Probability density for mass on a spring

[hotcommodity]

Homework Statement

A particle of mass m is attached to a spring with a spring constant k. The particle is moved a horizontal distance A from the equilibrium point and released from rest. We follow the motion for half a period, that is x \in [-A, A].

Show that if we take snapshots of the particle at random time intervals dt that the probability density for the particle is \rho (x) = \frac{1}{\pi \sqrt{A^2 - x^2}}

Homework Equations

x(t) = Acos(\omega t -\delta)

The Attempt at a Solution

There was a similar example in the textbook about dropping a ball from a given height h. I tried to follow that example, but I'm still getting hung up.

\frac{dx}{dt} = -A \omega sin(\omega t - \delta)

From the example in the book, the probability density is found from \frac{dt}{T}, where T is the total time traveled by the particle.

Rearranging gives, \frac{dt}{T} = \frac{dx}{T} \frac{1}{-A \omega sin(\omega t - \delta)}

I believe the idea is to somehow eliminate the time variable on the right hand side, but I can't get it to look anything like the given answer, nor do I understand where the pi comes from. Any help is appreciated.

 

[gabbagabbahey]

Hint: \sin^2 u+\cos^2 u=1...so \sin(\omega t-\delta)=_____?

 

[hotcommodity]

Oh! Thank you! You're a genius, haha. Maybe I was on summer vacation too long.

Ok so, now I have:

\frac{dt}{T} = \frac{dx}{-T \omega} \frac{1}{\sqrt{A^2 - x^2}}

Does the -T \omega somehow turn into pi? I know omega is 2pi/T, but that leaves a "2" and a negative sign...

 

[gabbagabbahey]

Let's start with the negative sign...As x varies from -A to A, what does the argument of the cosine vary from/to? Is \sin(\omega t -\delta) positive or negative over that interval? (Remeber, \sin^2 u+ \cos^2 u=1 \implies \sin u=\pm\sqrt{1-\cos^2 u} so you need to look at the argument of the cosine in order to fix the sign)

As for the factor of 2, you seem to be confusing the interval of the motion over which you are averaging the probability density (the T in the equation \rho=\frac{dt}{T}) with the period of oscillation (2\pi\omega)...in this case, the interval T is half a period of oscillation, so T=\pi \omega

 

[Feldoh]

Just as a side note, do you go to CWRU?

But T = \frac{\pi}{\omega}