/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 15 A pendulum consists of a mass \(... [FREE SOLUTION] | 91Ó°ÊÓ

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A pendulum consists of a mass \(m\) suspended by a massless spring with unextended length \(b\) and spring constant \(k\). Find Lagrange's equations of motion.

Short Answer

Expert verified
The equation of motion for the mass-spring system with mass \(m\), unextended length \(b\), spring constant \(k\), and displacement \(x\) is given by: \(m\ddot{x} + kx = 0\)

Step by step solution

01

Setup the problem

First, let's define the position of the mass m using a coordinate x, where x represents the displacement from the unextended position of the spring. So, at x = 0, the spring is neither compressed nor elongated. As the mass m moves, the spring is either extended or compressed depending on the direction.
02

Find the kinetic and potential energy of the system

The kinetic energy of the system is only due to the mass m and can be given by the equation: \(T = \frac{1}{2}mv^2\) where v is the velocity of the mass. The potential energy of the mass-spring system is given by: \(V = \frac{1}{2}kx^2\) where x is the displacement of the mass from the unextended spring position, and k is the spring constant.
03

Write the Lagrangian

The Lagrangian (L) is the difference between the kinetic and potential energy of the system: \(L = T - V\) Replacing T and V with their expressions from Step 2: \(L = \frac{1}{2}mv^2 - \frac{1}{2}kx^2\) Since v can also be written as the time derivative of x (v = \(\dot{x}\)): \(L = \frac{1}{2}m\dot{x}^2 - \frac{1}{2}kx^2\)
04

Apply Lagrange's equation

Lagrange's equation for the coordinate x is given by: \(\frac{d}{dt}(\frac{\partial L}{\partial \dot{x}}) - \frac{\partial L}{\partial x} = 0\) Now we will compute the necessary partial derivatives: \(\frac{\partial L}{\partial \dot{x}} = m\dot{x}\) \(\frac{d}{dt}(\frac{\partial L}{\partial \dot{x}}) = m\ddot{x}\) \(\frac{\partial L}{\partial x} = -kx\) Now, substitute these derivatives back into the Lagrange's equation: \(m\ddot{x} + kx = 0\) This is the equation of motion for the mass-spring system using Lagrange's equations.

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Key Concepts

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

Lagrange's equations
Lagrange's equations provide a powerful tool for analyzing mechanical systems, particularly when dealing with complex systems or non-Cartesian coordinates. They are derived from the principle of least action and give equations of motion based on energy functions rather than forces directly. This is particularly helpful as it simplifies problems considerably by focusing on scalar quantities like kinetic and potential energy.

The general form of Lagrange's equations is:
  • For a system with N degrees of freedom, the generalized coordinate is represented as \( q_i \).
  • The Lagrangian (\( L \)) of the system is defined as \( L = T - V \), where \( T \) is the kinetic energy and \( V \) is the potential energy.
  • The equations are \( \frac{d}{dt} \left( \frac{\partial L}{\partial \dot{q}_i} \right) - \frac{\partial L}{\partial q_i} = 0 \).
In this exercise, using Lagrange’s equations allowed us to formulate the motion of a mass on a spring by focusing on how energy changes in the system and provided a clean path to the equations of motion.
kinetic energy
Kinetic energy represents the energy a body has due to its motion. It is a crucial part of mechanical analysis as it reflects how energy is transferred through movement.

For a mass \( m \) moving with velocity \( v \), the expression for kinetic energy \( T \) is given by:
  • \( T = \frac{1}{2}mv^2 \)
In the problem at hand, the kinetic energy equation shows how the speed of the mass contributes to its total energy. It's a straightforward expression because our focus is on linear, translational motion. A higher velocity means more kinetic energy, which directly impacts the Lagrangian and consequently the equations of motion derived used for identifying dynamics of the system.
potential energy
Potential energy is stored energy, determined by the position or state of an object. In a spring-mass system, potential energy arises from the spring's deformation.

In our case, the potential energy \( V \) of the spring system is given by:
  • \( V = \frac{1}{2}kx^2 \)
Here, \( x \) is the displacement from the spring's natural length, and \( k \) is the spring constant—a measure of the spring's stiffness. Potential energy increases with the square of displacement, meaning further compression or extension increases the stored energy. This potential energy plays a vital role in forming the Lagrangian, providing a balancing term against kinetic energy, and thus guiding the subsequent motion equations derived through Lagrange's method.
spring-mass system
A spring-mass system is one of the simplest mechanical systems used often in physics to model behaviors of forces and energy transfer.

It typically involves:
  • A mass \( m \) attached to a spring with a specific constant \( k \),
  • The spring having an unextended length \( b \),
  • Movements characterized by compression or extension of the spring,
  • Forces and energies that predictably model harmonic oscillation or simple harmonic motion.
In our exercise, the spring-mass system provides a practical scenario to apply Lagrange's equations. By understanding this system, students can see firsthand how kinetic and potential energies interplay to produce dynamic behavior—represented through differential equations of motion. It's a foundational setup for exploring more complex systems in classical mechanics.

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Most popular questions from this chapter

A particle of mass \(m\) is constrained to move on a circle of radius \(R\) The circle rotates in space about one point on the circle, which is fixed. The rotation takes place in the plane of the circle and with constant angular speed \(\omega .\) In the absence of a gravitational force, show that the particle's motion about one end of a diameter passing through the pivot point and the center of the circle is the same as that of a plane pendulum in a uniform gravitational field. Explain why this is a reasonable result.

A massless spring of length \(b\) and spring constant \(k\) connects two particles of masses \(m_{1}\) and \(m_{2} .\) The system rests on a smooth table and may oscillate and rotate. (a) Determine Lagrange's equations of motion. (b) What are the generalized momenta associated with any cyclic coordinates? (c) Determine Hamilton's equations of motion.

A sphere of radius \(\rho\) is constrained to roll without slipping on the lower half of the inner surface of a hollow cylinder of inside radius \(R\). Determine the Lagrangian function, the equation of constraint, and Lagrange's equations of motion. Find the frequency of small oscillations.

Consider any two continuous functions of the generalized coordinates and momenta \(g\left(q_{k}, p_{k}\right)\) and \(h\left(q_{k}, p_{k}\right) .\) The Poisson brackets are defined by $$[g, h]=\sum_{k}\left(\frac{\partial g}{\partial q_{k}} \frac{\partial h}{\partial p_{k}}-\frac{\partial g}{\partial p_{k}} \frac{\partial h}{\partial q_{k}}\right)$$ Verify the following properties of the Poisson brackets: (a) \(\frac{d g}{d t}=[g, H]+\frac{\partial g}{d t}\) where \(H\) is the Hamiltonian. If the Poisson bracket of two quantities vanishes, the quantities are said to commute. If the Poisson bracket of two quantities equals unity, the quantities are said to be canonically conjugate. (e) Show that any quantity that does not depend explicitly on the time and that commutes with the Hamiltonian is a constant of the motion of the system. Poisson-bracket formalism is of considerable importance in quantum mechanics. (b) \(\dot{q}_{j}=\left[q_{i}, H\right], \quad j_{j}=\left[p_{j}, H\right]\) (c) \(\left[p_{l}, p_{j}\right]=0,\left[q_{l}, q_{j}\right]=0\) (d) \(\left[q_{l}, p_{j}\right]=\delta_{j}\)

A double pendulum consists of two simple pendula, with one pendulum suspended from the bob of the other. If the two pendula have equal lengths and have bobs of equal mass and if both pendula are confined to move in the same plane, find Lagrange's equations of motion for the system. Do not assume small angles.

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