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A pendulum is made from a massless spring (force constant \(k\) and unstretched length \(l_{\mathrm{o}}\)) that is suspended at one end from a fixed pivot \(O\) and has a mass \(m\) attached to its other end. The spring can stretch and compress but cannot bend, and the whole system is confined to a single vertical plane. (a) Write down the Lagrangian for the pendulum, using as generalized coordinates the usual angle \(\phi\) and the length \(r\) of the spring. (b) Find the two Lagrange equations of the system and interpret them in terms of Newton's second law, as given in Equation (1.48). (c) The equations of part (b) cannot be solved analytically in general. However, they can be solved for small oscillations. Do this and describe the motion. [Hint: Let \(l\) denote the equilibrium length of the spring with the mass hanging from it and write \(r=l+\epsilon .\) "Small oscillations" involve only small values of \(\epsilon\) and \(\phi,\) so you can use the small-angle approximations and drop from your equations all terms that involve powers of \(\epsilon\) or \(\phi\) (or their derivatives) higher than the first power (also products of \(\epsilon\) and \(\phi\) or their derivatives). This dramatically simplifies and uncouples the equations.]

Short Answer

Expert verified
The Lagrangian is \( \mathcal{L} = \frac{1}{2} m \dot{r}^2 + \frac{1}{2} m r^2 \dot{\phi}^2 + mgr\cos\phi - \frac{1}{2} k (r - l_{\mathrm{o}})^2 \). For small oscillations, \( \epsilon \) and \( \phi \) oscillate independently.

Step by step solution

01

Kinetic Energy of the Pendulum

The kinetic energy consists of two parts: one due to the radial motion and the other due to angular motion. The radial part is \\[ T_r = \frac{1}{2} m \dot{r}^2 \] \and the angular part is \\[ T_\phi = \frac{1}{2} m r^2 \dot{\phi}^2. \] \Thus, the total kinetic energy is \\[ T = \frac{1}{2} m \dot{r}^2 + \frac{1}{2} m r^2 \dot{\phi}^2. \]
02

Potential Energy of the System

The potential energy of the system is due to the gravitational potential energy and the elastic potential energy of the spring. The gravitational potential energy is \\[ V_g = -mgr\cos\phi. \] \The elastic energy stored in the spring is \\[ V_s = \frac{1}{2} k (r - l_{\mathrm{o}})^2. \] \Therefore, the total potential energy is \\[ V = -mgr\cos\phi + \frac{1}{2} k (r - l_{\mathrm{o}})^2. \]
03

Lagrangian of the System

The Lagrangian \( \mathcal{L} \) is defined as the difference between the kinetic and potential energies. Thus, \\[ \mathcal{L} = T - V = \frac{1}{2} m \dot{r}^2 + \frac{1}{2} m r^2 \dot{\phi}^2 + mgr\cos\phi - \frac{1}{2} k (r - l_{\mathrm{o}})^2. \]
04

Lagrange's Equations

The Lagrangian's equations are derived by applying the Euler-Lagrange equation for each generalized coordinate \( q \), given by \\[ \frac{d}{dt} \left( \frac{\partial \mathcal{L}}{\partial \dot{q}} \right) - \frac{\partial \mathcal{L}}{\partial q} = 0. \] \Let's first apply it to \( r \): \\[ m \ddot{r} - mr\dot{\phi}^2 + mg\cos\phi - k(r - l_{\mathrm{o}}) = 0. \] \Now apply it to \( \phi \): \\[ \frac{d}{dt}(mr^2\dot{\phi}) + mgr\sin\phi = 0. \]
05

Simplification for Small Oscillations

Considering small oscillations where \( r = l + \epsilon \) and adopting small-angle approximations \( \sin\phi \approx \phi \) and \( \cos\phi \approx 1 \), the Lagrangian equations simplify. For the radial equation: \\[ m \ddot{\epsilon} - ml\dot{\phi}^2 + (k - \frac{mg}{l})\epsilon = 0. \] \For the angular motion: \\[ \frac{d}{dt}(ml^2\dot{\phi}) + mgl\phi \approx mgl\phi = 0. \] \This implies harmonic oscillations in both \( r \) and \( \phi \).
06

