Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 5: Problem 10

Period of an anharmonic oscillator The simple harmonic oscillator crops up in many places. Its behavior can be studied readily using analytic methods and it has the important property that its period of oscillation is a constant, independent of its amplitude, making it useful, for instance, for keeping time in watches and clocks. Frequently in physics, however, we also come across anharmonic oscillators, whose period varies with amplitude and whose behavior cannot usually be calculated analytically. A general classical oscillator can be thought of as a particle in a concave potential well. When disturbed, the particle will rock back and forth in the well: The harmonic oscillator corresponds to a quadratic potential \(V(x) \propto x^{2}\). Any other form gives an anharmonic oscillator. (Thus there are many different kinds of anharmonic oscillator, depending on the exact form of the potential.) One way to calculate the motion of an oscillator is to write down the equation for the conservation of energy in the system. If the particle has mass \(m\) and position \(x\), then the total energy is equal to the sum of the kinetic and potential energies thus: $$ E=\frac{1}{2} m\left(\frac{\mathrm{d} x}{\mathrm{~d} t}\right)^{2}+V(x) $$ Since the energy must be constant over time, this equation is effectively a (nonlinear) differential equation linking \(x\) and \(t\). Let us assume that the potential \(V(x)\) is symmetric about \(x=0\) and let us set our anharmonic oscillator going with amplitude \(a\). That is, at \(t=0\) we release it from rest at position \(x=a\) and it swings back towards the origin. Then at \(t=0\) we have \(\mathrm{d} x / \mathrm{d} t=0\) and the equation above reads \(E=V(a)\), which gives us the total energy of the particle in terms of the amplitude. a) When the particle reaches the origin for the first time, it has gone through one quarter of a period of the oscillator. By rearranging the equation above for \(\mathrm{d} x / \mathrm{d} t\) and then integrating with respect to \(t\) from 0 to \(\frac{1}{4} T\), show that the period \(T\) is given by $$ T=\sqrt{8 m} \int_{0}^{a} \frac{\mathrm{d} x}{\sqrt{V(a)-V(x)}} $$ b) Suppose the potential is \(V(x)=x^{4}\) and the mass of the particle is \(m=1 .\) Write a Python function that calculates the period of the oscillator for given amplitude a using Gaussian quadrature with \(N=20\) points, then use your function to make a graph of the period for amplitudes ranging from \(a=0\) to \(a=2\). c) You should find that the oscillator gets faster as the amplitude increases, even though the particle has further to travel for larger amplitude. And you should find that the period diverges as the amplitude goes to zero. How do you explain these results?

Short Answer

Expert verified
The period is determined by the integral and increases with amplitude. At zero amplitude, the period diverges.

Step by step solution

01

- Understanding the Energy Conservation

Start by using the energy conservation equation: \[E = \frac{1}{2} m \left(\frac{\mathrm{d} x}{\mathrm{~d} t}\right)^{2} + V(x)\]Here, the kinetic energy is \(\frac{1}{2} m \left(\frac{\mathrm{d} x}{\mathrm{~d} t}\right)^{2}\) and the potential energy is \(V(x)\).
02

- Setting Initial Conditions

At time \(t = 0\), the particle is at rest and positioned at amplitude \(a\). Hence, the velocity \(\frac{\mathrm{d} x}{\mathrm{~d} t}\) is zero, and the total energy is purely potential: \[E = V(a)\]
03

- Rearranging for Velocity

Using the energy conservation equation, rearrange for \(\frac{\mathrm{d} x}{\mathrm{~d} t}\): \[\frac{1}{2} m \left(\frac{\mathrm{d} x}{\mathrm{~d} t}\right)^{2} = V(a) - V(x)\]\[\frac{\mathrm{d} x}{\mathrm{~d} t} = \sqrt{\frac{2(V(a) - V(x))}{m}}\]
04

