeing its interference pattern sir?

John Rennie

Do you mean one slit admits a transverse way and the other admits a longitudinal wave?

M. Guru Vishnu

11:31 AM

I think no difference.

John Rennie

Or do you mean you compare a YSDE for a transverse wave with a YSDE for a longitudinal wave?

M. Guru Vishnu

@JohnRennie No sir. Not like that. I have two waves of same frequency, velocity etc. But they vary only in their nature (longitudinal or transverse). I let them pass through the double slit individually.

@JohnRennie Precisely yes sir.

John Rennie

If the waves have the same wavelength you will get the same difraction pattern from both. So you won't be able to tell the difference from the diffraction pattern.

M. Guru Vishnu

@JohnRennie Ok sir. Thank you. May I ask one more doubt sir?

John Rennie

@M.GuruVishnu yes ...

M. Guru Vishnu

11:35 AM

Can two waves of different types (longitudinal and transverse) of same kind (electric or magnetic or mechanical) interfere with each other?

John Rennie

You need to step back a bit and think about what we mean by a wave.

M. Guru Vishnu

@JohnRennie Energy transferred from one place to another without any huge shift of a particle from source to the destination is wave.

John Rennie

Let's take a light wave as an example. This is an electromagnetic wave so it is a combination of oscillating electric and magnetic fields. For simplicity ignore the magnetic field and treat the wave as an oscillating electric field. OK so far?

M. Guru Vishnu

@JohnRennie Yes sir.

John Rennie

You can have static electric fields as well. For example around a point charge we get a static electric field $\mathbf E = kQ/r^2 \hat{r}$. Yes?

M. Guru Vishnu

11:39 AM

@JohnRennie Yes sir.

John Rennie

In general we can have an electric field $\mathbf E(t,\mathbf r)$ where the value of the field is a function of both position $\mathbf r$ and time $t$.

And a light wave is just a configuration of this field of the form $\mathbf E = \mathbf E_0 e^{i(\omega t - \mathbf k \cdot \mathbf r)}$

So the light wave isn't a separate object, it is just a particular configuration of the electric field.

M. Guru Vishnu

@JohnRennie Ok sir.

John Rennie

A simple analogy would be to look at the surface of a pond on a still day, then create a wave by waggling your hand in the water. The wave isn't some separate object, it's just a variation in height of the water.

Likewise a light wave is just a variation in the value of the electric field.

M. Guru Vishnu

Ok sir. So, are we allowed to apply the superposition principle for two waves of different kinds sir? I think yes based on you previous messages.

John Rennie

Now consider what we mean when we say two light waves interfere. It means we have our electric field oscillating in two different ways, and in some places those oscillations reinforce each other and in other places the oscillations cancel each other.

M. Guru Vishnu

11:47 AM

@JohnRennie Yes sir.

John Rennie

@M.GuruVishnu the oscillations can reinforce or cancel because they are changes in the value of the same field.

yuvraj singh

@JohnRennie hi, Alesha sorry to interrupt, can we create a standing electromagnetic wave.

John Rennie

@yuvrajsingh yes

That's what an etalon does.

@M.GuruVishnu but, for example, could an electric wave interfere with a magnetic wave?

M. Guru Vishnu

@JohnRennie No sir.

John Rennie

And the answer is no because they are not the same field. They are two different fields.

M. Guru Vishnu

11:49 AM

@JohnRennie Dimensionally invalid too!

John Rennie

Yes.

yuvraj singh

@JohnRennie if I understand your last comment,, you mean we can create standing wave in light.

John Rennie

Actually electric and magnetic fields are really not different things - they are different parts of the electromagnetic field, but that starts getting complicated so we'll ignore this for now :-)

M. Guru Vishnu

@JohnRennie So moral of the story: It would be no problem to have interference of two waves of different kind?

John Rennie

@M.GuruVishnu in general two waves of different kinds will not interfere.

@yuvrajsingh yes

yuvraj singh

11:51 AM

@JohnRennie can you explain how can we create that..

Like in string we know.

M. Guru Vishnu

@JohnRennie I'm sorry. The different kind I referred to were transverse and longitudinal waves. (I should have been a bit more specific)

John Rennie

@yuvrajsingh read this:

Fabry–Pérot interferometer

In optics, a Fabry–Pérot interferometer (FPI) or etalon is an optical cavity made from two parallel reflecting surfaces (i.e: thin mirrors). Optical waves can pass through the optical cavity only when they are in resonance with it. It is named after Charles Fabry and Alfred Perot, who developed the instrument in 1899. Etalon is from the French étalon, meaning "measuring gauge" or "standard".Etalons are widely used in telecommunications, lasers and spectroscopy to control and measure the wavelengths of light. Recent advances in fabrication technique allow the creation of very precise tunable Fabry...

yuvraj singh

Or will it be automatically created if I interfere tow different light wave of different wavelength.@JohnRennie

John Rennie

@M.GuruVishnu light only has transverse waves. You can't have a logitudinal light wave so you can't have transverse and longitudinal waves interfereing for light.

yuvraj singh

@JohnRennie, it should be in a vacuum?

John Rennie

11:55 AM

But in solids you can have both transverse and longitudinal waves. The longitudinal waves are sound waves and the transverse waves are called shear waves.

M. Guru Vishnu

@JohnRennie Let us not consider light. What if we have a metal rod (for example), we could have both transverse and longitudinal waves of same nature (mechanical). I asked whether interference of two such waves could take place.

John Rennie

@M.GuruVishnu that's what I am describing ...

Both shear and longitudinal waves are a displacement of the molecules of the solid from their equilibrium position. Yes?

M. Guru Vishnu

@JohnRennie Yes sir.

John Rennie

But the displacement vectors are at 90° to each other.

So when you combine a longitudinal and transverse wave you are adding two vectors that are normal.

OK so far?

M. Guru Vishnu

Yes sir. Now understood. We are just adding two displacement vectors. So no problem in interference. Thank you very much :-)

As interference is nothing but superposition which is nothing but vector addition

John Rennie

11:59 AM

@M.GuruVishnu because if two vectors are normal they will never add to give zero unless both vectors are zero.

M. Guru Vishnu

@JohnRennie Yes sir. I think the resultant wave will be a bit complicated however.

John Rennie

Yes, it will be a complicated wave and won't have a static diffraction pattern like you get with a light wave.

M. Guru Vishnu

@JohnRennie Thank you sir :-)

yuvraj singh

@JohnRennie, my last comment?

John Rennie

@yuvrajsingh the reason you get a standing wave ona string is because the wave reflects off the ends of the string. So we get two waves moving in opposite directions along the string. The two waves add to give the standing wave. OK so far?

yuvraj singh

12:25 PM

Yup.

John Rennie

@yuvrajsingh well we get a light standing wave in exactly the same way. An etalon has two parallel mirrors and the light bounces between the mirrors. So in between the mirrors we have two light waves moving in opposite directions, and those two waves add to give a standing wave.

Transcript for

Problem Solving Strategies