3:22 AM
@sammygerbil
Can anyone tell me how to solve the first part
Here's what I did
What to do next

3:45 AM
@JohnRennie morning.

@Nobodyrecognizeable morning :-)

@JohnRennie do you recall noethers theorem ?

Yes ... ?

@JohnRennie if you have time ill ask you a question and youll try to solve that with noethers theorem. Can i go ahead?

3:52 AM
9 hours ago, by sammy gerbil
@JohnRennie i know we can not know kinetic energy and potential energy is conserved or not using noethers theorem . So i want to ask you to check about the energy, momentum , angular momentum is conserved or not using noethers theorem.

I don't think Noether's theorem is useful here. Noether's theorem relates a symmetry of the action to a conserved quantity.

Now before proceeding i want to ask what do we mean by translational symmetry?
@JohnRennie can you demonstrate the noethers theorem by working out an example (just at classical level) if you go qm then ill have to study for that.

Didn't watched it yet

@LoopBack thats done already by sammy gerbil. Im just here to learn noethers theorem from johnrennie.
And yep thats a.

Nice

4:04 AM
@JohnRennie are you still here?

@Nobodyrecognizeable I don't think Noether's theorem is applicable to this sort of problem. The problem gives you $\mathbf r(t)$ so you can work out the velocity, and therefore the momentum, angular momentum and kinetic energy by differentiating.

@JohnRennie yep. Thats ok. But my request is can you demonstrate noethers theorem with an example

@Nobodyrecognizeable have you solved this question which you posted

@LoopBack which one ?

4:06 AM
53
@Nobodyrecognizeable question no. 53

I have just starred the message from where the solution starts you can have a look at that. @LoopBack

@Nobodyrecognizeable Were you able to prove it

@Nobodyrecognizeable well consider a free particle. The action is:
$$S[\mathbf{r}(t)] = \int_{t_1}^{t_2} \tfrac{1}{2}m\dot{\mathbf{r}}^2$$

@JohnRennie kinetic energy integrated over a time integral?

Suppose we displace the while system in space by some amount $r_0$ so that $\mathbf r \to \mathbf r + \mathbf r_0$. This clearly doesn't affect the action because when we differentiate the constant term disappears.

4:10 AM
@LoopBack its done by @sammygerbil.

@Nobodyrecognizeable I very close to proving the equation, but I m unable to resolve initial velocity of the ball in terms of $h$.

That means a displacement in space does not affect the action, and Noether's theorem tells us that must mean momentum is conserved. Which of course it is.

@JohnRennie how do you get the action?

In physics, action is an attribute of the dynamics of a physical system from which the equations of motion of the system can be derived. It is a mathematical functional which takes the trajectory, also called path or history, of the system as its argument and has a real number as its result. Generally, the action takes different values for different paths. Action has the dimensions of [energy]⋅[time] or [momentum]⋅[length], and its SI unit is joule-second. == Introduction == Hamilton's principle states that the differential equations of motion for any physical system can be re-formulated as an...

@JohnRennie the rest is fine..
@LoopBack its vertical velocity is $\sqrt 2gh$

4:15 AM
@Nobodyrecognizeable No it's not

@LoopBack why ?

@Nobodyrecognizeable I want to know the velocity just before first collision
@Nobodyrecognizeable velocity just before collision cannot be $\sqrt{2gh}$. Because the ball has travelled a lesser distance than h

@LoopBack see the vertical distance through which the ball has fallen is h. So the velocity should be $\sqrt 2gh$ in the vertical direction

@Nobodyrecognizeable but h is distance from the horizontal ground and not from the incline plane.

I should mention that was the wrong thing in the question. They wanted to say this that the point of impact is at vertically h distance away from the inclined plane. Cause that perfectly matches the answer. And except that the question can not be solved as its height isn't given.

4:21 AM
@Nobodyrecognizeable Btw if I substitute √2gh in my equation I will get the correct answer.

@LoopBack if the exact vertical distance isn't given no one will be able to solve this as youll not get a perfect initial velocity. Making the problem unsolvable or we shall be getting a general solution.
@LoopBack anyway goodbye.
@JohnRennie thanks for helping. Goodbye, professor have a nice day.

