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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

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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 \(k = \sin(\alpha / 2)\).

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

The system \\[ x^{\prime}=-y, \quad y^{\prime}=-\gamma y-x(x-0.15)(x-2) \\] results from an approximation to the Hodgkin-Huxley \(^{6}\) equations, which model the transmission of neural impulses along an axon. a. Find the critical points, and classify them by investigating the approximate linear system near each one. b. Draw phase portraits for \(\gamma=0.8\) and for \(\gamma=1.5\) c. Consider the trajectory that leaves the critical point (2, 0). Find the value of \(\gamma\) for which this trajectory ultimately approaches the origin as \(t \rightarrow \infty .\) Draw a phase portrait for this value of \(\gamma\).

an autonomous system is expressed in polar coordinates. Determine all periodic solutions, all limit cycles, and determine their stability characteristics. $$ d r / d t=r(1-r)(r-2), \quad d \theta / d t=-1 $$

(a) A special case of the Lienard equation of Problem 8 is $$ \frac{d^{2} u}{d t^{2}}+\frac{d u}{d t}+g(u)=0 $$ where \(g\) satisfies the conditions of Problem 6 . Letting \(x=u, y=d u / d t,\) show that the origin is a critical point of the resulting system. This equation can be interpreted as describing the motion of a spring-mass system with damping proportional to the velocity and a nonlinear restoring force. Using the Liapunov function of Problem \(6,\) show that the origin is a stable critical point, but note that even with damping we cannot conclude asymptotic stability using this Liapunov function. (b) Asymptotic stability of the critical point \((0,0)\) can be shown by constructing a better Liapunov function as was done in part (d) of Problem 7 . However, the analysis for a general function \(g\) is somewhat sophisticated and we only mention that appropriate form for \(V\) is $$ V(x, y)=\frac{1}{2} y^{2}+A y g(x)+\int_{0}^{x} g(s) d s $$ where \(A\) is a positive constant to be chosen so that \(V\) is positive definite and \(\hat{V}\) is negative definite. For the pendulum problem \([g(x)=\sin x]\) use \(V\) as given by the preceding equation with \(A=\frac{1}{2}\) to show that the origin is asymptotically stable. Hint: Use \(\sin x=x-\alpha x^{3} / 3 !\) and \(\cos x=1-\beta x^{2} / 2 !\) where \(\alpha\) and \(\beta\) depend on \(x,\) but \(0<\alpha<1\) and \(0<\beta<1\) for \(-\pi / 2

In this problem we prove a part of Theorem 9.3 .2 relating to instability. (a) Show that if \(a_{11}+a_{22}>0\) and \(a_{11} a_{22}-a_{12} a_{21}>0\), then the critical point \((0,0)\) of the linear system (i) is unstable. (b) The same result holds for the almost linear system (i). As in Problems 10 and 11 construct a positive definite function \(V\) such that \(V(x, y)=x^{2}+y^{2}\) and hence is positive definite, and then invoke Theorem 9.6.2.

This problem extends Problem 22 to a damped pendulum . The equations of motion are $$ d x / d t=y, \quad d y / d t=-4 \sin x-\gamma y . $$ where \(\gamma\) is the damping coefficient, with the initial conditions \(x(0)=0, y(0)=v\) (a) For \(\gamma=1 / 4\) plot \(x\) versus \(t\) for \(v=2\) and for \(v=5 .\) Explain these plots in terms of the motions of the pendulum that they represent. Also explain how they relate to the corresponding graphs in Problem 22 (a). (b) Estimate the critical value \(v_{c}\) of the initial velocity where the transition from one type of motion to the other occurs. (c) Repeat part (b) for other values of \(\gamma\) and determine how \(v_{c}\) depends on \(\gamma\).

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