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Chapter 6: Problem 1

Use Lagrange's Equations to derive the equation of motion for a simple massspring-damper system.

Short Answer

Expert verified
Answer: The equation of motion is (m-c)¨x + kx = 0.

Step by step solution

01

Identify the system and its components

A simple mass-spring-damper system consists of a mass (m) attached to a spring with spring constant (k) and a damper with damping coefficient (c). The system is designed to oscillate in one dimension.
02

Construct the Lagrangian

The Lagrangian (L) of a system represents the difference between its kinetic energy (T) and its potential energy (V), that is: L = T - V. For a mass-spring-damper system, the kinetic energy is given by: T = \frac{1}{2}m\dot{x}^2 Where x is the displacement of the mass and \dot{x} represents its first time derivative, indicating the velocity of the mass. The potential energy in the system comes from the spring as a result of Hooke's Law: V = \frac{1}{2}kx^2 Finally, the system has damping, which we introduce as a Rayleigh dissipation function (F): F = \frac{1}{2}c\dot{x}^2 Now, the Lagrangian becomes: L = T - V - F L = \frac{1}{2}m\dot{x}^2 - \frac{1}{2}kx^2 - \frac{1}{2}c\dot{x}^2
03

Apply Lagrange's Equation

To derive the equation of motion, we will apply Lagrange's Equation, which is given by: \frac{d}{dt}(\frac{\partial L}{\partial \dot{x}}) - \frac{\partial L}{\partial x} = 0 Let's calculate the required partial derivatives for our system: \frac{\partial L}{\partial \dot{x}} = m\dot{x} - c\dot{x} \frac{\partial L}{\partial x} = -kx Now, take the time derivative of the first partial derivative: \frac{d}{dt}(\frac{\partial L}{\partial \dot{x}}) = m\ddot{x} - c\ddot{x} Now, plug the derivatives back into Lagrange's Equation: m\ddot{x} - c\ddot{x} + kx = 0
04

Write the final equation of motion

We can now simplify the equation of motion to its final form by collecting the terms with the same variables: (m-c)\ddot{x} + kx = 0 This is the equation of motion for a simple mass-spring-damper system derived using Lagrange's Equations.

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Most popular questions from this chapter

Use Lagrange's Equations to derive the equations of motion for the constrained hook and ladder shown. The spring is untorqued when \(\theta=0\) and the tip of the ladder is subjected to a downward vertical force \(F_{2}\) as indicated.

Two wheels, each of mass \(m\) and radius \(r_{w}\), are connected by an elastic coupler of effective stiffness \(k\) and undeformed length \(L\). The system rolls without slip around a circular track of radius \(R\), as shown. Derive the equations of motion of the wheel system.

A rigid rod of length \(L\), and mass \(m_{a}\) is connected to a rigid base of mass \(m_{b}\) through a torsional spring of stiffness \(k_{T}\) as shown. The base sits on an elastic support of stiffiness \(k\) as indicated. Derive the equations of motion of the system using Lagrange's Equations.

Use Lagrange's Equations to derive the equations of motion for the triple pendulum whose bobs are subjected to horizontal forces \(F_{1}, F_{2}\) and \(F_{3}\), respectively.

Derive the equations of motion for the crankshaft system shown in Figure P6.6 using Lagrange's Equations. The spring is undeformed when the connecting pin \(A\) is directly above or below the hub of the wheel.
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