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].
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)?
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]
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.