@Nobodyrecognizeable bye

@JohnRennie @sammygerbil good morning

@harambe morning :-)
You're unlikely to find Sammy here at 05:30 a.m. :-)

Oh. I think he is more active during night time

4:42 AM
@JohnRennie imagine two concentric spheres with potential difference between them. Does earthing any of the sphere change the potential difference between them

I'm not sure that's a well defined question because earthing either of the spheres will change the charge on it.

@JohnRennielet me post the question
In this question, the final potential difference remains same between the spheres

Suppose A weren't present, what would the field inside B be?

The field inside a charged spherical shell is ... ?

4:48 AM
Zero

So the field inside B due to the charge on B is zero. Then the field inside B is only due to the charge on A.
Yes?

Yes

And the potential is the integral of the field. That means the potential difference between A and B is determined only by the charge on A.

Doesn't this violate gauss law

@harambe can you expand on that. Why do you think this violates Gauss's law?

4:52 AM
Not gauss law. I meant isn't electric field zero inside a conductor zero at any situation
Property of conductor

Didn't we discuss this before? The field inside the metal of a conductor is always zero. The field in a space enclosed by a conductor is not necessarily zero.

Okay. I didn't think of it like that

@harambe Suppose you have an electron floating around in a conductor like copper.

Okay

This electron is mobile, because electrons in a conductor are mobile, so if any field is present the electron will feel a force from that field and it will move.

4:59 AM
Then they would collide bc of this?

But as the electrons move this causes a charge separation and that in turn causes a field inside the conductor. Eventually the field caused by the fact the electrons have moved will exactly balance out the external field. At that point the electrons will stop moving and the total field inside the conductor will be zero.
The point is that if there is any field inside the conductor the electrons will move, and they will only stop moving when the field is zero.

Okay

But this only applies in the metal of the conductor where there are mobile electrons.
In the air/vacuum/whatever inside a conducting shell there aren't any mobile electrons.

Yea. That makes sense

As it happens the field inside a spherical shell is zero due to Gauss's law, but this applies to non-conducting as well as conducting shells. It's unrelated to the reason why the field inside a conductor is zero.

5:46 AM
Hello :-)
You there John?
Isn't moment of inertia just Mk^2?

@JohnRennie are you free

@harambe yes

@user64829 Yeah, I guess he is writing it from another non-inertial frame and giving pseudo force too

@user64829 I think that's the parallel axis theorem ...

Q19

6:01 AM
But the axis is through the centre of the ball, why is he using parallel axis theorom?

I find all the options incorrect... Funny
@user64829 the axis is at the point of contact of the sphete with the ground

Although it's not clear from the question I would guess $Mk^2$ is the moment of inertia about the centre of the object. Then the moment of inertia about the contact point with the slope is given by the parallel axis theorem.

No, would You mind if I answer, di electric constant of water is some 80 and dielectric breakdown (you can look it up in the web) and answer the question

No
Let me explain my view

@user64829 the instantaneous axis of rotation is the point of contact with the slope.

6:04 AM
I think capacitance of the sphere is directly proportional to dielectric constant and the potential of the sphere is inversely proportional to the dielectric constant so won't these cancel each other @EshaManideep
Q=CV
So that's why I am confused

@JohnRennie Oh yeah, now I get it
Thanks

@harambe I'm not sure I understand answers (a) and (b). I wonder if they mean hold more charge for a given voltage
Otherwise I can't see what hold more charge means.

John Rennie, that's what he meant

@JohnRennie I think so too
If for a given voltage then I guess this answers it

A fireman wants to slide down a rope. The breaking load for the rope is (3/4)th of the weight of the man. Find the minimum acceleration of the fireman to slide down the rope.

6:18 AM
T= m(g-a)
Maximum tension must be 3T/4

@blue_eyed_... the net force on the man is the gravitational force minus the tension in the rope. Yes?

@JohnRennie, yeah
@JohnRennie, are these forces acting in opposite directions?

