How do I solve this problem? The period of the free oscillations of the system shown here if mass M1 is pulled down a little and force constant of the spring is k and masses of the fixed pulleys are negligible, is

The answer is 2π √(M2+4M1/k)

Here M2 is mass of movable pulley and M1 is mass of the block

I deduced that when M1 moves down by x, M2 moves up by x/2 and therefore spring stretches by x/2,but I'm not sure what to do next.

@JohnRennie What is "outside radius of spiral's turns"

@Abcd don't know. What is the context?

@JohnRennie leave it. I will just take it to be radius and solve and see if i get the answer.

@JohnRennie Could you give me a mathematically accurate and rigorous view of azeotropes??

@Abcd that would be a long and complicated discussion.

...Okay.

The vapour pressure is the pressure at which the chemical potential of the component in the vapour and the solution is the same, so you'd have to consider how the chemical potential varied in both phases.

I dont know chemical potential as well :O

If you were doing a chemistry degree then this would be a very interesting thing to pursue because it would teach you lots about thermodynamics. However I think it would be a somewhat pointless diversion right now.

... Okay ...

@JohnRennie A plane spiral with a great number N of turns wound tightly to one another is located in a uniform magnetic field perpendicular to the spiral's plane. The outside radius of the spiral's turns is a. The magnetic induction varies with time as $B = B_O \sin (\omega t)$, where $B_o$ and $\omega$ are constants. Find the amplitude of induced emf.

I'm working for a few minutes ..

Hi

@Abcd isn't it just $d/dt$ of $N$ times $B$ times the coil area?

@JohnRennie Its not a coil

Its a spiral

It's effectively a coil. A coil is just a spiral compressed along its axis.

Huh?

Is that from the question?

@JohnRennie I got the wrong answer by considering it a solenoid. This is the shape that author wants^

@JohnRennie No

Ah ......

a plane spiral

So it's your diagram but compressed into a plane

Oh

So the EMF is the same as you'd get from a straight wire of the same length

No way

It's $$\dfrac{d\left(B(t)\displaystyle\int_0^a\pi r^2 \dfrac{N}{a}dr\right)}{dt}$$

@JohnRennie Please explain??

@JohnRennie That's^ the answer given

Isn't it just this?

@JohnRennie But it says "tightly wound"

That integral just gives you the length of the wire

So I dont understand how area is associated with the loop

It's not a loop

It's effectively just a long straight wire

@JohnRennie I cant understand your method. Please explain.

Suppose we had a long straight wire of length $\ell$ with a field $B$ normal to it, then the EMF developed between the ends of the wire would be $d/dt ( B \ell )$. Yes?

Well how??

Emf is $d\phi/dt$

How can a non closed figure have flux associated with it.

@JohnRennie Please reply

I've been Googling for a nice article. This one is the best I've found so far.

No

Thats an entirely different thing

Thats motional emf

When conductor is moving

In this case you have a stationary wire and a changing field. The same applies.

This is what solution author says:

But ... it isn't made up of loops

....And I am getting confused amidst all this.

I can see what the solution is doing. If you take a loop of radius $r$ then the flux linked is $\pi r^2 B$ and $d/dt$ of this gives you the EMF. You treat the spiral as being made up of those loops and integrate to sum up the EMFs.

(approximating the sum as an integral because $N$ is large)

I'm just not sure I understand why that is a valid treatment.

The width is infinitesimaly small so we have loops?

@JohnRennie Which loops??

Its more like a disc

We dont treat disc like loops.

Well, yes ...

It's not like a disk since (I assume) the wire is insulated so current can flow only along the wire not sideways between adjacent parts of the wire.

@JohnRennie $?^\infty$

@JohnRennie How can area be associated with any loop despite the presence of so Many loops infront of it.

I have to work now for half an hour or so

Presumably the question means something like this:

@Abcd superposition principle

Maybe

No

@JohnRennie Are you there?

Yes

@JohnRennie What does this mean: A long solenoid of cross sectional radius a has a thin insulated wire ring tightly put on its winding

How is Swatz theorem applicable in thermodynamics

There is something called Contiunity that needs to be checked for scwatz therorem

However I am unable to arrtive at the particular justification

@Abcd I'd guess it means the radius of the ring is only just big enough for the ring to fit round the solenoid. The question is probably going to assume the flux through the ring is the same as the flux through the core of the solenoid.

