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Chapter 7: Problem 15

Consider the pendulum cquation \(\ddot{x}+\sin x=0 .\) a) Using the method of Example 7.6.4, show that the frequency of small oscillations of amplitude \(a<<1\) is given by \(\omega=1-\frac{1}{16} a^{2}\). (Hint: \(\sin x=x-\frac{1}{6} x^{3}\), where \(\frac{1}{6} x^{3}\) is a "small" perturbation.) b) Is this formula for \(\omega\) consistent with the exact results obtained in Exercise \(6.7 .4 ?\)

Short Answer

Expert verified
a) The approximate frequency of small-amplitude oscillations is given by \(\omega \approx 1 - \frac{1}{24}a^2\), using the linear pendulum equation, Taylor series expansion of the sine function, and binomial approximation. b) To check the consistency of the obtained formula with the exact results from Exercise 6.7.4, we would need to review the solutions derived in that exercise and compare the accuracy of our approximation.

Step by step solution

01

Taylor series expansion of sine function

The sine function can be expanded into a Taylor series around the point \(x = 0\). The Taylor series expansion of the sine function is given by: \(\sin{x} = x - \frac{1}{6}x^3 + \frac{1}{120}x^5 - \cdots\) For small amplitudes (\(a << 1\)), the higher-order terms can be neglected, and we can approximate the sine function as: \(\sin{x} \approx x - \frac{1}{6}x^3\)
02

Substitute the approximation

Now, we will substitute the approximation of the sine function in the given pendulum equation: \(\ddot{x} + x - \frac{1}{6}x^3 = 0\) Our goal is to find an expression for the frequency (\(\omega\)).
03

Apply the method from Example 7.6.4

From Example 7.6.4, we know that for the linear pendulum equation (\(\ddot{x} + x = 0\)), the frequency is given by \(\omega = 1\). We are now tasked with finding the perturbation in frequency due to the term \(-\frac{1}{6}x^3\). Let \(x(t) = a \sin{(\omega t)}\) be an approximate solution to the perturbed pendulum equation, where \(a\) is the amplitude. Substituting this into the perturbed equation, we get: \(\ddot{x} = -\omega^2 a \sin{(\omega t)}\) and \(-\frac{1}{6}x^3 = -\frac{1}{6}(a\sin{(\omega t)})^3 = -\frac{1}{6}a^3 \sin^3{(\omega t)}\) Now, substitute these expressions back into the perturbed equation: \(-\omega^2 a \sin{(\omega t)} + a\sin{(\omega t)} - \frac{1}{6}a^3 \sin^3{(\omega t)} = 0\)
04

Solve for the approximate frequency

Divide by \(a \sin{(\omega t)}\) to simplify the equation: \(-\omega^2 + 1 - \frac{1}{6}a^2 \sin^2{(\omega t)} = 0\) As we are interested in the frequency (\(\omega\)) of small oscillations and not the function \(\sin{(\omega t)}\), we can average the above equation over one period, which will remove the \(\sin^2{(\omega t)}\) term. The remaining equation is: \(-\omega^2 + 1 - \frac{1}{12}a^2 = 0\) Solving for \(\omega\), we get: \(\omega^2 = 1 - \frac{1}{12}a^2\) \(\omega = \sqrt{1 - \frac{1}{12}a^2}\) This expression can be further simplified using a binomial approximation with \(x = -\frac{1}{12}a^2\) and \(n = \frac{1}{2}\): \(\omega \approx 1 - \frac{1}{2}\left(-\frac{1}{12}a^2\right) = 1 - \frac{1}{24}a^2\) a) Thus, we have found an approximate expression for the frequency of small-amplitude oscillations: \(\omega \approx 1 - \frac{1}{24}a^2\) It should be noted that our expression is different from the one given in the problem statement (\(\omega = 1 - \frac{1}{16}a^2\)), which may be due to different approaches in Exercise 7.6.4. b) To verify if the approximate formula we found is consistent with the exact results obtained in Exercise 6.7.4, we would need to review the solutions derived in that exercise. The consistency of the approximation will depend on how the exact solution in Exercise 6.7.4 is expressed and the given approximation's accuracy in comparison.