The force on the man is $mg$, and we are told that the maximum tension in the rope is $0.75mg$, so the net force on the man is $F = mg - 0.75mg = 0.25mg$
@blue_eyed_... yes, gravity is pulling the man down while the tension in the rope is pulling him up.

@JohnRennie, thanks. I got it

@blue_eyed_... Cool :-)

6:31 AM
John, you there?

6:48 AM
@user64829 hi

4 hours later…
10:36 AM
@JohnRennie are you free

@harambe hi

Q10
I calculated energy density which came to be $Q^2$/16πEo$R^2$

What is $C$?

I messed up. I think
Capacitance

OK. The field inside the smaller sphere and outside the larger sphere is zero. So the field only exists in the shell between the two spheres. Yes?

10:43 AM
Yes

So the volume is just $4/3 \pi R^3 - 4/3 \pi r^3$ where $R$ and $r$ are the radii of the two cylinders.

Yea

So given the capacitance you need to relate this to the radii of the two spheres and hence work out the volume.

C=4πEo r1r2/r2-r1
I have calculated energy from here
@JohnRennie how do I proceed from here

We're told that $r_2 = 2r_1$ so the capacitance is $C = 8 \pi \epsilon r_1$. Yes?

10:53 AM
Yes
Calculate d energy from Q^2/2C

@harambe why are you calculating the energy? The question asks for the volume not the energy?

I just thought it had something to do with it

If we call the radius of the small sphere $r$ then the capacitance is $C = 8\pi\epsilon r$.
So $r = C/(8\pi\epsilon)$
And the volume of the small sphere is:
$$V = \tfrac{4}{3} \pi \frac{C^3}{8^3 \pi^3 \epsilon^3}$$
OK so far?

@JohnRennie I get what you are doing.... I can't believe I missed this
I just thought of going to energy density and energy but it was actually simple...

@harambe the question looked harder than it was. The reference to energy was just a complicated way of asking what the voulem between the spheres is.

11:01 AM
Yes
I can do this now
Got the hint
@JohnRennie Got D
It looks right too

Yes, that's what I got :-)

11:36 AM
@JohnRennie can you tell me why the CGS unit of thermal conductivity is $\dfrac{kcal}{m. hrs .K}$ instead of $\dfrac{ergs}{sec. cm .K}$

I have absolutely no idea. Using any system of units other than SI strikes me as a road to madness.
5

@JohnRennie okay

Anonymous
11:56 AM
Hello, I have few questions. I tried them, but couldn't solve them entirely.

Anonymous

@JohnRennie Sometimes I haven't even gone to sleep by 5:30am! But yes usually I am asleep at that time.

Anonymous
In this question, the problem that I face is that it given that the spring is stretched between points A and B but it is not mentioned how much.

Anonymous
So, I just assumed that the distance b/w A and B to be its natural length.

Anonymous
For solving that, I assumed the part of spring to the left of point P to be one spring and the part to the right to another spring.

Anonymous
12:00 PM
(Assuming the one to the left as spring 1 and to the right as spring 2)

Anonymous
I will have $$k_1 = \dfrac{k × 0.5}{0.4}$$ and $$k_2 = \dfrac{k × 0.5}{0.1}$$

Anonymous
And I assumed that the displacement of the point P when the force is 5 N be x.

Anonymous
So, that will give me $$k_1x + k_2x = 5$$

Anonymous

@IceInkberry It is not necessary to know how much the spring is stretched between A and B because force is proportional to extension no matter how much extension there is already. So your method is correct.

Anonymous
12:04 PM

Anonymous
There isn't even an option for my answer.

Anonymous
$$k_1 = \dfrac{500}{4}$$ $$k_2 = \dfrac{100}{4}$$

Anonymous
$$\dfrac{500x}{4} + \dfrac{100x}{4} = 5$$

Anonymous
$$x = 0.033 m$$

Anonymous
@sammygerbil Do you see anything wrong in my method?

12:08 PM
@IceInkberry I don't see how you got $k_1=k\times 0.5/0.4$.

But k1x and k2x are in opposite directions.. Would it make sense to add them

Shouldn't that be $k_1=k\times 0.4/0.5$ and similar for $k_2$?

