How Does a Pendulum Bob's Speed Relate to Its Displacement Angle?

[ubergewehr273]

Homework Statement

Show that a simple pendulum bob which has been pulled aside from its equilibrium position through an angle $\theta$ and then released will pass through the equilibrium position with speed $v = \sqrt{2gl(1-cos\theta)}$, where $l$ is the length of the pendulum.

Homework Equations

$K.E = \frac{1}{2}mv^2$
$P.E = mgh$

The Attempt at a Solution

I tried a lot in finding the start of the solution but I have no idea from where to start.

 

[Satvik Pandey]

Hi Ashes. You have to use conservation of energy here. Try it.

 

[ehild]

Draw a picture first.

 

[ubergewehr273]

Here is the diagram.

 

[Satvik Pandey]

Here is the diagram.

The diagram is correct. Try to apply law of conservation of energy. What are the values of initial and final kinetic and potential energies?

 

[Maged Saeed]

You may find some ambiguity of the term
L(1-cos theta)

But this term is the horizontal distance from the equilibrium position of the bob to any position the bob can occupy .

:)

 

[ehild]

No, it is not the horizontal distance. What is the height of the bob above the deepest position, when it deflects by angle theta from the vertical? See picture. Find h .

 

[Maged Saeed]

Oh sorry ,,
it is the height though..

 

[ubergewehr273]

No, it is not the horizontal distance. What is the height of the bob above the deepest position, when it deflects by angle theta from the vertical? See picture. Find h .

I got it!
$cos\theta=\frac{l-h}{l}$
$\Rightarrow l-h=lcos\theta$
$\Rightarrow h=l(1-cos\theta)$

Since, the pendulum starts moving from the non - equilibrium position, its initial velocity would be zero.
So, the final velocity $v$ is given by,
$v=\sqrt{2gh}$ where $g$ is acceleration due to gravity.
Also, the maximum velocity is achieved at the equilibrium position.
So, substituting for $h$ we get,
$v=\sqrt{2gl(1-cos\theta)}$
Thanks a lot ehild.

 

[ehild]

So, substituting for $h$ we get,
$v=\sqrt{2gl(l-cos\theta)}$
Thanks a lot ehild.

There is a typo in the final formula. It has to be $v=\sqrt{2gl(1-cos\theta)}$

Anyway, you understood and you did it, congrats! :)

 

[ubergewehr273]

Thanks a lot for the clue ehild.
P.S I got the typo fixed.