Problem Solving Strategies

4:59 AM

@GaurangTandon ... There you go...

5:51 AM

hi

Gaurang Tandon

Thanks Nehal! @Rick where are you stuck here?

6:12 AM

@GaurangTandon I got the first two parts, but I'm not getting the last one by writing the loop equation

schrodinger_16

6:38 AM

How do we find the kinetic energy of a rigid body rotating about more than one axes?

7:15 AM

@JohnRennie ... You there ?

@NehalSamee I am, though I'll be busy for the next half hour or so. I can ping you when I'm free ...

8:53 AM

@JohnRennie Are you there now?

9:52 AM

@Abcd @NehalSamee I'm around now for a couple of hours

hi

Why there is a difference in the speed of light when it passes through different medium?

sir?

Sir John Rennie?

please respond if you are there

@JohnRennie They asked some very difficult questions yesterday...Let me show:

A: What causes light to refract?

@Akash.B I've answered that question on the main site. Let me look for a link ...

5

The reason that light travels more slowly in a dielectric is because it interacts with the electrons in that dielectric. Light has an oscillating electric field, and if any charged particle is in the path of the light that particle will feel an oscillating force due to the oscillating electric f...

@Abcd yes?

@JohnRennie Do you know what a binary star system is?

@JohnRennie why space keeps on expanding?

10:06 AM

@Akash.B Great that you understood the reason of refraction of light in merely 2 minutes!

@Abcd No

@Abcd the first question looks straightforward ...

@JohnRennie Let me share my thoughts:

@Abcd a binary star system is just two stars orbiting each other.

Option A is correct coz tension is perpendicular to displacement at any instant.

Option B seems correct...very unsure

Option C doesn't look right to me.

Option D is correct as W= $\int F.dx$

@JohnRennie I need help with option B to C (and D if my reasoning is wrong)

10:11 AM

The geometry isn't clear to me. The hand is pulling the block round the track using a string. Yes?

Yes

Is the string pulling the block along a tangent to the track? From the diagram there's an angle $\theta$ and I'm not sure what the significance of the angle is.

@JohnRennie Even I didn't understand this during my exam ...

OK I guess we leave the angle for the moment ...

Option A doesn't look right. The string exerts a force on the block, and the block moves so work is being done.

I agree D is correct

@JohnRennie But when force is perpendicular to displacement, work is zero right?

10:17 AM

I don't understand why you say the force is perpendicular to the displacement.

The question says the string remains horizontal, so it's in the same plane as the track.

Okay

I suspect we're both being confused by the weird diagram ...

hmm

@Abcd If the string was along the tangent to the track then it would be easy because the tension in the string, the velocity of the block and the frictional force are all in line.

@JohnRennie What about option C?

10:21 AM

I think the diagram differs from that only in that the string is being pulled at an angle

The hand does work on the string, and the string does the same work on the block, and the block does the same work on the track. And that's the negative of the work the track does on the block (i.e. the frictional force).

So I think C is true

@JohnRennie "the string does the same work on the block"?

Well the string has no net work done in it. It doesn't stretch, or change its kinetic energy.

So the work you do on the string has to go somewhere ...

Got it. Okay.

Next question now...

Option B is surely correct I am sure about that

Option D is correct because linear speeds of rim particles is changing

Option A is surely incorrect

I just need help with option C

JD_PM

Hi everyone. I already asked this question before, but as I do not have it clear yet... why there is no torque in the following perfect inellastic collision ball-rod? Thanks

imgur.com/gallery/0jj3Q

@JD_PM Newton's third law.

So two forces of same magnitude and opposite direction

Every action has an equal and opposite reaction.

10:29 AM

@Abcd on the basis of energy balance I agree. At the end of the day all the work goes into friction - there isn't anywhere else for the work to go. So I don't see how the work done by the hand/string can depend on the angle.

One provides a torque in $+\hat{k}$ direction

Another in $-\hat{k}$ direction. Get it @JD_PM ?

@JohnRennie I was talking about the question in the next picture. Yes, it doesn't depend on that. Got it.

23 mins ago, by Abcd

JD_PM

@Abcd you mean gravitational force goes downward and normal goes upwards and both are contrarested by each other

Okey thanks

@JD_PM well I wasn't talking about gravitational force

I was talking about the equal and opposite impulsive forces

@JohnRennie I am listening.