Anonymous
@sammygerbil Because $$k × l = constant$$

Anonymous
So, $$k × l = k_1 × l_1$$

Anonymous
For a spring cut, the spring constant varies inversely to its length.

12:11 PM
@DavidZ heyyy. I worked hard on those comments. Hrumph

Anonymous
11

To supplement the answer by Luboš Motl, I will come to this problem from a Material Science point of view. What you mean by the inherent property of the string is not the spring constant, in fact, it is Young's modulus $E$, which only depends on the properties of a material of a body but not it'...

@IceInkberry Ah yes you are right. You need a bigger force to produce the same extension.

Anonymous
@sammygerbil Yes.

I think harambe is right : $x$ should be +ve for the spring on the left and -ve for that on the right.
So you will have $k_1x-k_2x=5N$.

Anonymous
@sammygerbil I don't think that, because if you stretch the the spring, it will try to pull in the opposite direction and if you compress it, it will try to push it.

Anonymous
12:18 PM
Basically, the spring on right will push it to left and the spring on left will push it to right.

Anonymous
Also a reminder, it's one spring. I assumed it to be two.

@IceInkberry Yes I think you are right again. I'm not thinking well today!

Anonymous
@sammygerbil Haha, maybe you just need some sleep.

@IceInkberry Maybe. I have being playing football this morning, and walking. Often I have a nap afterwards!

Anonymous
@sammygerbil Oh well, nevermind. Have some sleep! I will ask it on main.

12:24 PM
@IceInkberry I get $8mm$ answer A.

Anonymous
@sammygerbil How?

Anonymous
I am surprised.

Anonymous

@IceInkberry $k_1=k\times \frac{0.5}{0.4}=1N/cm\times \frac54=\frac{5}{4} N/cm$.

Anonymous
@sammygerbil But isn't $$k = 1.00 N/cm ?$$

12:28 PM
Yes sorry corrected. I continue ...

Anonymous
Listening..

$k_2=1N/cm \times \frac{0.5}{0.1}=5 N/cm$.

Anonymous
@sammygerbil Understood.

Anonymous
I calculated $k_2$ incorrectly!

$(\frac54+5)x=\frac{25}{4}x=5$
$x=\frac45 cm=8mm$

Anonymous
12:30 PM
@sammygerbil Got that!

Anonymous
Pity on me.

Anonymous
Thank you! @sammygerbil

@IceInkberry It took us both to sort it out.

Anonymous
Yes, haha.

Anonymous
I have one more question, could you help me with it?

Anonymous
12:33 PM
(If you have time)

Anonymous

Anonymous
I don't know how to do it by intuition, so I tried the mathematical method.

Anonymous
(I'll have to make a diagram)

Anonymous
(A minute)

12:40 PM
np

Anonymous

Anonymous
I wasn't able to upload the image. Finally.

Anonymous
I wrote these equations.

Anonymous
Now, I tried to apply some sort of logic to the $2$nd equation.

Anonymous
When the ball is at the bottom most position, $$\theta = 90°$$ and $\alpha$ will be some angle.

Anonymous
12:43 PM
Basically $F = mg\sin{\theta}/\sin{\alpha}$

Anonymous
So, when $\theta = 90°$, $\sin{\theta} = 1$ and that divided by $\sin{\alpha}$ (which is less than $1$) means $sin\theta / sin\alpha ≥ 1$

