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Chapter 9: Problem 20

(a) By solving the equation for \(d y / d x,\) show that the equation of the trajectories of the undamped pendulum of Problem 19 can be written as $$ \frac{1}{2} y^{2}+\omega^{2}(1-\cos x)=c $$ where \(c\) is a constant of integration. (b) Multiply Eq. (i) by \(m L L^{2} .\) Then express the result in terms of \(\theta\) to obtain $$ \frac{1}{2} m L^{2}\left(\frac{d \theta}{d t}\right)^{2}+m g L(1-\cos \theta)=E $$ where \(E=m L^{2} c .\) (c) Show that the first term in Eq. (ii) is the kinetic energy of the pendulum and that the second term is the potential energy due to gravity. Thus the total energy \(E\) of the pendulum is constant along any trajectory; its value is determined by the initial conditions.

Short Answer

Expert verified
#Short Answer# The kinetic energy of the undamped pendulum is given by \(\frac{1}{2}mL^2\left(\frac{d\theta}{dt}\right)^2\), while the potential energy due to gravity is given by \(-m\omega^2 L^2\left(1-\cos\theta\right)\). The total energy of the pendulum remains constant along any trajectory, and its value is determined by the initial conditions.

Step by step solution

01

Solve the equation for \(\frac{dy}{dx}\)

First, we need to solve the equation for \(\frac{dy}{dx}\), where \(y\) represents the vertical displacement and \(x\) represents the horizontal displacement. In this case, the equation we are given is the trajectory equation for an undamped pendulum: $$ \frac{1}{2} y^{2}+\omega^{2}(1-\cos x)=c $$ Differentiate both sides of the equation with respect to \(x\): $$ \frac{d}{dx}\left(\frac{1}{2} y^{2}+\omega^{2}(1-\cos x)\right)=\frac{dc}{dx} $$ Now, apply the chain rule to differentiate \(y^2\) with respect to \(x\): $$ y\frac{dy}{dx} + \omega^{2}\sin x = 0 $$ So, we have found that \(\frac{dy}{dx} = -\frac{\omega^2}{y}\sin x\).
02

Rewrite the equation of trajectories

Now we rewrite the trajectory equation, substituting the value of \(\frac{dy}{dx}\) we found previously: $$ \frac{1}{2} y^{2}+\omega^{2}(1-\cos x)=c $$ This is the required equation for the trajectories of the undamped pendulum.
03

Multiply by \(mLL^2\) and express the result in terms of \(\theta\)

We are given a transformation from \(y\) to \(\theta\) in the problem: \(y = mL\frac{d\theta}{dt}\). Now, we must multiply the equation by \(mLL^2\) and express the result in terms of \(\theta\): $$ mL^2\left(\frac{1}{2} y^{2}+\omega^{2}(1-\cos x)\right)=mL^2c $$ Substitute the expression for \(y\) in terms of \(\theta\): $$ \frac{1}{2}mL^2\left(\frac{d\theta}{dt}\right)^2 + m\omega^2 L^2 (1-\cos\theta) = mL^2E $$ where \(E = mc\).
04

Identify the kinetic and potential energy terms

Now, let's show that the first term in the resulting equation is the kinetic energy, and the second term is the potential energy due to gravity. The kinetic energy of the pendulum is given by: $$ K.E. = \frac{1}{2}mL^2\left(\frac{d\theta}{dt}\right)^2 $$ And the potential energy due to gravity is given by: $$ P.E. = -m\omega^2 L^2\left(1-\cos\theta\right) $$ Thus, the total energy of the pendulum is: $$ E = K.E. + P.E. = \frac{1}{2}mL^2\left(\frac{d\theta}{dt}\right)^2 - m\omega^2 L^2\left(1-\cos\theta\right) $$ Since this total energy \(E\) is constant along any trajectory, its value is determined by the initial conditions.

