5:20 AM
@JohnRennie sir please tell me the source from where i can learn Boltzmann distribution law
As when i searched myself it says statistical mechanics
Needed

@PrateekMourya are you preparing for jee?
If so any jee book is suffice .

5:39 AM
@PrateekMourya hi :-)

6:07 AM
Hello sir
@Dirichlet some extra stufy

What do you want to ask about, specific heats, Boltzmann distribution or conical pendulums? Or all three :-)

2 hours later…
7:49 AM
@JohnRennie sir please can you draw a diagram of that path
All!
But specific heat first

@PrateekMourya with a regular pendulum the motion is along only one axis, and the restoring force is proportional to the displacement. OK so far?
Like this. Yes?
But we can also make the pendulum move like this:
So now the bob doesn't just move along a line. Instead it moves in a loop in the xy plane.
@PrateekMourya what's the problem with the specific heat. What are you not sure about?

8:16 AM
Sorry for late joining
Ok
I was asking if there was much better method to describe heat transwrs and temperature chamge
It is $m(integral)sdt$

@PrateekMourya so to describe the motion in the xy plane we need to know the forces Fx and Fy. Yes?

Thus just integrating heat change vs temperature change curve and integrating

Well yes, though the specific heat is usually constant to a good approximation so we don't need the integral.

I view this as the slope of temperature vs heat energy

But you're quite correct that where the specific heat is a function of temperature we need to use integration.

8:21 AM
Yes

The integral sign is \int in MathJax. So you can write $m\int s(T) dT$
Let's move this to the problem solving room ...

Ok now i know
Is mathjax a lengthy language to lwarn?

12 messages moved from The h Bar
@PrateekMourya it's easy to learn the basics and for most purposes that's all you need. MathJax is an absolutely enormous language with thousands of symbols, but you only need about a dozen of them for most equations.

Yes i was thinking for here it will be helpful to learn it
Ok back to the pendulum

Pretty much all you need to remember is \frac to do fractions, \int for integrals and \sqrt for square root.
Anyhow, the pendulum:
22 mins ago, by John Rennie
Suppose the bob is at some position (x,y) as I've drawn. Then the distance from the origin is r, and the angle with the x axis is θ. OK so far?

8:27 AM
Ok

Because the pendulum is displaced a distance $r$ from the origin the restoring force is $F = -kr$, and that force points along the line $r$ i.e. from the bob to th origin. OK so far?

The x component is $F_x = F\cos\theta$ and $\cos\theta = x/r$
So $F_x = (-kr) . (x/r) = -kx$
And likewise $F_y = -ky$
OK so far?

Ok
Pretty easy so far

So the motion in the x and y directions is independent. We can write the equation of motion as:
$$x(t) = A\sin(\omega t + \phi_x)$$
$$y(t) = B\sin(\omega t + \phi_y)$$

8:32 AM
I don't know SHM so much but I know that the pendulum is periodic motion

You need to understand SHM because this is just two separate SHM motions in the x and y directions.

....

SHM is easy to learn. We can go through it if you want.

Right now my coaching is teaching thermodynamics and heat
No sir i have other subjects too
I come her in brake time

OK, let me know when you want to talk about SHM and we can discuss it.

8:34 AM
How lkng will it take though?

10 minutes

To lean shm?

Yes. It's really simple.

Can we continue after 2 hours?

Yes, I'll be here.

8:35 AM
Thanks alot sir

:-)

7 hours later…
3:50 PM
Deleted

1 hour later…
5:00 PM
@PrateekMourya hi :-)

Sorry for being super late
Meanwhile i read some of the portion of shm

How did you get on?

Like shm is defined as to and fro motion in which a mean position seeking force which is proportional to its position from mean position
And just the velocity as a function of time and distance from
Mean position

Yes. All that matters for SHM is that the restoring force is given by an equation F = -kx where x is the displacement and k is a constant.

Yes
Is x^2 be too a shm

5:03 PM
No, if the force is F = -kx² then that is an anharmonic oscillator and solving the equation of motion is much harder.

Ok
Is the information i learned enough for pendulum problem

Well, do you understand how to describe the motion of a simple 1D pendulum?
i.e. how you solve the equation of motion?

Honestly no
Since i haven't practiced anything

Well we know that F = -kx. Yes?

I will watch a video on YouTube

5:08 PM
We can do it now - it's really simple.

So we start from F = -kx. Yes?