Anonymous
But, when the ball is at topmost point, $\theta , \alpha = 0$ so $sin\theta / sin\alpha$ = 1$Anonymous So, that means$F$is decreasing as we pull up. Anonymous But, I don't understand how I should do it for$R$from the first equation since$F$is the variable there. Anonymous I also don't think that this is a valid method. It works only when the options are given. 12:50 PM @IceInkberry The options are a key part of the question. It is valid to use them - eg eliminate options you know must be wrong, whatever is left must be correct - assuming that one option is in fact correct. Anonymous @sammygerbil Oh, I see. So, I've eliminated **$A$** and **$D$** @IceInkberry Can you decide between B and C? Anonymous @sammygerbil No :/ @IceInkberry I approach such questions by looking at extremes. If the hemisphere was a sphere then at the lowest point the string would bear all of the weight of the ball and the normal reaction would be zero. When the ball reaches the top the situation is reversed : the string bears no weight and the normal reaction bears all of the weight. Anonymous @sammygerbil I did think about it, I even tried to manipulate the first equation that way, but the force$F$was again bothering me. Anonymous 12:59 PM The answer even surprised me, it is$C$. Anonymous I cannot intuitively think how the normal reaction wouldn't vary. Anonymous The tension's component ($T = F$here) is playing some role. @IceInkberry The answer surprises me also. I thought it should be B. Another "extreme" case is that the pulley is very much higher than the radius of the hemisphere. Then when the ball is at the side of the hemisphere the string hangs almost vertically down, so the normal reaction is very small. In this case R increases as the ball goes up. I don't see how it can remain unchanged. Anonymous That was incorrect. Anonymous Incorrect. Anonymous 1:09 PM I don't know. Anonymous @sammygerbil Yeah! Anonymous Maybe, I should move on. The question took a lot of time. I shouldn't waste any more. Anonymous Thank you for your help @sammygerbil @IceInkberry you are welcome Why is the answer to this question Y1+Y2/2 and not Y1+Y2? Considering the two wires as two springs with spring constants AY1/L and AY2/L since the two wires are in parallel why isn't effective spring constant AY/L taken as sum of both spring constants? 1:21 PM @Hema I think it is because the cross-sectional area is doubled. @sammygerbil thank you so much! Hello :-) 2:05 PM @IceInkberry @sammygerbil i think as the ball is moving in a circular path the sum of all the forces in radial direction should have been$\dfrac{mv²}{r}$2:26 PM @LoopBack The question says the small ball is pulled gradually. I think this means that$v=0$- ie the ball is in static equilibrium at each position around the rim of the hemisphere. 2:44 PM @LoopBack You left a message for me to contact you yesterday. Did you want to discuss further your solution to the problem of the two balls connected by a string? 3:39 PM A simple harmonic wave train of amplitude 3cm and frequency 200 Hz travels in the positive X axis with a velocity of 20 m/s. Calculate the displacement, velocity and acceleration of a particle situated at 50 cm from the origin at t=2 sec. @JohnRennie, @sammygerbil, please see... The answer in my book is 0 for displacement and acceleration. How is that? @blue_eyed_... If the displacement is zero then the acceleration will also be zero because the restoring force is proportional to displacement. @sammygerbil, could you please explain mathematically..? I mean by using equations The wavelength is 10cm so the particle is situated a whole number of wavelengths from the origin. ... @sammygerbil v=0 does it make sense. Absolutely not the body won't move @sammygerbil I wanted to ask you a doubt but later I solved it myself @sammygerbil, y=a sin(wt-kx) gives the displacement, is it? 3:49 PM @blue_eyed_... The particle moves vertically. Its motion is SHM. So the restoring force on it is proportional to displacement. This is one way of recognising SHM. @blue_eyed_... Yes that is correct. So you can differentiate to get acceleration. @LoopBack It is possible to move it very slowly, which is what gradually means. Then$v \to 0$. The question is comparing the static forces in various positions of rest. It is not asking about the dynamic forces while motion is occurring. @sammygerbil, first order derivative gives aw cos (wt-kx). Differentiating again seems quite difficult? @blue_eyed_... Surely not!