Problem Solving Strategies

@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.

Prateek Mourya

Yes i was thinking for here it will be helpful to learn it

Ok back to the pendulum

John Rennie

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?

Prateek Mourya

8:27 AM

Ok

John Rennie

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?

Prateek Mourya

Ok

John Rennie

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?

Prateek Mourya

Ok

Pretty easy so far

John Rennie

So the motion in the x and y directions is independent. We can write the equation of motion as:

$$ x(t) = A\sin(\omega t + \phi_x) $$
$$ y(t) = B\sin(\omega t + \phi_y) $$

Prateek Mourya

8:32 AM

I don't know SHM so much but I know that the pendulum is periodic motion

John Rennie

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

Prateek Mourya

....

Bad luck

John Rennie

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

Prateek Mourya

Right now my coaching is teaching thermodynamics and heat

No sir i have other subjects too

I come her in brake time

John Rennie

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

Prateek Mourya

8:34 AM

How lkng will it take though?

John Rennie

10 minutes

Prateek Mourya

To lean shm?

John Rennie

Yes. It's really simple.

Prateek Mourya

Can we continue after 2 hours?

John Rennie

Yes, I'll be here.

Prateek Mourya

8:35 AM

Thanks alot sir

John Rennie

:-)

John Rennie

3:50 PM

Deleted

John Rennie

5:00 PM

@PrateekMourya hi :-)

Prateek Mourya

Sorry for being super late

Meanwhile i read some of the portion of shm

John Rennie

How did you get on?

Prateek Mourya

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

John Rennie

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.

Prateek Mourya

Yes

Is x^2 be too a shm

John Rennie

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.

Prateek Mourya

Ok

Is the information i learned enough for pendulum problem

John Rennie

Well, do you understand how to describe the motion of a simple 1D pendulum?

i.e. how you solve the equation of motion?

Prateek Mourya

Honestly no

Since i haven't practiced anything

John Rennie

Well we know that F = -kx. Yes?

Prateek Mourya

I will watch a video on YouTube

John Rennie

5:08 PM

We can do it now - it's really simple.

Prateek Mourya

Ok

John Rennie

So we start from F = -kx. Yes?

Prateek Mourya

Ok

John Rennie

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?

Prateek Mourya

Ok

John Rennie

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.

Prateek Mourya

Ok

John Rennie

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

Prateek Mourya

Ok

John Rennie

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?

Prateek Mourya

Ok

John Rennie

5:16 PM

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

Prateek Mourya

Ok

John Rennie

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.

Prateek Mourya

Wait

Lemme switch to laptop

ok

John Rennie

You can see all the MathJax now?

Prateek Mourya

yes

John Rennie

5:23 PM

Does it all make sense?

Prateek Mourya

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

John Rennie

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

Prateek Mourya

ok

John Rennie

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

Prateek Mourya

yes

John Rennie

5:26 PM

Do you want to look at that 2D pendulum now?

Prateek Mourya

yes

John Rennie

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?

Prateek Mourya

ok

John Rennie

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?

Prateek Mourya

ok

wait

isnt the mg downward

you mean componet of tension

John Rennie

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?

Prateek Mourya

yes

John Rennie

And $\cos\theta = x/r$

So that means $F_x = -kr(x/r) = -kx$

And likewise $F_y = -ky$. Yes?

Prateek Mourya

is f tension here

John Rennie

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

Prateek Mourya

or we just rotated our coordinate axes a bit

so that its 2d pendulum

just in a new plane

John Rennie

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.

Prateek Mourya

@JohnRennie ok

ok

John Rennie

What we have found is that the position (x,y) of the bob in the xy plane is determined by the two equations:

$$ \frac{d^2x}{dt^2} = - \frac{k}{m} x $$

$$ \frac{d^2y}{dt^2} = - \frac{k}{m} y $$

Prateek Mourya

ok

John Rennie

So the solutions are going to be:

$$ x(t) = A \sin\left(\sqrt{\frac{k}{m}} t + \phi_x\right) $$
$$ y(t) = B \sin\left(\sqrt{\frac{k}{m}} t + \phi_y\right) $$

Prateek Mourya

ok

John Rennie

5:38 PM

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

Prateek Mourya

ok

John Rennie

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

Prateek Mourya

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

John Rennie

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.

Prateek Mourya

then given velocity

perp

to the plane

so sing angle is pi/2

John Rennie

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.

Prateek Mourya

gravity in z axis

John Rennie

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.

Prateek Mourya

ok from diagam

i understand the oversll structurer

John Rennie

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

Prateek Mourya

ya i was about to say th same

but for chester

John Rennie

5:53 PM

It's about 6 p.m. here

But I started work at 5 a.m. :-)

So I'm getting tired now.

Transcript for

Problem Solving Strategies