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Chapter 9: Problem 27

In this problem we derive a formula for the natural period of an undamped nonlinear pendulum \([c=0 \text { in Eq. }(10) \text { of Section } 9.2]\). Suppose that the bob is pulled through a positive angle \(\alpha\) and then released with zero velocity. (a) We usually think of \(\theta\) and \(d \theta / d t\) as functions of \(t\). However, if we reverse the roles of \(t\) and \(\theta,\) we can regard \(t\) as a function of \(\theta,\) and consequently also think of \(d \theta / d t\) as a function of \(\theta .\) Then derive the following sequence of equations: $$ \begin{aligned} \frac{1}{2} m L^{2} \frac{d}{d \theta}\left[\left(\frac{d \theta}{d t}\right)^{2}\right] &=-m g L \sin \theta \\ \frac{1}{2} m\left(L \frac{d \theta}{d t}\right)^{2}=& m g L(\cos \theta-\cos \alpha) \\ d t &=-\sqrt{\frac{L}{2 g}} \frac{d \theta}{\sqrt{\cos \theta-\cos \alpha}} \end{aligned} $$ Why was the negative square root chosen in the last equation? (b) If \(T\) is the natural period of oscillation, derive the formula $$ \frac{T}{4}=-\sqrt{\frac{L}{2 g}} \int_{\alpha}^{0} \frac{d \theta}{\sqrt{\cos \theta-\cos \alpha}} $$ (c) By using the identities \(\cos \theta=1-2 \sin ^{2}(\theta / 2)\) and \(\cos \alpha=1-2 \sin ^{2}(\alpha / 2),\) followed by the change of variable \(\sin (\theta / 2)=k \sin \phi\) with \(k=\sin (\alpha / 2),\) show that $$ T=4 \sqrt{\frac{L}{g}} \int_{0}^{\pi / 2} \frac{d \phi}{\sqrt{1-k^{2} \sin ^{2} \phi}} $$ The integral is called the elliptic integral of the first kind. Note that the period depends on the ratio \(L / \mathrm{g}\) and also the initial displacement \(\alpha\) through \(k=\sin (\alpha / 2) .\) (d) By evaluating the integral in the expression for \(T\) obtain values for \(T\) that you can compare with the graphical estimates you obtained in Problem 21 .

Short Answer

Expert verified
Answer: The natural period of an undamped nonlinear pendulum with a starting angle \(\alpha\) can be found using the formula: $$ T = 4\sqrt{\frac{L}{g}}\int_{0}^{\frac{\pi}{2}}\frac{d\phi}{\sqrt{1-k^2\sin^2\phi}} $$ where \(T\) is the natural period, \(L\) is the length of the pendulum, \(g\) is the acceleration due to gravity, and \(k = \sin(\alpha / 2)\). The integral in the expression for \(T\) is an elliptic integral of the first kind, which can be evaluated using numerical methods, tables, or software.

Step by step solution

01

Derive the equations with \(\theta\) as a function of \(t\)

To derive the equations, we will reverse the roles of \(t\) and \(\theta\), and consider \(t\) as a function of \(\theta\). Differentiate the given equation with respect to \(\theta\): $$ \frac{1}{2} m L^{2} \frac{d}{d \theta}\left[\left(\frac{d\theta}{dt}\right)^{2}\right] = -m g L \sin \theta $$
02

Find the equation for the kinetic energy

We can rewrite the above equation as a conservation of energy: $$ \frac{1}{2}m\left(L\frac{d\theta}{dt}\right)^{2} = m g L(\cos \theta - \cos \alpha) $$
03

Find the equation for \(dt\)

Rearrange the conservation of energy equation and find the equation for \(dt\): $$ dt =-\sqrt{\frac{L}{2g}}\frac{d\theta}{\sqrt{\cos\theta-\cos\alpha}} $$ The negative square root is chosen in the last equation because we are considering the bob being pulled through a positive angle \(\alpha\) and then released with zero velocity, meaning that the pendulum is moving in a negative direction relative to the angle.
04

Derive the formula for \(T\)

To find the formula for the natural period \(T\) of oscillation, integrate both sides of the equation we derived for \(dt\): $$ \frac{T}{4} = -\sqrt{\frac{L}{2g}} \int_{\alpha}^{0} \frac{d\theta}{\sqrt{\cos\theta-\cos\alpha}} $$
05

Rewrite the equation using trigonometric identities

Using the trigonometric identities \(\cos \theta = 1 - 2\sin^2(\theta / 2)\) and \(\cos \alpha = 1 - 2\sin^2(\alpha / 2)\) and the change of variables \(\sin(\theta / 2) = k\sin\phi\) with \(k = \sin(\alpha / 2)\), rewrite the equation for \(T\) as: $$ T = 4\sqrt{\frac{L}{g}}\int_{0}^{\frac{\pi}{2}}\frac{d\phi}{\sqrt{1-k^2\sin^2\phi}} $$
06

Evaluate the integral for T

To obtain values for \(T\), we need to evaluate the integral in the expression for \(T\). Since this is an elliptic integral of the first kind, its evaluation is beyond the scope of high school math. However, one can use numerical methods, tables, or software to approximate the integral and obtain the values of \(T\). The resulting \(T\) values can be compared with the graphical estimates obtained in Problem 21. Note that the period depends on the ratio \(L/g\) and also the initial displacement \(\alpha\) through \(k = \sin(\alpha / 2)\).