- Setting up the Integral

Separate variables and integrate over one quarter period (from 0 to \(\frac{T}{4}\)): \[\frac{\mathrm{d} t}{\mathrm{d} x} = \frac{1}{\sqrt{\frac{2(V(a)-V(x))}{m}}} = \sqrt{\frac{m}{2}} \cdot \frac{1}{\sqrt{V(a) - V(x)}}\]Integrate from 0 to \(a\): \[\frac{T}{4} = \sqrt{\frac{m}{2}} \int_{0}^{a} \frac{1}{\sqrt{V(a) - V(x)}} \mathrm{d} x\]
05

- Solving for Period

Multiply both sides by 4 to get the full period: \[T = 4 \sqrt{\frac{m}{2}} \int_{0}^{a} \frac{1}{\sqrt{V(a) - V(x)}} \mathrm{d} x\]Simplify to: \[T = \sqrt{8 m} \int_{0}^{a} \frac{\mathrm{d} x}{\sqrt{V(a) - V(x)}}\]
06

- Python Function for Potential \(V(x) = x^4\)

For \(V(x) = x^4\) and \(m = 1\), write a Python function using Gaussian quadrature to evaluate the period. Here's a sample code:```pythonimport numpy as npfrom scipy.integrate import quad# Define the integranddef integrand(x, a): return 1 / np.sqrt(a**4 - x**4)# Calculate period using Gaussian quadraturedef period(a, N=20): result, _ = quad(integrand, 0, a, args=(a,), limit=N) return np.sqrt(8) * result# Plotting the period for varying amplitudesa_values = np.linspace(0.1, 2, 100)T_values = np.array([period(a) for a in a_values])import matplotlib.pyplot as pltplt.plot(a_values, T_values)plt.xlabel('Amplitude a')plt.ylabel('Period T')plt.title('Period of Anharmonic Oscillator vs Amplitude')plt.show()```
07