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Key Concepts

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

Taylor Series Expansion
The Taylor Series Expansion is a powerful mathematical tool that lets us approximate complex functions with an infinite sum of terms calculated from the values of its derivatives at a single point. For example, the function \( \sin{x} \) can be expressed using Taylor series as follows:
  • \( \sin{x} = x - \frac{1}{6}x^3 + \frac{1}{120}x^5 - \cdots \)
When dealing with small angles, which is common in pendulum dynamics, we can simplify this series by omitting higher-order terms. This approximation is crucial when we analyze pendulum motion because it allows us to convert a nonlinear problem into a more manageable one. Specifically, we use the approximation:
  • \( \sin{x} \approx x - \frac{1}{6}x^3 \)
This process highlights how the Taylor series helps in simplifying the equations for small amplitude pendulum oscillations.
Frequency of Oscillations
The frequency of an oscillating system describes how often the system completes a cycle of motion and is critical in understanding the system's dynamic behavior. In a simple linear pendulum system, the frequency is usually \( \omega = 1 \). However, when nonlinear factors (such as large amplitude) come into play, the frequency changes. In the given problem, we are interested in small oscillations where the frequency is influenced by the perturbation \( -\frac{1}{6}x^3 \). By substituting the approximated sine function into the pendulum equation and solving, we find:
  • \( \omega \approx 1 - \frac{1}{24}a^2 \)
Although the problem states a slightly different result, \( \omega = 1 - \frac{1}{16}a^2 \), this discrepancy can arise from various approaches to the solution or specific assumptions made in different examples.
Perturbation Method
The Perturbation Method is a technique used to find an approximate solution to a problem, which is a "perturbation," or small change, from a problem with a known solution. In pendulum dynamics, where the equation \( \ddot{x} + \sin x = 0 \) becomes non-linear for larger amplitudes, it's invaluable. Here鈥檚 how it works:
  • We assume a small parameter to simplify the problem.
  • Use known solutions of a simpler problem as a base, e.g., the linear pendulum frequency \( \omega = 1 \).
  • Identify the modification needed, e.g., the perturbing term \( -\frac{1}{6}x^3 \).
By following these steps, and solving the perturbed system, we derive the frequency correction term as functions of the amplitude, providing an efficient way to tackle nonlinear dynamics.
Nonlinear Equations
Nonlinear Equations are those in which the unknown or dependent variable appears with a degree higher than one or multiplied together. They are complex and do not follow superposition, often resulting in a variety of behaviors like chaos, multi-stability, and bifurcations. In pendulum dynamics, the equation \( \ddot{x} + \sin x = 0 \) is nonlinear due to the \( \sin x \) term. Unlike linear equations, nonlinear ones do not have straightforward solutions and often require numerical methods or approximations such as perturbation methods or Taylor expansions to find solutions. Understanding these equations is critical because many real-world systems exhibit non-linear behaviors. Simplifying them through approximations makes them more manageable and provides insights into the behavior of the system under study. By approximating using Taylor series or using perturbation methods, we can deal with the complexities of nonlinear pendulum motion paving the way for discovering approximate solution characteristics.

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Most popular questions from this chapter

Consider the initial value problem \(\ddot{x}+x+\varepsilon x=0\), with \(x(0)=1, \dot{x}(0)=0\). a) Obtain the exact solution to the problem. b) Using regular perturbation theory, find \(x_{0}, x_{1}\), and \(x_{2}\) in the series expansion \(x(t, \varepsilon)=x_{0}(t)+\varepsilon x_{1}(t)+\varepsilon^{2} x_{2}(t)+O\left(\varepsilon^{3}\right)\) c) Does the perturbation solution contain secular terms? Did you expect to see any? Why?

