Question

(a) Find the probability density function $f\left(x\right)$for the position x of a particle which is executing simple harmonic motion on $(-a,a)$along the x axis. (See Chapter $7$ , Section $2$ , for a discussion of simple harmonic motion.) Hint: The value of x at time t is $x=acos\omega t$. Find the velocity $\frac{dx}{dt}$ ; then the probability of finding the particle in a given $dx$ is proportional to the time it spends there which is inversely proportional to its speed there. Don’t forget that the total probability of finding the particle somewhere must be $1$.

(b) Sketch the probability density function $f\left(x\right)$found in part (a) and also the cumulative distribution function $f\left(x\right)$ [see equation $(6.4)$].

(c) Find the average and the standard deviation of x in part (a).

Short answer

The required values are mentioned below.

$\begin{array}{c}f\left(x\right)=\frac{1}{\pi \sqrt{a^2-x^2}}\\ \mu =0\\ \mathrm{Var}{\left(\mathrm{x}\right)}_{-\mathrm{a}}^{\mathrm{a}}=\frac{a^2}{2}\\ \sigma =\frac{a}{\sqrt{2}}\end{array}$

Step-by-step solution

Given Information

A particle is executing a simple harmonic motion.

Definition of the probability density function.

a continuous random variable,whose integral across an interval offers the likelihood that the variable's value falls inside the same interval.

Solve part (a).

Function x is given as .

$\begin{array}{c}v=\frac{dx}{dt}\\ =-a\omega \sin \left(\omega t\right)\\ =-a\omega \sqrt{1-{\cos }^2\left(\omega t\right)}\end{array}$

The value of $\frac{dx}{dt}$is given below.

$\begin{array}{c}\frac{dx}{dt}=-a\omega \sqrt{1-{\left(\frac{x}{a}\right)}^2}\\ =-\omega \sqrt{a^2-x^2}\end{array}$

$\begin{array}{c}dt=\frac{dx}{v}\\ =-\frac{dx}{\omega \sqrt{a^2-x^2}}\end{array}$

The probability of function is given below.

$\begin{array}{c}f\left(x\right)dx=\frac{dx}{\sqrt{a^2-x^2}}\\ f\left(x\right)=\frac{C}{\sqrt{a^2-x^2}}\end{array}$

Find the value of constant.

$\begin{array}{c}{\int }_{-a}^a\frac{Cdx}{\sqrt{a^2-x^2}}=C{\left[\arccos \left(\frac{x}{a}\right)\right]}_{-a}^a\\ =C\pi \\ C\pi =1\\ C=\frac{1}{\pi }\end{array}$

Substitute the value of C the function becomes as follows.

$f\left(x\right)=\frac{1}{\pi \sqrt{a^2-x^2}}$

Solve part (b).

The graphs are shown below.

Solve part (c).

The average of the function is mentioned below

$\begin{array}{c}\mu ={\int }_{-\infty }^{\infty }xf\left(x\right)dx\\ ={\int }_{-\infty }^{\infty }\frac{xdx}{\pi \sqrt{a^2-x^2}}\\ ={\left[-\frac{\sqrt{a^2-x^2}}{\pi }\right]}_{-\infty }^{\infty }\\ =0\end{array}$

The variance is given below.

$\begin{array}{c}\mathrm{Var}\left(x\right)={\int }_{-\infty }^{\infty }{x}^{2}f\left(x\right)dx\\ ={\int }_{-a}^a\frac{x^{2}dx}{\pi \sqrt{a^2-x^2}}\\ ={\left[\frac{a^2\arcsin \left(\frac{x}{\left|a\right|}\right)-x\sqrt{a^2-x^2}}{2\pi }\right]}_{-a}^a\\ =\frac{a^2}{2}\end{array}$

The standard deviation is given below.

$\begin{array}{c}\sigma =\sqrt{\mathrm{var}\left(x\right)}\\ =\sqrt{\frac{a^2}{2}}\\ =\frac{a}{\sqrt{2}}\end{array}$

The required values are mentioned below.

$\begin{array}{c}f\left(x\right)=\frac{1}{\pi \sqrt{a^2-x^2}}\\ \mu =0\\ \mathrm{Var}\left(\mathrm{x}\right){}_{-\mathrm{a}}^{\mathrm{a}}=\frac{a^2}{2}\\ \sigma =\frac{a}{\sqrt{2}}\end{array}$