And from Newton's second law F = ma and a = d²x/dt². So we can substitute for F to get our equation of motion:
$$\frac{d^2x}{dt^2} = - \frac{k}{m} x$$
OK so far?

5:11 PM
And we guess that the solution has the form $x(t) = A\sin(\omega t + \phi)$, where $A$, $\omega$ and $\phi$ are constants that we need to determine.

So: $$\frac{dx}{dt} = A\omega\cos(\omega t + \phi)$$

And $$\frac{d^2x}{dt^2} = -A\omega^2\sin(\omega t + \phi)$$
But we started out with $x(t) = A\sin(\omega t + \phi)$, so if we substitute for $A\sin(\omega t + \phi)$ we get:
$$\frac{d^2x}{dt^2} = -\omega^2 x$$
OK so far?

5:16 PM
And our original equation was $$\frac{d^2x}{dt^2} = - \frac{k}{m} x$$

So $x(t) = A\sin(\omega t + \phi)$ is a solution as long as $\omega = \sqrt{k/m}$
i.e. we have solved the equation to get $$x(t) = A \sin\left(\sqrt{\frac{k}{m}} x + \phi\right)$$
The constants $A$ and $\phi$ we get from the initial conditions.
And that's it. That's all there is to SHM.

Wait
Lemme switch to laptop
ok

You can see all the MathJax now?

5:23 PM
Does it all make sense?

before i was thinking that i will be able to follow without it

You could, but it's much easier when you can see it rendered.

So are you happy you understand how we solve for the equation of motion in SHM?

5:26 PM
Do you want to look at that 2D pendulum now?

Let me grab the diagram ...
9 hours ago, by John Rennie
On the right is a perspective view of the pendulum, and on the left we have just the bob moving in the xy plane. OK so far?

Now we have displaced the bob a distance $r$ from the origin, so we have a force $F = -kr$ pointing towards the origin i.e. along the vector $\mathbf r$. OK so far?

ok
wait
isnt the mg downward
you mean componet of tension

5:31 PM
We want the two components of this force $F_x$ and $F_y$, because we are going to write $d^2x/dt^2 = F_x/m$ and $d^2y/dt^2 = F_y/m$.
If you look at the left diagram $F_x = F \cos\theta$. Yes?

And $\cos\theta = x/r$
So that means $F_x = -kr(x/r) = -kx$
And likewise $F_y = -ky$. Yes?

is f tension here

$F$ is the component of the tension in the xy plane.

or we just rotated our coordinate axes a bit
so that its 2d pendulum
just in a new plane

5:35 PM
The motion is effectively two dimensional i.e. the bob moves in the xy plane. So we are considering the motion in the xy plane.

@JohnRennie ok
ok

What we have found is that the position (x,y) of the bob in the xy plane is determined by the two equations:
$$\frac{d^2x}{dt^2} = - \frac{k}{m} x$$
$$\frac{d^2y}{dt^2} = - \frac{k}{m} y$$

So the solutions are going to be:
$$x(t) = A \sin\left(\sqrt{\frac{k}{m}} t + \phi_x\right)$$
$$y(t) = B \sin\left(\sqrt{\frac{k}{m}} t + \phi_y\right)$$

5:38 PM
where $A$, $B$, $\phi_x$ and $\phi_y$ are constants that we will get from the initial conditions.

I can't remember what you said the initial conditions were ...

the bob is diplced theta angle form mean poistion in any ranom plane

in The h Bar, 24 hours ago, by Prateek Mourya
A small ball is suspended from a point O by a light thread of length l. Then the ball is drawn aside so that the thread deviates through an angle θ from the vertical and set in motion in a horizontal direction at right angles to the vertical plane in which the thread is located. The initial velocity that has to be imparted to the ball so that it could deviate through the maximum angle
2
π

in the process of motion is v
0

=
cosθ
xgl

. Find x.

then given velocity
perp
to the plane
so sing angle is pi/2

5:42 PM
Well let's suppose that the bob starts at the position $(A,0)$ with the velocity $(0,v)$. Is it OK to start from here?
The bob is only moving in the xy plane i.e. only moving horizontally.
So the z component of its position and velocity is always zero.

gravity in z axis

We don't care what forces are acting in the z direction, because the bob is held by the string so it cannot move up or down. It can only move in the x and y directions.

ok from diagam
i understand the oversll structurer

I need to go now, and it must be getting late in India.

ya i was about to say th same
but for chester

5:53 PM