Probability of observing harmonic oscillator at a particular position Consider a classical harmonic oscillator whose Hamiltonian is
$$H=\frac{p^2}{2m} +\frac{1}{2}mw^2x^2$$
where $w$ is the oscillating frequency.
I wish to find the probability of observing the harmonic oscillator at a particular position $x$. This should be inversely proportional to the speed of the oscillator at $x$, i.e.
$$P(x)\propto \frac{1}{v(x)}.$$
If the amplitude of the oscillation is $x_0$, the speed of the harmonic oscillator at $x$  would be
$$v(x)=w(x_0^2 - x^2)^{\frac{1}{2}}.$$
Hence the probability of the oscillator being observed at $x$ is
$$P(x) \propto {1 \over w(x_0^2 - x^2)^{\frac{1}{2}}}.$$
There is one problem though. The probability to find $x=x_0$ would be infinite. Is there any way to work around this infinity?
 A: When dealing with continuous distributions, one should make difference between probability and probability density. Probability of hitting any specific point of a continuous distribution is zero, since the number of points in a continuum is infinite. Probability of getting a point in an interval $[x,x+\Delta x]$ is finite:
$$
P(x<X<x+\Delta x)=\int_{x}^{x+\Delta x}dxp(x)\approx p(x)dx
$$
Now, the oscillator probability density in the OP takes infinite values at the turning points, $x= \pm x_0$, however, it diverges at these points as $\sim \frac{1}{\sqrt{x_0-x}}$, that is the integral of this probability density is convergent.
One could also use the cumulative distribution function
$$
F(x)=\int_{-\infty}^xdxp(x),
$$
which gives the probability of hitting any point smaller than $x$. The probability of hitting an interval is then
$$
P(x<X<x+\Delta x)=F(x+\Delta x) - F(x).
$$
In our case
$$
F(x)=\int_{-x_0}^xdxp(x)=\frac{1}{2} + \frac{1}{\pi}\arcsin\left(\frac{x}{x_0}\right)
$$
(The normalization constant $w$ is fixed to $w=\pi$ by requirement that the full probability of hitting a point between $-x_0$ and $x_0$ is $F(x_0)=1$.)