- Explanation of Results

As the amplitude increases, the potential energy contribution becomes more significant, allowing the particle to move faster through the larger oscillations. Conversely, as the amplitude approaches zero, the particle's motion slows significantly, causing the period to diverge.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Nonlinear Differential Equations
A linear differential equation has solutions that can be superimposed to form additional solutions. However, for an anharmonic oscillator, we deal with nonlinear differential equations. These equations are more complex and their solutions can't be simply summed up to form new solutions. Nonlinear differential equations often don't have straightforward analytical solutions. This means we frequently have to resort to numerical techniques or approximations to understand the behavior of the system.
Writing the nonlinear differential equations for an anharmonic oscillator begins with the energy conservation principle: \[E = \frac{1}{2} m \left(\frac{\mathrm{d} x}{\mathrm{~d} t}\right)^{2} + V(x)\]
Here, the total energy is the sum of kinetic and potential energies. Rearranging for velocity leads to:
\[ \frac{\mathrm{d} x}{\mathrm{~d} t} = \sqrt{\frac{2(V(a) - V(x))}{m}} \]
This equation links the position derivative with time, making it a nonlinear relationship between position, time, and velocity.
Conservation of Energy
The principle of conservation of energy is pivotal in physics. It states that energy in a closed system is constant over time. For oscillators, the total energy is split between kinetic and potential energy. At amplitude, all the energy is potential: \[E = V(a)\]
As the particle oscillates through its path, energy transforms between kinetic and potential, but the total remains the same. For an anharmonic oscillator, this means the potential energy is not just \(x^2\) but takes a different form, such as \(x^4\).
We use the energy conservation equation to describe the motion. When rearranged, it provides the velocity: \[E = \frac{1}{2} m \left(\frac{\mathrm{d}\, x}{\mathrm{~d}\, t}\right)^{2} + V(x)\]
Transposing terms helps us set up the integral needed to calculate the period of the oscillation.
Gaussian Quadrature
Numerical integration is necessary for solving complex integrals. Gaussian quadrature is a method of approximating the integral of a function. It works by selecting specific points and weights for these points to give an accurate approximation of the integral. This method is efficient and accurate, especially for smooth functions. For our anharmonic oscillator with potential \(V(x) = x^4\), we need to evaluate:
\[T = \sqrt{8 m} \int_{0}^{a} \frac{\mathrm{d}\, x}{\sqrt{V(a) - V(x)}}\]
Using Gaussian quadrature, we approximate this integral in Python. The 'quad' function from the scipy library is helpful here:
```python
import numpy as np
from scipy.integrate import quad
# Define the integrand
def integrand(x, a): return 1 / np.sqrt(a**4 - x**4)
# Calculate period using Gaussian quadrature
def period(a, N=20): result, _ = quad(integrand, 0, a, args=(a,), limit=N) return np.sqrt(8) * result
`
This code snippet sets up the integrand and uses Gaussian quadrature for integration.
Period of Oscillation
For a harmonic oscillator, the period of oscillation is independent of the amplitude. But for an anharmonic oscillator, the period varies with amplitude. To derive the period \(T\), we use the energy conservation equation. Start by isolating the velocity term:
\[ \frac{1}{2} m \left(\frac{\mathrm{d} x}{\mathrm{~d} t}\right)^{2} = V(a) - V(x) \]
Simplify to:
\[ \frac{\mathrm{d} t}{\mathrm{d} x} = \sqrt{\frac{m}{2}} \cdot \frac{1}{\sqrt{V(a) - V(x)}}\]
Integrate from 0 to \(a\):
\[ \frac{T}{4}= \sqrt{\frac{m}{2}} \int_{0}^{a} \frac{1}{\sqrt{V(a) - V(x)}} \mathrm{d} x \]
Multiply by 4 to find the full period:
\[T=\sqrt{8m} \int_{0}^{a} \frac{\mathrm{d} x}{\sqrt{V(a) - V(x)}}\]
This integral provides the framework for determining how the period changes with varying amplitude.
Potential Energy
Potential energy describes the stored energy in a system due to its position, shape, or configuration. For an oscillator, this is tied to its displacement from equilibrium. Harmonic oscillators have potential energy proportional to the square of displacement: \(V(x) \propto x^2\). In contrast, an anharmonic oscillator like the one in our exercise may have a potential energy of the form \(x^4\).
For potential \(V(x) = x^4\), we see:
\[V(x) = x^4\]
The conservation of energy helps us determine the equations governing its motion. Initially, when the particle is released from rest at amplitude \(a\), the total energy is: \(E = V(a) = a^4\)