Solution :

on the phone. Back in a mo ..

JD_PM

10:36 AM

@Abcd but without gravitational force there’re not impulsive force going downwards right?

@JohnRennie Why did they consider Vc= V due to dipole?

Shouldn't be their potential due to the induced charges on the sphere?

Phone call ended. Where did we get to?

@JohnRennie I am having a severe headache, can we discuss them tomorrow?

@Abcd yes, that's fine. Just ping me.

Okay

10:48 AM

@MadhuchhandaMandal if the sphere is conducting then the potential is constant everywhere in/on the sphere.

That's because if there were any differences in potential the charges would move to the region with a lower potential.

You're quite correct that there will be induced charges, and those charges will arrange themselves so they exactly cancel out the potential due to the dipole.

@JD_PM I'm not sure I understand what the diagram is supposed to show? Does the ball fall and hit the rod then stick to it?

@JohnRennie Hi, can you help me with an LC oscillation related question?

@Rick my knowledge of circuits is very rusty. I guess you're asking about the diagram of the inductor and two capacitors?

Yes

I had a quick look at it but I can't remember anything about inductors. However it should be obvious what happens. The inductor has energy due to the magnetic field it has stored. When you close the switch the inductor uses that energy to charge the capacitor.

Ok, yes, that's what I did, I conserved energy, but I wasn't able to get it by writing the loop equation

11:03 AM

@JohnRennie "cancel out potential"?? I think you meant Electric field

?

@MadhuchhandaMandal Same difference. The field is zero inside the sphere because the potential is constant inside the sphere.

@JohnRennie Ofcourse potential is constant inside the sphere. No doubt in that. My question was : Why VA=VC= V due to dipole at C?

(Note : I agree that VA=VC)

According to me VC=V due to dipole at C+ V due to induced charges on the sphere

@JohnRennie ... Available ?

@MadhuchhandaMandal you've just agreed that the potential is constant everywhere in the sphere, and that includes the surface of the sphere. Since the point A is on the surface if the sphere it is connected to the rest of the sphere and must be at the same potential as the rest of the sphere.

@NehalSamee Yes ...

@JohnRennie ...My question : A body moves up an inclined plane with constant velocity and reaches the highest peak . The inclined plane consists of friction . I was told to calculate the work done by the body to move up the inclined plane . First , the body gained potential energy . Then , my book calculates work done against friction $Fs$ , as $s$ is given . Finally , they add $mgh + Fs$ ... I am confused of the addition part ... Why do they add ? Shouldn't we subtract ?

I've figured out that in order to maintain constant velocity , force equal to that of friction is to be applied and so we add it right ?

11:11 AM

@JohnRennie I just found out that it's a non charged sphere .. So , (V induced)= Sum(k dQinduced / R) = Sum(dQinduced) *(k/R) and by Conservation of Charge we get Sum(dQinduced)=0 so potential due to induced charges at Centre = 0

@NehalSamee When you say the work done by the body presumably there is a motor or something similar inside the body to propel it up the incline? So is it the work done by this motor or whatever that we are trying to calculate?

@JohnRennie ... , what will happen in case of an accelerating body ? Will we subtract then ?

@JohnRennie ... Yes...

@NehalSamee OK, so if the incline was flat (i.e. mgh=0) the motor would have to do work against the friction $W_f = \mu M g d$. Yes?

@JohnRennie Okay.. I am not able to convey my question correctly

Yes ... @JohnRennie

11:14 AM

@JohnRennie Potential at the Surface and A is exactly same as potential at C

There was no doubt in that

@NehalSamee now make the incline not flat, but set the friction to zero, so the motor has to do work $W_h = mgh$. Yes?

@MadhuchhandaMandal OK, I thought that was what you were asking ...

My doubt was : Why potential at C = Potential due to dipole at C?

@JohnRennie ... OK...

Because there will be induced charges on the surface of the sphere and we need to account for their potentials at C also

@NehalSamee so if the incline isn't flat and the friction isn't zero the motor has to do work both against the friction, $W_f$, and to go up the incline, $W_h$, so the total work done by the motor is $W_f + W_h$. That is, you add the two works together.

11:17 AM

@JohnRennie ... OK , now if it is accelerating / decelerating , then what ?

I.e. decelerating due to friction ...