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Key Concepts

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

Kinetic Energy
When analyzing a pendulum, its kinetic energy is associated with its motion. Kinetic energy is defined as the energy possessed by an object due to its motion. In the context of a pendulum, this type of energy is derived from its angular velocity as it swings back and forth.
The mathematical representation for kinetic energy, denoted as \( K.E. \), in the case of a pendulum, is given by the equation:
  • \( K.E. = \frac{1}{2}mL^2\left(\frac{d\theta}{dt}\right)^2 \)

This formula showcases that kinetic energy depends on the mass \(m\), the length of the pendulum \(L\), and the angular velocity \(\frac{d\theta}{dt}\). The angular velocity represents how fast the angle \(\theta\) is changing as time \(t\) progresses.
It's important to note that the kinetic energy reaches its maximum value when the pendulum is at the lowest point of its swing, where the velocity is the greatest. Conversely, the kinetic energy is zero at the highest points of the swing, because at those points, the pendulum momentarily stops before changing direction.
Potential Energy
Potential energy in a pendulum is related to its position relative to its rest position. This energy is stored due to the height from which the pendulum moves under the influence of gravity. Potential energy, denoted as \( P.E. \), can be expressed as the energy associated with the angle of displacement \(\theta\) from the vertical resting position.
The potential energy of a pendulum is given by the formula:
  • \( P.E. = m g L(1 - \cos\theta) \)

Here, \( m \) is the mass of the pendulum, \( g \) represents the acceleration due to gravity, \( L \) is the length of the pendulum, and \( \theta \) is the angle of displacement from the vertical.
The potential energy is highest when the pendulum is at its furthest point from the rest position (the peak of its swing), as this is where the height and thus the energy stored is greatest. Conversely, potential energy is zero when the pendulum is in the vertical position because the height above the ground is minimal.
Energy Conservation in Pendulum Motion
In pendulum motion, the law of conservation of energy plays a crucial role. This principle states that energy cannot be created or destroyed in an isolated system, only transformed from one form to another. For a pendulum, the total mechanical energy, which is the sum of kinetic and potential energy, remains constant throughout its motion.
The total energy \( E \) of a pendulum is expressed as:
  • \( E = \frac{1}{2}mL^2\left(\frac{d\theta}{dt}\right)^2 + m g L (1 - \cos\theta) \)

At the highest point of its swing, all of the pendulum's energy is potential. As it descends, potential energy is transformed into kinetic energy. When it reaches the lowest point of its swing, the potential energy is at its minimum and kinetic energy is at its maximum.
The pendulum then ascends, converting kinetic energy back into potential energy, repeating this cycle indefinitely. This transformation between kinetic and potential energy explains the swing motion of a pendulum. However, in the real world, factors like air resistance and friction at the pivot prevent perpetual motion, causing the pendulum to eventually stop swinging unless an external force is applied to maintain its motion.

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Most popular questions from this chapter

(a) Find all the critical points (equilibrium solutions). (b) Use a computer to draw a direction field and portrait for the system. (c) From the plot(s) in part (b) determine whether each critical point is asymptotically stable, stable, or unstable, and classify it as to type. $$ \text { The van der Pol equation: } \quad d x / d t=y, \quad d y / d t=\left(1-x^{2}\right) y-x $$

For certain \(r\) intervals, or windows, the Lorenz equations exhibit a period- doubling property similar to that of the logistic difference equation discussed in Section \(2.9 .\) Careful calculations may reveal this phenomenon. (a) One period-doubling window contains the value \(r=100 .\) Let \(r=100\) and plot the trajectory starting at \((5,5,5)\) or some other initial point of your choice. Does the solution appear to be periodic? What is the period? (b) Repeat the calculation in part (a) for slightly smaller values of \(r .\) When \(r \cong 99.98\), you may be able to observe that the period of the solution doubles. Try to observe this result by performing calculations with nearby values of \(r\). (c) As \(r\) decreases further, the period of the solution doubles repeatedly. The next period doubling occurs at about \(r=99.629 .\) Try to observe this by plotting trajectories for nearby values of \(r .\)

(a) By solving Eq. (9) numerically show that the real part of the complex roots changes sign when \(r \cong 24.737\). (b) Show that a cubic polynomial \(x^{3}+A x^{2}+B x+C\) has one real zero and two pure imaginary zeros only if \(A B=C\). (c) By applying the result of part (b) to Eq. (9) show that the real part of the complex roots changes sign when \(r=470 / 19\).

(a) Determine all critical points of the given system of equations. (b) Find the corresponding linear system near each critical point. (c) Find the eigenalues of each linear system. What conclusions can you then draw about the nonlinear system? (d) Draw a phase portrait of the nonlinear system to confirm your conclusions, or to extend them in those cases where the linear system does not provide definite information about the nonlinear system. $$ d x / d t=x+x^{2}+y^{2}, \quad d y / d t=y-x y $$

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