# Pendulum motion and pythagorean theorem

1. Mar 12, 2013

### sunnyday01

1. The problem statement, all variables and given/known data

Show that the relation between the horizontal and vertical components of the ball's position is given by the equation: y = L - [(L^2 - x^2)^1/2]

http://www.flickr.com/photos/94066958@N08/8553595522/in/photostream/

2. Relevant equations

y = L - [(L^2 - x^2)^1/2]

3. The attempt at a solution

I know the solution must involve Pythagorean theorem and drawing a second triangle.
The first triangle has hypotenuse of length L, and other sides are L-y and x. That identity is given by L^2 = (L-y)^2 - x^2.
Drawing a second triangle, the sides are x and y but I don't know what the identity of the hypotenuse is. that equation would be x^2 + y^2 = hypotenuse^2

I don't know where to go from there...
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

#### Attached Files:

• ###### pendulum equation question.jpg
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2. Mar 12, 2013

### SammyS

Staff Emeritus
Hello sunnyday01. Welcome to PF !

If a & b are legs of a right triangle with hypotenuse, c, then
c2 = a2 + b2

In your equation, L^2 = (L-y)^2 - x^2, why do you have the sign between. (L-y)2 and x2 as a minus sign ?

3. Mar 12, 2013

### sunnyday01

formula correction

It should be:
L2 = (L-y)2 + x2

4. Mar 12, 2013

### SammyS

Staff Emeritus
Yes.

... and that should be all you need to get the desired result.

5. Mar 12, 2013

### sunnyday01

I think I figured it out!

so, I think:
L2 = (L-y)2 + x2
L2-x2 = (L-y)2
(L2-x2)1/2 = L - y
-(L2-x2)1/2 = - L + y
L -(L2-x2)1/2 = y

I think that's it! Having to crop a picture for this question made me focus only on the variables I needed in the diagram.

I see the solution now is really simple, I feel silly for not seeing it sooner. Thank you for being so nice about it!