@NehalSamee If the velocity changes then the total work done by the motor has to include the change in kinetic energy.

OK , then will I write :

mgh-W_{friction}=∆K ...?

@NehalSamee If $W_t$ is the total work then: $$ W_t = W_f + W_h + (KE_f - KE_i) $$

where $KE_f$ is the final KE and $KE_i$ is the initial KE

@MadhuchhandaMandal Hmm. The induced charge will be symmetric i.e. the charge in one direction will have an equal and opposite charge in the other direction.

@JohnRennie ... However , I watched in lecture of Walter Lewin that during a body falls from top to bottom in frictioned incline plane , he wrote: $mgh-W_{friction}=\Delta K$

@NehalSamee When you're going up an incline the frictional force and the gravitational force are acting in the same direction. When you're going down an incline they are acting in different directions.

Remember that a frictional force always acts opposite to the direction of motion. It doesn't have a fixed direction like gravity does.

11:31 AM

@JohnRennie...So , is the equation correct for downward fall , write ?

@NehalSamee yes, the equation is correct. The mgh term tends to increase the KE while the friction tends to reduce the KE.

@JohnRennie...It is : $mgh-W_{friction} or $\Delta k$ or $mgh-W_{friction} - \Delta K$ ...?

For downward movement...

Hmm, your equation isn't rendering ...

?

11:35 AM

@JohnRennie... there any problem?

Did you mean $mgh-W_{friction} = \Delta k$

@JohnRennie... Yes...But what about the third one ?

$mgh-W_{friction} - \Delta K$

Shouldn't there be an = in there somewhere?

@JohnRennie... know... But I just used the case of upward movement you provided , changing sign of friction , is it permissible ?

Yes, because the direction of the frictional force changes

11:39 AM

@JohnRennie... Should I write work done is mgh-W_{friction} or work done is ∆K...

?

Cause both are equal ... ?

@NehalSamee Both are perfectly true statements

@JohnRennie ... Last but not the least , isn't work done to overcome friction , irrespective of upward or downward movement , so why do we subtract in downward and add in upward ?

Work is force times distance. If you are interested in the work done by the body (or on the body) then calculate the net force on the body.

If the body is moving upwards the friction and gravtational forces both act in the same direction i.e. down the slope. So they havethe same sign and simply add together.

If the body is moving downwards the the gravitational force acts down the slope and the frictional force acts in the other direction up the slope. So they have different signs.

@JohnRennie .. so , isn't work done to overcome friction in both cases to maintain motion ...?

I'm not sure that is a helpful way to look at the problem.

11:50 AM

@JohnRennie See this

I'm unable to understand from where they got the factor of (1/2)

@MadhuchhandaMandal I don't know to be honest. Electrodynamics is not one of my strong subjects.

@JohnRennie Okay it's absolutely fine !!!

@JohnRennie ... By the way , thank you...

3:03 PM

in The h Bar, 45 mins ago, by Abcd

The question is: The displacement of the particle at $x=0$ of a stretched string carrying a wave in the positive $x$ direction is given by $f(t)= A\sin (x/a)$, where $A,a$ are constants. The wave speed is $v$. Write the wave equation....

@JohnRennie Could you please explain what this question is trying to convey?

@Abcd The wave equation is $f(t)= A\sin (kx - \omega t)$. You just need to figure out what the angular frequency and wave vector are.

I have mistyped.

@JohnRennie ... On which one of these [ wavelength , color , frequency ] does refractive index of a medium depends ?

Q: Why does the refractive index depend on wavelength?

@NehalSamee wavelength IIRC

@NehalSamee Please research before posting :)

9

Why do different wavelength get impeded more or less when in different materials? Moving with the same speed, but a longer physical distance would imply that the fields oscillate less times in the material, but I don't know why a difference in the number of oscillations would impede the wave- I d...

@JohnRennie This is the one^

@NehalSamee wavelength, colour and frequency are all related, so the refractive index depends on them all. I would probably say it's frequency dependent as it's really down to the photon energy and that depends on the frequency.

3:12 PM

Part a is simple, I need help with part b

@NehalSamee Are you appearing for JEE 2018?

@Abcd the wave velocity $v$ is the phase velocity, and it is related to the angular frequency and the wave vector by $v = \omega/k$.

phase velocity?