@JohnRennie A long straight solenoid of cross sectional diameter d= 5cm and with n = 20 turns per cm has a round turn of copper wire tightly put on its winding. The cross sectional area of copper turn is $S= 1mm^2$

@JohnRennie Just look at the cross sectional area!^^ as compared to the area of solenoid!!

I mean thee theroem that proves symmetry of the partial derivatives.

@Abcd I have no idea what that question means

Okay :/

I am consulted various resources but I am not able to decipher that

@JohnRennie

@JohnRennie Are you still here?

Yes

@JohnRennie Very hard please see^

@sammygerbil Please see if you have any idea too ...

Presumably this is because the voltage difference across the high resistance part and low resistance part is different ...

The total EMF round the ring will just be $bA$, where $A$ is the area of the ring ($A = \pi a^2$ presumably).

@JohnRennie its induced electric field

@JohnRennie yes

So the voltage drop across the high resistance half will be $bA\eta/(1+\eta)$

And the voltage drop across the low resistance half will be $bA/(1+\eta)$

@JohnRennie I am listening.

I'm not sure what further to say because I don't know what the question means by the magnitude of the electric field strength in the ring.

@Abcd it's not a disk

@AvnishKabaj Keep 10000 loops around each other with negligible separation and tell me from far away if they dont look like a disc

@Abcd lol

It's not a disk

If you make a pizza out of chocolates doesn't mean it's become a pizza

It's still a chocolate

@JohnRennie good morning

@harambe morning :-)

@JohnRennie can you help me with some collision problems

Yes. What's the problem?

@JohnRennie imgur.com/a/e2r16k7

No idea how to solve such questions

@harambe That's quite a hard question because the two disks will move off at an angle and you have to work out what angle they move off at.

If you work in the COM frame it's not impossibly hard, but not trivial either.

Figured but I don't get why they move at a angle. Any explanation

Take a simple example (diagram incoming):

@harambe suppose you have a collision like this:

Okay

When the disks touch they touch like this:

This looks like oblique collision. I don't even understand how to proceed with these problems

The force between them acts at the contact point and normal to their surface, so it acts at an angle as shown by the red arrows. That means the force pushes the disks sideways.

So after the collision you get:

These are all drawn in the COM frame.

Okay

@JohnRennie the force acts angle to what

The calculation isn't as hard as you think. Momentum is always conserved so the total vertical components of the momenta must add up to zero as well as the total horizontal components.

And energy is conserved, or if the coefficient of restitution isn't unity you can still calculate what the final energy is.

The hard bit is that for an oblique collision the momentum change, the impulse, is a vector that lies along the line connecting the two centres of the disks.

Yea so the momentum conservation fails there right

@JohnRennie I am still confused

@harambe no, momentum conservation doesn't fail.

But impulse is an external force

@harambe no it isn't

Okay so in oblique collision momentum gets conserved and we can more or less use energy equation right?

@harambe each disk delivers an equal and opposite impulse to the other disk, so the total impulse adds up to zero.

Got it

So the total momentum starts at zero, is changed by zero during the collision, and ends up at zero.

Okay

One thing can you explain to me why did the spheres go sideways after collision

I deliberately drew the diagram so at the moment of the collision the line joining the centres of the disks was at 45º.

That means the impulse has equal horizontal and vertical components.

i.e. the change in the horizontal component of the momentum is equal to the change in the vertical component of the momentum.

In your problem with the big disk and the two little disks you'd have to work out the angle of the impulse vector. It's not that hard - it's just geometry - but adds an extra complication.

I need to work I'm afraid so I'll have to call it a day now.

Okay. Cya later sir

@Blue how do we find angle of impulse

@Blue :o

@Blue the wires ain't gonna touch

Are they?

It's a pizza string

I got it. Basically the impulse is 45 with the line of collision and thus will have equal components along and perpendicular to the line of collision

And impulse is change of momentum so the changes will be equal in x and y components

@harambe yes

Okay. I think I can solve my question now

@Blue hmmmm

Ok

@JohnRennie what to do. After I have calculated impulse angle

I get final velocity of the balls to be 3u/4√2