Needle probability distribution problem

[Elwin.Martin]

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.

 

[anathema]

Thank you for your post. Your approach for part b is on the right track, but there are a few errors in your calculations. Let's go through it step by step:

1. First, you correctly identified that the x-coordinate of the needle point can be represented as x = rcos(theta), where r is the length of the needle.

2. Next, you found the inverse relationship between x and theta, which is arccos(x) = theta.

3. However, your next step of finding d(theta) is incorrect. Remember that theta is a function of x, not the other way around. So the correct expression for d(theta) is d(theta) = dx/(-r*sin(theta)).

4. Now, to find the probability function, we need to use the formula p(x) = p(theta) * |d(theta)/dx|. Here, p(theta) is the probability function for theta that you found in part a, which is 1/(2*pi).

5. Plugging in the correct expression for d(theta) and using the chain rule, we get p(x) = (1/(2*pi)) * |-1/(r*sin(arccos(x)))|.

6. Simplifying this, we get p(x) = (1/(2*pi)) * (1/(r*sqrt(1-x^2))).

7. Finally, we need to multiply this by a factor of 1/r to account for the length of the needle, giving us the final expression for p(x) = (1/(2*pi*r^2)) * (1/sqrt(1-x^2)).

I hope this helps clarify your approach. Keep in mind that this is only one possible solution and there may be other valid approaches as well. Best of luck with your studies.