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## Homework Statement

part a(complete)

The needle on a broken car speedometer is free to swing, and bounces

perfectly off the pins at either end, so that if you give it a flick it is equally likely to

come to rest at any angle between 0 and [tex]\pi[/tex].

part b(attempting)

We consider the same device as the previous problem, but this time

we are interested in the x-coordinate of the needle point--that is, the "shadow", or

"projection", of the needle on the horizontal line.

1. what is [tex]\rho[/tex](x)?

## Homework Equations

x=rcos([tex]\theta[/tex])

## The Attempt at a Solution

assuming that my solution for the probability function for [tex]\theta[/tex] is correct, (1/[tex]\theta[/tex])

(i'm going to cheat and use a unit needle for a second)

I took x=cos([tex]\theta[/tex])

and found arccos(x)=[tex]\theta[/tex]

and d[tex]\theta[/tex]=-dx/([tex]\sqrt{1-x^{2}}[/tex]

but I am unsure of the usefulness of this relationship

so I took dx=-sin([tex]\theta[/tex])d[tex]\theta[/tex]

and got p(x)dx=(-dx/[tex]\pi[/tex])(sin(arccos(x)))^-1

is this reasonable? please help me see where I've made some sort of foolish assumption.

thank you for your time, I know that many of you going through the homework forums have better things to do. This is not actually homework, but for the purposes of this forum I figured it would best be placed here, I apologize if it was placed in error.

I believe that this works out for a needle of length r if I just substitute a (1/r) in front of everything but I am not positive. If any of you recognize this, it's from one of the earlier versions of Griffith's QM, problem 1.4.