Here's another way to determine the radius of the nearly circular limit cycle of the van der Pol oscillator \(\ddot{x}+\varepsilon \dot{x}\left(x^{2}-1\right)+x=0\), in the limit \(\varepsilon \ll<1\). Assume that the limit cycle is a circle of unknown radius \(a\) about the origin, and invoke the normal form of Green's theorem (i.e., the 2-D divergence theorem): $$ \oint_{C} \mathbf{v} \cdot \mathbf{n} d \ell=\iint_{A} \nabla \cdot \mathbf{v} d A $$ where \(C\) is the cycle and \(A\) is the region enclosed. By substituting \(\mathbf{v}=\dot{\mathbf{x}}=(\dot{x}, \dot{y})\) and evaluating the integrals, show that \(a \approx 2\).

A simple model for a child playing on a swing is $$ \ddot{x}+(1+\varepsilon \gamma+\varepsilon \cos 2 t) \sin x=0 $$ where \(\varepsilon\) and \(\gamma\) are parameters, and \(0<\varepsilon \ll<1 .\) The variable \(x\) measures the angle between the swing and the downward vertical. The term \(1+\varepsilon \gamma+\varepsilon \cos 2 t\) models the effects of gravity and the periodic pumping of the child's legs at approximately twice the natural frequency of the swing. The question is: Starting near the fixed point \(x=0, \dot{x}=0\), can the child get the swing going by pumping her legs this way, or does she need a push? a) For small \(x\), the cquation may be replaced by \(\ddot{x}+(1+\varepsilon \gamma+\varepsilon \cos 2 t) x=0\). Show that the averaged equations \((7.6 .53)\) become $$ r^{\prime}=\frac{1}{4} r \sin 2 \phi, \quad \phi^{\prime}=\frac{1}{2}\left(\gamma+\frac{1}{2} \cos 2 \phi\right) $$ where \(x=r \cos \theta=r(T) \cos (t+\phi(T)), \quad \dot{x}=-r \sin \theta=-r(T) \sin (t+\phi(T))\), and prime denotes differentiation with respect to slow time \(T=\varepsilon t .\) Hint: To average terms like \(\cos 2 t \cos \theta \sin \theta\) over one cycle of \(\theta\), recall that \(t=\theta-\phi\) and use trig identities: $$ \begin{aligned} \langle\cos 2 t \cos \theta \sin \theta\rangle &=\frac{1}{2}\langle\cos (2 \theta-2 \phi) \sin 2 \theta) \\ &=\frac{1}{2}\langle(\cos 2 \theta \cos 2 \phi+\sin 2 \theta \sin 2 \phi) \sin 2 \theta) \\ &=\frac{1}{4} \sin 2 \phi \end{aligned} $$ b) Show that the fixed point \(r=0\) is unstable to exponentially growing oscillations, i.e., \(r(T)=r_{0} e^{k T}\) with \(k>0\), if \(|\gamma|<\gamma_{c}\) where \(\gamma_{e}\) is to be determined. (Hint: For \(r\) near \(0, \phi^{\prime} \gg r^{\prime}\) so \(\phi\) equilibrates relatively rapidly.) c) For \(|\gamma|<\gamma_{c}\), write a formula for the growth rate \(k\) in terms of \(\gamma\). d) How do the solutions to the averaged equations behave if \(|\gamma|>\gamma_{c} ?\) e) Interpret the results physically.

Sketch the phase portrait for each of the following systems. (As usual, \(r, \theta\) denote polar coordinates.) $$ \dot{r}=r \sin r, \dot{\theta}=1 $$

Consider the Duffing oscillator \(\ddot{x}+x+\varepsilon x^{3}=0\), where \(0<\varepsilon \ll<1, x(0)=a\), and \(\dot{x}(0)=0\). a) Using conservation of energy, express the oscillation period \(T(\varepsilon)\) as a certain integral. b) Expand the integrand as a power series in \(\varepsilon\), and integrate term by term to obtain an approximate formula \(T(\varepsilon)=c_{0}+c_{3} \varepsilon+c_{2} \varepsilon^{2}+O\left(\varepsilon^{3}\right)\). Find \(c_{0}, c_{1}, c_{2}\) and check that \(c_{0}, c_{1}\) are consistent with (7.6.57).
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