The wave vector is related to the wavelength by $k = 2\pi/\lambda$

@Abcd ... I'm from Bangladesh ... Just far away from JEE...

@NehalSamee Cool!

@JohnRennie Wave vector. My books call $k$, the "wave number"

3:16 PM

@Abcd same thing by different names

@JohnRennie Okay, how to solve that question then...

Anyhow, since you know the wavelength you know $k$, and then $\omega = kv$ so you can find $\omega$.

And the wave equation is just $\sin(kx - \omega t)$

Wavelength isn't given...

Dude! $g(x) = A\sin(x/a)$

One wavelength is when $x/a = 2\pi$

Actually $k = 1/a$ falls straight out. Then $\omega = v/a$.

@JohnRennie And the question only talks about the first wavelength right?

Also, how is that a wavelength? (first time solving questions of this chapter)

3:28 PM

The wavelength is the distance over which the phase of the wave changes by $2\pi$. That's because $\sin$ is periodic with a period $2\pi$.

You're given that at time zero the wave varies with $x$ as $g(x) = A\sin(x/a)$

So at $x=0$ the wave $g(x) is zero.

Yes

Suppose $x = \pi a/2$. If we substitute this into the equation for $g$ we get $g = \sin(\pi a/2 /a) = \sin(\pi/2) = 1$

So the wave has gone from zero up to one.

When $x = \pi a$ we get $g = \sin(\pi a/a) = 0$ so the wave has gone back down to zero i.e. half a wave.

@JohnRennie i think thats half a wave

@Abcd Oops, just a spelling error :-)

Lol, okay..

@JohnRennie then?

3:34 PM

Anyhow, it should be obvious that $\pi a$ is half a wavelength

So a full wavelength is $2\pi a$

If you drew a graph of $g(x)$ it would be obvious

Yes, it's obvious.

Wavelength is the separation between two particles in same phase

so $k=1/a$

$\omega = v/a$

@JohnRennie Why is "A" there in both equations?

Both equations tell something different, so how can A be there in both?

$A$ is just the maximum amplitude of the wave i.e. $-A \le g(x,t) \le +A$

@JohnRennie So is the interpretation of $x$ different in both equations?

My book says that $g(x)$ gives the shape of the wave.

So will I have to substitute $x=1,2,3$ etc. to obtain the shape of the wave?

I can't understand the difference between $g(x)$ and $f(x)$

If you write $g(t,x) = A \sin(kx - \omega t)$ then your function depends both on time and position. You can ask, suppose I take a moment in time - e.g. take a picture - what does the wave look like, and you get this simply by setting $t$ equal to a constant.

If we make that constant zero, i.e. choose the moment in time when $t=0$, we get $g(x) = \sin(kx)$. And that's what your question has done.

The question says: The shape of the string at t=0 is given by g(x) = etc

Gaurang Tandon

@MadhuchhandaMandal imgur.com/a/g9493 see if this helps (I swapped the radii, but the rest of the procedure is correct) @Rick sorry I wasnt myself sure of that question either. I'll take a closer look tomorrow and let you know. Ping me if I forget.

3:51 PM

@JohnRennie so if I put x=300, then? What will I get?

@Abcd I think we are mis-communicating somewhere. If you graph $g(x) = \sin(x/a)$ it's just a sine wave. Put $x=300$ and you get some point on that sine wave.

@JohnRennie okay, we'll get a point somewhere but that is not the shape of the string at $t=0$

The shape of the wave is the function $g(x)$

@JohnRennie Okay, but a sine wave isn't a pulse, is it?

Where did pulses come in?

3:59 PM

This is my understanding of pulses:

The question describes a 1D infinite wave

Yes, but that's not what the question is asking about

@JohnRennie No the question says "a wave pulse is travelling on a string...etc"

Ah OK, it does use the word pulse - I had missed that. But then it says the shape is $g(x) = A\sin(x/a)$ which is an infinte wave not a pulse.

4:01 PM

So the question is wrong?

Hmm, now you mention it the question does seem badly written.

However if you take ut the word pulse it then makes sense.

$g(x)= A\sin (x/a)$

$f(x)= A\sin(\dfrac{x-vt}{a})$

What's the difference between them?