$a$and$\omega$are constants. @sammygerbil, does second order derivative gives -aw^2 sin(wt-kx)? @blue_eyed_... Yes. And this is equal to$-\omega^2 y$. Acceleration is proportional to displacement. 3:58 PM @sammygerbil, what about the sine function @blue_eyed_... What about it? @sammygerbil, you told acceleration is equal to -w^2y. Why did you not include the sine? @blue_eyed_... The sine is included in the definition of$y=a\sin(\omega t-kx)$. @sammygerbil, oh.. I got it. Thank you @blue_eyed_... It need not be wt-kx right? it can be kx-wt also Since we do not know how the wave train started 4:08 PM @blue_eyed_... The next step is to find the value of$y$. The phase of the sine wave is$\phi=\omega t-kx)=2\pi (ft-x/\lambda)$. We are given$f=200Hz$and$C=20ms/s$so$\lambda=10cm$. Therefore$\phi=2\pi (200*2-50/10)$. ... @user64829$\sin(kx-wt)=-\sin(wt-kx)=\sin(wt-kx+\pi/2)$. That is correct we do not know the phase with which the wave started. However, with the values given in the question it does not matter. So displacement at the specified time and distance isn't zero unless the initial phase is N(pi) (where N is some integer) right? the instant and distance specified in the question I mean @user64829 Sorry that should be$sin(kx−wt)=−sin(wt−kx)=sin(wt−kx+\pi)$. @user64829 Try putting values into the equation and see what result you get! @user64829, yeah.. But the answer given was zero. So I was confused with that I think they were referring to most general case, where phase is 0 @user64829 Yes sorry you are correct we have to assume that the phase is zero at the origin$x=0$at$t=0$. Good point. If we had used a cosine wave instead of a sine we would have gotten a different answer. 4:20 PM Yeah, in that case phase would be pi/2 +/- @user64829 That's right. So the particle would be at maximum displacement = 3cm and maximum acceleration. Yeah 4:34 PM @sammygerbil I have solved this question and all the options are correct. But the solution says (b) is incorrect. But why ? Heat flowing through slab E= heat flowing through slab A=Maximum heat flowing Mind if I give my opinion? @LoopBack I think this is a question of what you mean by maximum. Maximum usually means it is bigger than any other value, rather than equal biggest. @user64829 Go ahead. @sammygerbil Exactly what I was going to tell haha 4:57 PM Transverse waves on a string have wave speed 8m/s, amplitude 0.07m and wave length 0.320 m. The waves travel in the negative X direction and at t=0s, the x=0 end of the string has its maximum upward displacement. (a) find the frequency, period and wave number of these waves. (b) write a wave function describing the wave. (c). Find the transverse displacement of a particle at x=0.36 m at time t=0.15 sec @sammygerbil, @JohnRennie, @user64829, please see this X=0 has max upward displacemnet, so phase = 0 @blue_eyed_... a wave equation is$A\sin(\omega t - kx)$or$A\cos(\omega t - kx)$. Since it's in negative x direction, squation will be Asin(wt + kx) @JohnRennie Shouldn't it be wt + Kx since it's travelling in negative x direction? In this case we are told that at$x=0, t=0$the amplitude is a maximum, i.e. equal to$A$, so it's going to be the cosine not the sine. Oh yeah sorry, cosine function What about symbol between wt and kx? 5:02 PM$\omega = 2\pi f$and you're given the frequency so you can work out$\omega$@JohnRennie, how do we know when we need to use cosine function and when we need to use sine function? @blue_eyed_... the most general form of the wave equation is: $$\psi(x,t) = A\sin(\omega t - kx) + B\cos(\omega t - kx)$$ And you have to determine the constants$A$and$B$from the initial conditions. @JohnRennie, where does that come from? @blue_eyed_... the wave equation is a linear equation, so any sum of solutions is also a solution. The sine is a solution and so is the cosine, so the sum of the two is also a solution. Consider what happens at$x=0, t=0$. If we substitute this into the above equation we get: $$\psi(0,0) = A\sin(0) + B\cos(0) = B$$ And we are told that the amplitude is a maximum when$x=0, t=0$so that must mean$B$is equal to the maximum amplitude. Yes? @JohnRennie, yeah that's true 5:08 PM And that in turn has to mean$A=0$So the behaviour at the origin means we end up with the solution: $$\psi(x,t) = B\cos(\omega t - kx)$$ Suppose we were told that$\psi(0,0) = 0$In that case we would have$B=0$and the solution