Question

(a) Find the probability density function \(f(x)\) 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 \(x=a\) cos \(\omega t .\) Find the velocity \(d x / d t ;\) then the probability of finding the particle in a given \(d x\) 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 \(f(x)\) found in part (a) and also the cumulative distribution function \(F(x) \text { [see equation }(6.4)]\). (c) Find the average and the standard deviation of \(x\) in part (a).

Short answer

The probability density function is \( f(x) = \frac{1}{ \pi \sqrt{ a^2 - x^2 } } \). The mean is 0, and the standard deviation is \( \frac{a}{\sqrt{2}} \)

Step-by-step solution

Identify Position with Respect to Time

The position of the particle as a function of time in simple harmonic motion is given by: \[ x = a \, \text{cos}(\omega t) \]

Determine Velocity

The velocity of the particle is the time derivative of the position. Therefore, \[ \frac{dx}{dt} = -a \, \text{sin}(\omega t) \, \omega \]

Express Velocity in Terms of Position

Express the velocity in terms of position rather than time. Recall that \[ \text{cos}^2(\omega t) + \text{sin}^2(\omega t) = 1 \]. Thus, from \[ x = a \, \text{cos}(\omega t) \], we get: \[ \text{cos}(\omega t) = \frac{x}{a} \implies \text{sin}(\omega t) = \sqrt{1 - \left( \frac{x}{a} \right)^2 } \]. So, \[ \frac{dx}{dt} = -\omega \sqrt{ a^2 - x^2 } \]

Determine Time Spent in an Interval

The time spent in a small interval dx is inversely proportional to the speed at that point. \[ dt = \frac{dx}{ \left| \frac{dx}{dt} \right| } = \frac{dx}{ \omega \sqrt{ a^2 - x^2 } } \]

Find Probability Density Function

The probability density function is proportional to the time the particle spends in a given interval dx. Normalize by ensuring total probability over the range is 1. Set up the integral:\[ f(x) = \frac{1}{ T } \cdot \frac{1}{ \omega \sqrt{ a^2 - x^2 } } \]. Where T is chosen such that the integral over the interval \( -a \le x \le a \) equals 1:\[ \int_{-a}^{a} \frac{1}{ \omega \sqrt{ a^2 - x^2 } } dx = 1 \]. Solving this integral, \[ T = \frac{ \pi }{ \omega } \]. Therefore, \[ f(x) = \frac{1}{ \pi \sqrt{ a^2 - x^2 } } \]

Sketch the Probability Density Function

Plot the function \[ f(x) = \frac{1}{ \pi \sqrt{ a^2 - x^2 } } \] within the range \( -a \le x \le a \), with peaks at \( x = \pm a \)

Sketch the Cumulative Distribution Function

Integrate the probability density function f(x) to get the cumulative distribution function F(x) from x = -a to any x within \( -a \le x \le a \). Use the indefinite integral result and verify that \[ F(x) = \frac{1}{2} + \frac{1}{ \pi } \text{arcsin} \left( \frac{x}{a} \right) \]

Calculate the Average (Mean)

To find the mean, use the integral: \[ \langle x \rangle = \int_{-a}^{a} x f(x) dx \]. For symmetric functions, the mean is zero: \[ \langle x \rangle = 0 \]

Calculate the Standard Deviation

Find the standard deviation, which requires the variance: \[ \sigma^2 = \langle x^2 \rangle - \langle x \rangle^2 \]. Find \( \langle x^2 \rangle \): \[ \langle x^2 \rangle = \int_{-a}^{a} x^2 f(x) dx \]. With symmetry and normalization, results in:\[ \sigma^2 = \frac{ a^2 }{2} \implies \sigma = \frac{a}{\sqrt{2}} \]. Therefore, the standard deviation is \( \frac{a}{\sqrt{2}} \)

Key concepts

Simple Harmonic Motion

Simple harmonic motion (SHM) is a type of periodic motion where the restoring force is directly proportional to the displacement. Imagine a particle moving back and forth along the x-axis between positions -a and a. In this case, the position of the particle can be described by the function \[ x = a \cos (\omega t) \] Here, a is the amplitude (maximum displacement), and \( \omega \) is the angular frequency (rate of oscillation). The particle follows a sinusoidal path, reversing direction at the end points -a and a. The motion can be visualized as a smooth, continuous oscillation that repeats itself over time.

Probability Density Function

In the context of simple harmonic motion, the probability density function (PDF) \( f(x) \) describes the likelihood of finding the particle at a particular position x. The key idea is that the particle spends more time near the extreme positions (±a) where its speed is lower, and less time near the center (x=0) where its speed is higher. To derive the PDF for the position of the particle:

• First, determine the velocity of the particle using \( \frac{dx}{dt} = -a \sin(\omega t) \omega \).
• Express the velocity in terms of position: \( \frac{dx}{dt} = -\omega \sqrt{a^2 - x^2} \).
• The time spent in a small interval dx is inversely proportional to the speed: \( dt = \frac{dx}{ \omega \sqrt{a^2 - x^2} } \).
• To find the PDF, normalize this expression over the entire range \( -a \le x \le a \): \( f(x) = \frac{1}{ \pi \sqrt{a^2 - x^2} } \).

Cumulative Distribution Function

The cumulative distribution function (CDF) \( F(x) \) represents the probability that the particle is found at a position less than or equal to x. To obtain the CDF:

• Integrate the PDF from the lower bound of the interval to any position x: \( F(x) = \int_{-a}^{x} f(x') dx' \).
• The indefinite integral of the PDF results in: \( F(x) = \frac{1}{2} + \frac{1}{ \pi } \text{arcsin} \left( \frac{x}{a} \right) \).
• This function smoothly transitions from 0 to 1 as x moves from -a to a.

This cumulative representation helps us understand the likelihood of the particle being at or moving past any specific position.

Mean

The mean (or average) position of a particle undergoing SHM can be calculated using the integral: \( \langle x \rangle = \int_{-a}^{a} x f(x) dx \). For symmetric functions like SHM, the mean is generally zero due to the balance of positive and negative positions around the center. Thus, \( \langle x \rangle = 0 \). This result makes intuitive sense—the particle spends equal time on either side of the center position (zero), leading to an average position at the midpoint of the oscillation.

Standard Deviation

The standard deviation \( \sigma \) provides a measure of how much the particle's position varies from its mean position. It requires calculating the variance: \( \sigma^2 = \langle x^2 \rangle - \langle x \rangle^2 \).

• First, calculate \( \langle x^2 \rangle \) using: \( \langle x^2 \rangle = \int_{-a}^{a} x^2 f(x) dx \).
• For SHM, this results in \( \sigma^2 = \frac{a^2}{2} \).
• Hence, the standard deviation is \( \sigma = \frac{a}{\sqrt{2}} \).

This value tells us that the positions of the particle are more spread out as the amplitude increases, giving an intuitive way to understand the range of motion in SHM.