Interpretation of Small Oscillation Motion

The radial equation suggests that \( \epsilon \) oscillates harmonically with a modified spring constant \( k - \frac{mg}{l} \). The angular equation forms a simple harmonic motion in \( \phi \), suggesting a simple pendulum motion. Both \( \epsilon \) and \( \phi \) oscillate independently when small values are assumed.

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Key Concepts

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

Small Oscillations
In the context of physics, small oscillations refer to the slight deviations from an equilibrium position that a system undergoes. When a system oscillates, it moves back and forth. For small oscillations, these movements are minimal, allowing certain approximations to make complex calculations easier. In the case of our pendulum exercise, when the spring's length changes slightly around its equilibrium length, we describe this change as small oscillations, represented by \(\epsilon\). Likewise, small deviations in angle \(\phi\) are considered small oscillations.

For small oscillations, we use the small-angle approximation:
  • \(\sin\phi \approx \phi\)
  • \(\cos\phi \approx 1\)
These approximations are valid when the angles are nearly zero, allowing us to simplify complex trigonometric expressions into linear expressions. This becomes useful in uncoupling the equations of motion for the pendulum. By assuming \(\epsilon\) and \(\phi\) are small, we can eliminate higher order terms in those variables, simplifying our calculations.

This approach is not only applicable to pendulums but to any oscillating system that stays close to its equilibrium. It is a foundational concept that finds utility in various physics and engineering applications.
Simple Harmonic Motion
Simple harmonic motion (SHM) is a specific type of oscillatory motion where the restoring force is directly proportional to displacement, and acts in the opposite direction. In this exercise, the pendulum undergoes simple harmonic motion both in its radial and angular components.

For the radial motion, the equation derived:
  • \(m \ddot{\epsilon} + (k - \frac{mg}{l})\epsilon = 0\)
resembles the standard format of simple harmonic motion: \(m \ddot{x} + kx = 0\). Here \(x\) is replaced by \(\epsilon\), and the modified spring constant is \(k - \frac{mg}{l}\). This implies that the spring stretches and compresses symmetrically about its equilibrium position, demonstrating SHM.

Meanwhile, for the angular motion, the simplified equation:
  • \(mgl\phi = 0\)
indicates that \(\phi\) follows a pendulum-like SHM. The angle oscillates around its resting position, much like a conventional simple pendulum does under the influence of gravity.

Simple harmonic motion is characterized by:
  • Sinusoidal patterns (like sine or cosine waves)
  • A constant frequency and period
  • The system's total energy swaying between potential and kinetic forms
Studying SHM provides essential insights into more complex oscillating systems and underpins many physical processes from waves to oscillating circuits.
Euler-Lagrange Equation
The Euler-Lagrange equation is a central tool in Lagrangian mechanics, providing a formal way to derive equations of motion for a system. It is based on the principle of least action and is applicable in both classical and modern physics.

In this exercise, the Euler-Lagrange equation is used to find the equations of motion of the pendulum system. This is accomplished by treating \(\phi\) and \(r\) as generalized coordinates in the Lagrangian \(\mathcal{L}\), which is the difference between kinetic and potential energy:
\[\mathcal{L} = \frac{1}{2} m \dot{r}^2 + \frac{1}{2} m r^2 \dot{\phi}^2 + mgr\cos\phi - \frac{1}{2} k (r - l_{\mathrm{o}})^2\]

By applying the equation:
  • \(\frac{d}{dt} \left( \frac{\partial \mathcal{L}}{\partial \dot{q}} \right) - \frac{\partial \mathcal{L}}{\partial q} = 0\)
separately to both \(r\) and \(\phi\), we derive the system's equations of motion.