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Key Concepts

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

Natural Period
The natural period of a pendulum refers to the time it takes to complete one full oscillation, swinging back and forth. Understanding the natural period is crucial in topics involving oscillations and waves.

For a simple pendulum, which assumes small angles, the period is typically derived using the formula: \[T = 2\pi \sqrt{\frac{L}{g}}\] where:
  • \( L \) is the length of the pendulum.
  • \( g \) is the acceleration due to gravity.
However, when dealing with a nonlinear pendulum, the period depends significantly on the initial displacement angle \( \alpha \).

As the initial angle increases, the approximation for small oscillations fails. For larger \( \alpha \), the period is influenced by the gravitational pull and the length of the pendulum, but doesn't strictly follow the simple formula. This is why the nonlinear treatment requires more complex calculations involving elliptic integrals.
Elliptic Integral
An elliptic integral is a type of integral that arises when calculating more complex motions, such as those of a nonlinear pendulum. It's named this way because it was historically associated with calculating the arc length of an ellipse.

In the context of a pendulum, it appears in the formula derived for the natural period: \[T = 4\sqrt{\frac{L}{g}}\int_{0}^{\frac{\pi}{2}}\frac{d\phi}{\sqrt{1-k^2\sin^2\phi}}\]

  • The variable \( k \) represents \( \sin(\alpha/2) \), where \( \alpha \) is the initial angle of displacement.
  • The integral itself is termed as the elliptic integral of the first kind.
This integral doesn't have a simple analytic antiderivative. Thus, evaluating it often relies on numerical methods or precomputed tables. The presence of elliptic integrals makes analyzing nonlinear pendulums a more intricate task compared to their linear counterparts.
Conservation of Energy
Conservation of Energy is a fundamental principle that states energy within a closed system remains constant over time. For the pendulum, this principle is utilized to understand its motion and derive its equations.

Initially, all the pendulum's energy is gravitational potential energy at its highest point of displacement. As it swings down, this energy converts into kinetic energy. At the lowest point, potential energy is minimal, while kinetic energy is at its peak. Conversely, as it swings upwards, kinetic energy converts back into potential energy.

The expression \[\frac{1}{2}m\left(L\frac{d\theta}{dt}\right)^{2} = m g L(\cos \theta - \cos \alpha)\]exemplifies this principle. Here, the left side represents kinetic energy, and the right side signifies the difference in potential energy between two points, showing energy conservation. This transformation and conservation principle offers a way to derive the pendulum's dynamics accurately.
Trigonometric Identities
Trigonometric identities are essential tools when dealing with oscillatory systems such as pendulums. These identities allow for simplification and manipulation of equations, particularly when handling integrals.

For the nonlinear pendulum problem, identities like:
  • \( \cos \theta = 1 - 2\sin^{2}(\theta / 2) \)
  • \( \cos \alpha = 1 - 2\sin^{2}(\alpha / 2) \)
are employed to transform the expressions into a more manageable form. These transformations make it possible to rewrite the integral for the period \( T \) in terms of a variable \( \phi \), leading to an elliptic integral format.

Such trigonometric manipulations enable the conversion and solving of complex physics problems into forms that can be tackled using either analytical or numerical methods. They are invaluable for deriving accurate solutions in oscillatory and wave-related phenomena.

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Most popular questions from this chapter

In this problem we indicate how to show that the trajectories are ellipses when the eigen- values are pure imaginary. Consider the system $$ \left(\begin{array}{l}{x} \\\ {y}\end{array}\right)^{\prime}=\left(\begin{array}{ll}{a_{11}} & {a_{12}} \\\ {a_{21}} & {a_{22}}\end{array}\right)\left(\begin{array}{l}{x} \\\ {y}\end{array}\right) $$ (a) Show that the eigenvalues of the coefficient matrix are pure imaginary if and only if $$ a_{11}+a_{22}=0, \quad a_{11} a_{22}-a_{12} a_{21}>0 $$ (b) The trajectories of the system (i) can be found by converting Eqs. (i) into the single equation $$ \frac{d y}{d x}=\frac{d y / d t}{d x / d t}=\frac{a_{21} x+a_{22} y}{a_{11} x+a_{12} y} $$ Use the first of Eqs. (ii) to show that Eq. (iii) is exact. (c) By integrating Eq. (iii) show that $$ a_{21} x^{2}+2 a_{22} x y-a_{12} y^{2}=k $$ where \(k\) is a constant. Use Eqs. (ii) to conclude that the graph of Eq. (iv) is always an ellipse. Hint: What is the discriminant of the quadratic form in Eq. (iv)?