As it moves through its path, the energy mixes between kinetic and potential forms, but keeps the total constant. Understanding the potential energy form is crucial as it determines the system's behavior and how the period of oscillation changes with amplitude.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Diffraction gratings: Light with wavelength \(\lambda\) is incident on a diffraction grating of total width \(w\), gets diffracted, is focused with a lens of focal length \(f\), and falls on a screen: Theory tells us that the intensity of the diffraction pattern on the screen, a distance \(x\) from the central axis of the system, is given by $$ I(x)=\left|\int_{-w / 2}^{w / 2} \sqrt{q(u)} \mathrm{e}^{i 2 \pi x u / \lambda f} \mathrm{~d} u\right|^{2} $$ where \(q(u)\) is the intensity transmission function of the diffraction grating at a distance \(u\) from the central axis, i.e., the fraction of the incident light that the grating lets through. a) Consider a grating with transmission function \(q(u)=\sin ^{2} \alpha u\). What is the separation of the "slits" in this grating, expressed in terms of \(\alpha\) ? b) Write a Python function \(q(u)\) that returns the transmission function \(q(u)=\sin ^{2} \alpha u\) as above at position \(u\) for a grating whose slits have separation \(20 \mu \mathrm{m}\). c) Use your function in a program to calculate and graph the intensity of the diffraction pattern produced by such a grating having ten slits in total, if the incident light has wavelength \(\lambda=500 \mathrm{~nm}\). Assume the lens has a focal length of 1 meter and the screen is \(10 \mathrm{~cm}\) wide. You can use whatever method you think appropriate for doing the integral. Once you've made your choice you'll also need to decide the number of sample points you'll use. What criteria play into this decision? Notice that the integrand in the equation for \(I(x)\) is complex, so you will have to use complex variables in your program. As mentioned in Section 2.2.5, there is a version of the math package for use with complex variables called cmath. In particular you may find the exp function from cmath useful because it can calculate the exponentials of complex arguments. d) Create a visualization of how the diffraction pattern would look on the screen using a density plot (see Section 3.3). Your plot should look something like this: e) Modify your program further to make pictures of the diffraction patterns produced by gratings with the following profiles: i) A transmission profile that obeys \(q(u)=\sin ^{2} \alpha d \sin ^{2} \beta u\), with \(\alpha\) as before and the same total grating width \(w\), and \(\beta=\frac{1}{2} \alpha\). ii) Two "square" slits, meaning slits with \(100 \%\) transmission through the slit and \(0 \%\) transmission everywhere else. Calculate the diffraction pattern for non-identical slits, one \(10 \mu \mathrm{m}\) wide and the other \(20 \mu \mathrm{m}\) wide, with a \(60 \mu \mathrm{m}\) gap between the two.

Create a user-defined function \(f(x)\) that returns the value \(1+\frac{1}{2} \tanh 2 x\), then use a central difference to calculate the derivative of the function in the range \(-2 \leq x \leq 2\). Calculate an analytic formula for the derivative and make a graph with your numerical result and the analytic answer on the same plot. It may help to plot the exact answer as lines and the numerical one as dots. (Hint: In Python the tanh function is found in the math package, and it's called simply tanh.)

The diffraction limit of a telescope Our ability to resolve detail in astronomical observation is limited by the diffraction of light in our telescopes. Light from stars can be treated effectively as coming from a point source at infinity. When such light, with wavelength \(\lambda\), passes through the circular aperture of a telescope (which we'll assume to have unit radius) and is focused by the telescope in the focal plane, it produces not a single dot, but a circular diffraction pattern consisting of central spot surrounded by a series of concentric rings. The intensity of the light in this diffraction pattern is given by $$ I(r)=\left(\frac{J_{1}(k r)}{k r}\right)^{2} $$ where \(r\) is the distance in the focal plane from the center of the diffraction pattern, \(k=2 \pi / \lambda\), and \(J_{1}(x)\) is a Bessel function. The Bessel functions \(J_{m}(x)\) are given by $$ J_{m}(x)=\frac{1}{\pi} \int_{0}^{\pi} \cos (m \theta-x \sin \theta) \mathrm{d} \theta, $$ where \(m\) is a nonnegative integer and \(x \geq 0\). a) Write a Python function \(J(m, x)\) that calculates the value of \(J_{m}(x)\) using Simpson's rule with \(N=1000\) points. Use your function in a program to make a plot, on a single graph, of the Bessel functions \(J_{0}, J_{1}\) and \(J_{2}\) as a function of \(x\) from \(x=0\) to \(x=20\). b) Make a second program that makes a density plot of the intensity of the circular diffraction pattern of a point light source with \(\lambda=500 \mathrm{~nm}\), in a square region of the focal plane, using the formula given above. Your picture should cover values of \(r\) from zero up to about \(1 \mu \mathrm{m}\).