Both are infinitely long sine waves

$$ f(x,t)= A\sin(\dfrac{x-vt}{a}) $$

$f$ is a function of both position and time

1 min ago, by Abcd

Both are infinitely long sine waves

@JohnRennie Could you explain the difference between $g$ and $f$ using an example?

@Abcd Is this your first exposure to waves?

4:09 PM

@JohnRennie yes (I have read the theory given in my book')

@Abcd ah, OK, waves can be really confusing at first because generally speaking when students first encounter them they aren't used to functions of several variables.

OK, suppose I take a long bit of wire and bend it into the shape of a sine wave, then I place my wire on the $x$ axis.

yes

The distance of the wire from the $x$ axis will then be given by $g(x) = A \sin(kx)$ for some constants $A$ and $k$ that depend on how I bent the wire. OK so far?

@JohnRennie Why?

@Abcd well I bent the wire into the shape of a sine wave...

4:15 PM

@JohnRennie yes

OK, so if I sit on the $x$ axis at some point, e.g. $x=300$, then I'll just see a stationary bit of wire at a distance from the $x$ axis of $A\sin(300k)$

@JohnRennie you mean the part of the wire overhead right? yes...

@Abcd yes

then?

So does what I've said so far make sense? That is you can imagine my piece of bendy wire and you're happy its shape is described by a sine wave?

4:19 PM

Yes

Now suppose I give my piece of wire a velocity along the $x$ axis. I'm not changing anything about the wire - I'm not bending or reshaping it - I'm just moving the whole wire along the $x$ axis.

Yes

Now suppose you are sitting at $x=300$ and looking only at the piece of wire at that point. What do you see? Is the wire still stationary?

No, moving.

Right. So the distance of the piece of wire at $x=300$ is now changing in time.

4:23 PM

@JohnRennie perpendicular distance...

Yes

@GaurangTandon Never mind, I got it

In fact it is oscillating sinusoidally at some frequency that depends on how fast I am moving the wire along the $x$ axis.

Still with me?

@JohnRennie Didn't get this.

Start with the stationary wire, and suppose we've picked a point on the $x$ axis where the distance of the wire from the axis is zero.

If the wire isn't moving than the distance remains zero.

I wonder if I should draw a diagram ...

@JohnRennie No, I am able to imagine.

@JohnRennie then?

yes the distance changes continuously if its moving

4:28 PM

@Abcd aha! Yes!

So for a moving wire the distance depends on two things:
1. where on the $x$ axis you are
2. what the time is

Got it. Thanks.

@Abcd I kind of get the feeling that you've suffered from brain overload and decided to call it a day :-) If you want to come back to this I'm happy to attempt more (probably not very good) explanations.

4:50 PM

@JohnRennie okay, I am listening. I understand that the distance changes continuously, how do we "wavically" represent it?

@Abcd if the wire is moving at velocity $v$ then the distance it moves in a time $t$ is $vt$ (obviously)

yes

For our bit of wire to move up to a maximum, then down to a minimum then back to zero the wire has to have moved past us by one wavelength.

@JohnRennie what are you referring to by "our bit of wire"?

The bit of the wire at $x=300$ that we are looking at while we're sitting at $x=300$.

i.e. we're sitting a $x=300$ and measuring the distance of the wire normal to the $x$ axis.

4:55 PM

@JohnRennie Yes but its not the "same bit" always if I am not wrong

@Abcd yes, I'm being a bit loose with my terminology.

Okay

@JohnRennie yes

Are you happy that as the wire moves past us by one wavelength we see the distance go through one complete oscillation?

Can we call the distance $z$ just to avoid having to keep typing the distance the while time?

@JohnRennie yes

@JohnRennie I think by distance you mean amplitude.

@Abcd yes

The point of all this is that the time one wavelength of the wire takes to pass us, $\lambda/v$ is just the period of the oscillation.

5:06 PM

yeah

So $T = \lambda/v$, or rearranging gives $v = \lambda/T = \omega/k$, which is how we get the equation linking $v$, $\omega$ and $k$ that I mentioned earlier.

@JohnRennie Okay, and in a question they have given the shape of the string at $t=t_o$ insatead of t=0, what to do in such cases?

instead*

@Abcd if we go back to the example of the wire that I have been using, then at any time $t_0$ the wire is just a sine wave and all that changes is how far the wire has moved along the $x$ axis at the time $t_0$.

The distance moved since time zero is just $v t_0$

yes