Question

The equation of motion of classical harmonic oscillator is expressed by x=asin(omega)t.Show that probability of finding the particle between x and x+dx is given by,
P(x) dx=(dx)÷ (π√[d^2-x^2])

Answer 1

As per the question,

wave equation is given as $x=a\sin(\omega)t$

Let $dP=\dfrac{2dt}{T}-------------(i)$

Where T is the time period and dt is the time taken in x to x+dx

$x=a\sin(\omega)t$

$dx=a\omega \cos(\omega t)dt$

$\Rightarrow dt=\dfrac{dx}{a\omega \cos(\omega t)}$

now

from the equation (i),

$dP=\dfrac{2dx}{a\omega\cos(\omega t)T}$

we know that $\omega=\dfrac{2\pi}{T}$

$\Rightarrow dP=\dfrac{2dx}{a\dfrac{2\pi}{T}\cos(\omega t)T}=\dfrac{2dx}{2\pi a\cos(\omega t)}$

now we know that $x=a\sin(\omega)t$

$\sin(\omega t)=\dfrac{x}{a}$

$\cos(\omega t)=\sqrt{1-\sin^2(\omega t)}$

$\Rightarrow \cos(\omega t)=\sqrt{1-(\dfrac{x}{a})^2}$

$\Rightarrow dP=\dfrac{2dx}{2\pi a\cos(\omega t)}$

$\Rightarrow dP=\dfrac{2dx}{2\pi a\sqrt{1-(\dfrac{x}{a})^2}}$

$\Rightarrow dP=\dfrac{dx}{\pi \sqrt{a^2-(x)^2}}$

$\Rightarrow \dfrac{dP}{dx}=\dfrac{1}{\pi \sqrt{a^2-x^2}}$