The Stefan-Boltzmann constant The Planck theory of thermal radiation tells us that in the (angular) frequency interval \(\omega\) to \(\omega+d \omega\), a black body of unit area radiates electromagnetically an amount of thermal energy per second equal to \(I(\omega)\) d \(\omega\), where $$ I(\omega)=\frac{\hbar}{4 \pi^{2} c^{2}} \frac{\omega^{3}}{\left(\mathrm{e}^{\hbar \omega / k_{a} T}-1\right)} $$ Here \(h\) is Planck's constant over \(2 \pi, c\) is the speed of \(h\) ght, and \(k_{B}\) is Boltzmann's constant. a) Show that the total rate at which energy is radiated by a black body per unit area, over all frequencies, is $$ W=\frac{k_{B}^{4} T^{4}}{4 \pi^{2} c^{2} h^{3}} \int_{0}^{\infty} \frac{x^{3}}{\mathrm{e}^{x}-1} \mathrm{~d} x . $$ b) Write a program to evaluate the integral in this expression. Explain what method you used, and how accurate you think your answer is. c) Even before Planck gave his theory of thermal radiation around the turn of the 20 th century, it was known that the total energy \(W\) given off by a black body per unit area per second followed Stefan's law: \(W=\sigma T^{4}\), where \(\sigma\) is the StefanBoltzmann constant. Use your value for the integral above to compute a value for the Stefan-Boltzmann constant (in SI units) to three significant figures. Check your result against the known value, which you can find in books or on-line. You should get good agreement.

5.17 The gamma function: A commonly occurring function in physics calculations is the gamma function \(\Gamma(a)\), which is defined by the integral $$ \Gamma(a)=\int_{0}^{\infty} x^{n-1} \mathrm{e}^{-x} \mathrm{~d} x . $$ There is no closed-form expression for the gamma function, but one can calculate its value for given \(a\) by performing the integral above numerically. You have to be careful how you do it, however, if you wish to get an accurate answer. a) Write a program to make a graph of the value of the integrand \(x^{n-1} \mathrm{e}^{-x}\) as a function of \(x\) from \(x=0\) to \(x=5\), with three separate curves for \(a=2,3\), and 4 , all on the same axes. You should find that the integrand starts at zero, rises to a maximum, and then decays again for each curve. b) Show analytically that the maximum falls at \(x=a-1\). c) Most of the area under the integrand falls near the maximum, so to get an accurate value of the gamma function we need to do a good job of this part of the integral. We can change the integral from 0 to \(\infty\) to one over a finite range from 0 to 1 using the change of variables in Eq. (5.67), but this tends to squash the peak towards the edge of the \([0,1]\) range and does a poor job of evaluating the integral accurately. We can do a better job by making a different change of variables that puts the peak in the middle of the integration range, around \(\frac{1}{2}\). We will use the change of variables given in Eq. (5.69), which we repeat here for convenience: $$ z=\frac{x}{c+x} . $$ For what value of \(x\) does this change of variables give \(z=\frac{1}{2}\) ? Hence what is the appropriate choice of the parameter \(c\) that puts the peak of the integrand for the gamma function at \(z=\frac{1}{2}\) ? d) Before we can calculate the gamma function, there is another detail we need to attend to. The integrand \(x^{n-1} \mathrm{e}^{-x}\) can be difficult to evaluate because the factor \(x^{2-1}\) can become very large and the factor \(\mathrm{e}^{-x}\) very small, causing numerical overflow or underflow, or both, for some values of \(x\). Write \(x^{n-1}=\mathrm{e}^{(a-1) \ln x}\) to derive an alternative expression for the integrand that does not suffer from these problems (or at least not so much). Explain why your new expression is better than the old one. e) Now, using the change of variables above and the value of \(c\) you have chosen, write a user-defined function gamma (a) to calculate the gamma function for arbitrary argument \(a\). Use whatever integration method you feel is appropriate. Test your function by using it to calculate and print the value of \(\Gamma\left(\frac{3}{2}\right)\), which is known to be equal to \(\frac{1}{2} \sqrt{\pi} \simeq 0.886\). f) For integer values of \(a\) it can be shown that \(\Gamma(a)\) is equal to the factorial of \(a-\) 1. Use your Python function to calculate \(\Gamma(3), \Gamma(6)\), and \(\Gamma(10)\). You should get answers closely equal to \(2 !=2,5 !=120\), and \(9 !=362880\).
See all solutions

Recommended explanations on Physics Textbooks

Measurements

Read Explanation

Wave Optics

Read Explanation

Electrostatics

Read Explanation

Solid State Physics

Read Explanation

Medical Physics

Read Explanation

Force

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free