A particle performs harmonic oscillations along the `x` axis according to the law` x=a cos omega t`. Assuming the probablity `P` of the particle to fall withing an interval from `-a` to `+a` to be equal to unity, find how the probability density `dP//dx` depends on `x`. Here `dP` denotes the probablitity of the particle falling within an interval frm `x` to `x+ dx`. Plot `dP//dx` as a function of `x`.

Explanation

As the motion is periodic the particle repeatedly passes through any given region in the range `-a le xlea`. The probability that it lies in the range `(x,x+dx)` is defined as the fraction `(Deltat)/(t)` ( as `t rarr oo)` where `Deltat` is the time the particle lies in the range `(x,x+dx)` out of the total time `t`. Because of periodicity this is
`dP=(dP)/(dx)dx=(dt)/(T)=(2 dx)/(dT)`
where the factor 2 is needed to take account of the fact that the particl is in the range `(x, x+dx)` during both up and down phases of its motion. Now in a harmonic oscillator.
`v=x=omega cos omegat=omegasqrt(a^(2)-x^(2))`
Thus since `omegaT=2pi(T` is the time period )
We get
`dP=(dP)/(dx)dx=(1)/(pi)(dx)/(sqrt(a^(2)-x^(2)))`
Noe that
`int_(-a)^(+a)(dP)/(dx)dx=1`
so` (dP)/(dx)=(1)/(pi)(1)/(sqrt(a^(2)-x^(2)))` is properly normalized.