Consider the system of equations $$ d x / d t=y-x f(x, y), \quad d y / d t=-x-y f(x, y) $$ where \(f\) is continuous and has continuous first partial derivatives. Show that if \(f(x, y)>0\) in some neighborhood of the origin, then the origin is an asymptotically stable critical point, and if \(f(x, y)<0\) in some neighborhood of the origin, then the origin is an unstable critical point. Hint: Construct a Liapunov function of the form \(c\left(x^{2}+y^{2}\right) .\)

Each of Problems I through 6 can be interpreted as describing the interaction of two species with populations \(x\) and \(y .\) In each of these problems carry out the following steps. $$ \begin{array}{l}{\text { (a) Draw a direction field and describe how solutions seem to behave. }} \\ {\text { (b) Find the critical points. }} \\ {\text { (c) For each critical point find the corresponding linear system. Find the eigenvalues and }} \\ {\text { eigenvectors of the linear system; classify each critical point as to type, and determine }} \\ {\text { whether it is asymptotically stable, stable, or unstable. }}\end{array} $$ $$ \begin{array}{l}{\text { (d) Sketch the trajectories in the neighborhood of each critical point. }} \\ {\text { (c) Compute and plot enough trajectories of the given system to show clearly the behavior of }} \\ {\text { the solutions. }} \\ {\text { (f) Determine the limiting behavior of } x \text { and } y \text { as } t \rightarrow \infty \text { and interpret the results in terms of }} \\ {\text { the populations of the two species. }}\end{array} $$ $$ \begin{array}{l}{d x / d t=x(1.5-x-0.5 y)} \\ {d y / d t=y(2-y-0.75 x)}\end{array} $$

Each of Problems I through 6 can be interpreted as describing the interaction of two species with populations \(x\) and \(y .\) In each of these problems carry out the following steps. $$ \begin{array}{l}{\text { (a) Draw a direction field and describe how solutions seem to behave. }} \\ {\text { (b) Find the critical points. }} \\ {\text { (c) For each critical point find the corresponding linear system. Find the eigenvalues and }} \\ {\text { eigenvectors of the linear system; classify each critical point as to type, and determine }} \\ {\text { whether it is asymptotically stable, stable, or unstable. }}\end{array} $$ $$ \begin{array}{l}{\text { (d) Sketch the trajectories in the neighborhood of each critical point. }} \\ {\text { (c) Compute and plot enough trajectories of the given system to show clearly the behavior of }} \\ {\text { the solutions. }} \\ {\text { (f) Determine the limiting behavior of } x \text { and } y \text { as } t \rightarrow \infty \text { and interpret the results in terms of }} \\ {\text { the populations of the two species. }}\end{array} $$ $$ \begin{array}{l}{d x / d t=x(1.5-0.5 x-y)} \\ {d y / d t=y(2-y-1.125 x)}\end{array} $$

By introducing suitable dimensionless variables, the system of nonlinear equations for the damped pendulum [Frqs. (8) of Section 9.3] can be written as $$ d x / d t=y, \quad d y / d t=-y-\sin x \text { . } $$ (a) Show that the origin is a critical point. (b) Show that while \(V(x, y)=x^{2}+y^{2}\) is positive definite, \(f(x, y)\) takes on both positive and negative values in any domain containing the origin, so that \(V\) is not a Liapunov function. Hint: \(x-\sin x>0\) for \(x>0\) and \(x-\sin x<0\) for \(x<0 .\) Consider these cases with \(y\) positive but \(y\) so small that \(y^{2}\) can be ignored compared to \(y .\) (c) Using the energy function \(V(x, y)=\frac{1}{2} y^{2}+(1-\cos x)\) mentioned in Problem \(6(b),\) show that the origin is a stable critical point. Note, however, that even though there is damping and we can epect that the origin is asymptotically stable, it is not possible to draw this conclusion using this Liapunov function. (d) To show asymptotic stability it is necessary to construct a better Liapunov function than the one used in part (c). Show that \(V(x, y)=\frac{1}{2}(x+y)^{2}+x^{2}+\frac{1}{2} y^{2}\) is such a Liapunov function, and conclude that the origin is an asymptotically stable critical point. Hint: From Taylor's formula with a remainder it follows that \(\sin x=x-\alpha x^{3} / 3 !,\) where \(\alpha\) depends on \(x\) but \(0<\alpha<1\) for \(-\pi / 2
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