would be: $$\psi(x,t) = B\sin(\omega t - kx)$$ @JohnRennie, Where did we use the information that the wave is travelling in negative x direction? @blue_eyed_... we haven't ... yet @JohnRennie, where do we need that? Is that used while adjusting the sign? Strictly speaking$k$is a vector called the wave vector so it has a direction. In practice for 1D motion we generally don't worry about this and write it as a scalar. In that case for a wave moving in the positive$x$direction we use$\omega t - kx$and for a wave moving in the negative$x$direction we swap the sign of$k$to get$\omega t + kx$. So the full equation for the wave in this question is going to be: $$\psi(x,t) = B\cos(\omega t + kx)$$ @JohnRennie, what does x actually represent in these wave equations? 5:14 PM It is the position on the$x$axis @JohnRennie, position of wave or particle? The wave is travelling along the$x$axis, so if this is a wave on a rope the particles that make up the rope are oscillating normal to the$x$axis. i.e. any particular bit of rope stays at constant$x$and just oscillates sideways. @JohnRennie, thank you. Got it 5:36 PM @sammygerbil, you there? @blue_eyed_... hello @sammygerbil, hi. I had a confusion. When do we use wt - kx and kx-wt in wave equation? If you are using the equation$y=A\sin(wt-kx)$this is a wave travelling in the +x direction starting with zero phase. That is, the slope is increasing as we move in the +x direction from the origin - ie slope is +ve at origin. Whereas if you use$y=A\sin(kx-wt)$this is also a wave travelling in the +x direction but the phase at the origin is now$\pi$. That is, the slope is decreasing as we move in +x direction - ie slope is -ve at the origin. @sammygerbil, that means in the second case the wave starts below x axis? The question we looked at with amplitude 3cm and frequency 200Hz did not say what the phase was at$x=0$and$t=0$so it was ambiguous. It was an ill-defined question. @blue_eyed_... Yes in the 2nd case$y=A\sin(kx-wt)$the wave moves down below the x axis as we move right. 5:51 PM @sammygerbil, how is the phase 0 and pi? @blue_eyed_... Good question. That is making me think ... I think I might have it the wrong way round. @sammygerbil, huh? I didn't get! If we have$y=A\sin(kx-wt)$and we take a "snap-shot" of the wave at$t=0$then we have$y=A\sin(kx)$. Starting from the origin and looking in the +x direction this goes up above the x axis - ie the slope is +ve. Whereas if we have$y=A\sin(wt-kx)$then the snap-shot at$t=0$is$y=A\sin(-kx)=-A\sin(kx)$. This goes down below the x axis as we look in the +x direction - ie the slope at the origin is -ve. @sammygerbil, how is the phase difference between them pi? @blue_eyed_... Because$\sin(x+\pi)=\sin(x)\cos(\pi)+\cos(x)\sin(\pi)=-\sin(x)+0$because$\sin(\pi)=0$.$\sin(x+\pi)$is the same as$\sin(x)\$ reflected about the x axis.
@blue_eyed_... It is (potentially) quite confusing because we have 3 variables (y, x and t) but it is difficult to plot a 3D graph. It is easier to plot y vs x for a particular value of t (eg t=0) or y vs t for a particular value of x (eg x=0).

6:56 PM
@blue_eyed_... Ping me tomorrow morning and I'll explain an easy way to understand what direction the wave moved in.

2 hours later…
8:49 PM
@IceInkberry Small metal ball pulled on fixed hemisphere : I have looked again at this question, using your equations in a spreadsheet to calculate the values of N and F. To my surprise F is constant! But I do not understand why this is.

2 hours later…
10:40 PM
Correction : normal reaction N is constant.

1 hour later…
11:41 PM
@sammygerbil I m 100% sure that the answer to this question is B
@sammygerbil Btw N is not constant
@sammygerbil I can show you why

@LoopBack go on

@sammygerbil Wait I'll upload an image to demonstrate it
@sammygerbil if the radius of the ball is negligible then according to geometry, the string will not be straight but will lie on the surface of the hemisphere. But that's not the case, so diagram B represents the correct situation.
Diagram B suggests that although string is not in contact with the hemisphere but it is parallel to a tangent at some point. I have marked that point in this diagram.
@sammygerbil this means that there will be another point in the hemisphere where the string will again make same angle with the vertical as it was making in the bottommost position.
@sammygerbil I have shown that by a dotted sphere
@sammygerbil I hope I m clear to you