The genius of this method lies in its ability to handle complex systems in a systemic, clean approach, bypassing the need to directly apply Newton's second law. The Euler-Lagrange equation elegantly uses energy principles to yield motion descriptions. This abstraction is particularly powerful for multi-coordinate systems and forms the backbone of modern theoretical physics.

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Most popular questions from this chapter

Consider two particles moving unconstrained in three dimensions, with potential energy \(U\left(\mathbf{r}_{1}, \mathbf{r}_{2}\right) .\) (a) Write down the six equations of motion obtained by applying Newton's second law to each particle. (b) Write down the Lagrangian \(\mathcal{L}\left(\mathbf{r}_{1}, \mathbf{r}_{2}, \dot{\mathbf{r}}_{1}, \dot{\mathbf{r}}_{2}\right)=T-U\) and show that the six Lagrange equations are the same as the six Newtonian equations of part (a). This establishes the validity of Lagrange's equations in rectangular coordinates, which in turn establishes Hamilton's principle. since the latter is independent of coordinates, this proves Lagrange's equations in any coordinate system.

A mass \(m_{1}\) rests on a frictionless horizontal table and is attached to a massless string. The string runs horizontally to the edge of the table, where it passes over a massless, frictionless pulley and then hangs vertically down. A second mass \(m_{2}\) is now attached to the bottom end of the string. Write down the Lagrangian for the system. Find the Lagrange equation of motion, and solve it for the acceleration of the blocks. For your generalized coordinate, use the distance \(x\) of the second mass below the tabletop.

Lagrange's equations in the form discussed in this chapter hold only if the forces (at least the nonconstraint forces) are derivable from a potential energy. To get an idea how they can be modified to include forces like friction, consider the following: A single particle in one dimension is subject to various conservative forces (net conservative force \(=F=-\partial U / \partial x)\) and a nonconservative force (let's call it \(F_{\text {fric }}\) ). Define the Lagrangian as \(\mathcal{L}=T-U\) and show that the appropriate modification is $$\frac{\partial \mathcal{L}}{\partial x}+F_{\mathrm{fric}}=\frac{d}{d t} \frac{\partial \mathcal{L}}{\partial \dot{x}}.$$

The "spherical pendulum" is just a simple pendulum that is free to move in any sideways direction. (By contrast a "simple pendulum"- unqualified - is confined to a single vertical plane.) The bob of a spherical pendulum moves on a sphere, centered on the point of support with radius \(r=R\) the length of the pendulum. A convenient choice of coordinates is spherical polars, \(r, \theta, \phi,\) with the origin at the point of support and the polar axis pointing straight down. The two variables \(\theta\) and \(\phi\) make a good choice of generalized coordinates. (a) Find the Lagrangian and the two Lagrange equations. (b) Explain what the \(\phi\) equation tells us about the \(z\) component of angular momentum \(\ell_{z^{*}}\) (c) For the special case that \(\phi=\) const, describe what the \(\theta\) equation tells us. (d) Use the \(\phi\) equation to replace \(\dot{\phi}\) by \(\ell_{z}\) in the \(\theta\) equation and discuss the existence of an angle \(\theta_{\mathrm{o}}\) at which \(\theta\) can remain constant. Why is this motion called a conical pendulum? (e) Show that if \(\theta=\theta_{0}+\epsilon,\) with \(\epsilon\) small, then \(\theta\) oscillates about \(\theta_{\mathrm{o}}\) in harmonic motion. Describe the motion of the pendulum's bob.

A particle is confined to move on the surface of a circular cone with its axis on the \(z\) axis, vertex at the origin (pointing down), and half-angle \(\alpha\). The particle's position can be specified by two generalized coordinates, which you can choose to be the coordinates \((\rho, \phi)\) of cylindrical polar coordinates. Write down the equations that give the three Cartesian coordinates of the particle in terms of the generalized coordinates ( \(\rho, \phi\) ) and vice versa.

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