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Chapter 6: Problem 10

A \(32-\mathrm{lb}\) weight ( 1 slug) stretches a vertical spring 4 in. The resistance in the spring-mass system is equal to four times the instantaneous velocity of the mass. a. Find the equation of motion if it is released from its equilibrium position with a downward velocity of \(12 \mathrm{ft} / \mathrm{sec}\). b. Determine whether the motion is overdamped, critically damped, or underdamped.

Short Answer

Expert verified
Equation of motion is underdamped; it involves oscillations with reducing amplitude.

Step by step solution

01

Set Up the Problem

We know that the system involves a spring mass with a weight of 32 lb stretching it 4 inches. The spring constant can be found using Hooke鈥檚 Law: \[ F = kx \]where \( F = 32 \text{ lb} \) and \( x = \frac{4}{12} \text{ ft} = \frac{1}{3} \text{ ft} \). So, the spring constant \( k \) is 96 lb/ft.
02

Write the Differential Equation

The equation of motion for the spring-mass-damping system is given by:\[ m\frac{d^2y}{dt^2} + c\frac{dy}{dt} + ky = 0 \]where the mass \( m \) is 1 slug, the damping constant \( c = 4 \) (resistance four times velocity), and the spring constant \( k = 96 \).
03

Plug in Initial Conditions

The initial conditions are that the system is released from equilibrium, so \( y(0) = 0 \). The initial velocity condition is a downward velocity of 12 ft/s, so \( \frac{dy}{dt}(0) = -12 \). We apply these conditions to solve the equation for specific solution values.
04

Solve the Characteristic Equation

For a system described as:\[ m\frac{d^2y}{dt^2} + c\frac{dy}{dt} + ky = 0 \]The characteristic equation is:\[ mr^2 + cr + k = 0 \]Substituting the values, we have:\[ r^2 + 4r + 96 = 0 \]Applying the quadratic formula, or simply calculating discriminant, we get the solution to this equation.
05

Analyze Damping Characteristics

The discriminant \(\Delta\) of the quadratic equation is given by:\[ \Delta = c^2 - 4mk \]Substituting the values, \( \Delta = 4^2 - 4 \times 1 \times 96 = 16 - 384 = -368 \).Since \( \Delta < 0 \), this indicates the system is underdamped.

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Key Concepts

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

Spring-Mass System
In the realm of physics and engineering, a "spring-mass system" is a model used to understand how forces interact on a spring connected to a mass. Imagine you have a mass attached to a spring; when you pull or push this mass, the spring either contracts or stretches accordingly. This is a simple yet powerful way to study motion.Here are a few key points:
  • The force exerted by the spring is proportional to the displacement from its equilibrium position. This is described by Hooke's Law: \( F = kx \), where \( F \) is the force in pounds or newtons, \( k \) is the spring constant, and \( x \) is the displacement in feet or meters.
  • The spring constant \( k \) is a crucial factor鈥攊t tells us how stiff the spring is. A high \( k \) means a stiff spring, while a low \( k \) suggests a more flexible spring.
  • The motion can be described by a second-order linear differential equation. This equation considers the mass of the object, the damping effect, and the spring's force.
Understanding the behaviors and laws governing a spring-mass system is fundamental to solving many mechanical and dynamic systems problems.
Damping in Systems
Damping refers to the forces that reduce motion over time. Think of it as a way to slow down or calm the repetitive bouncing or oscillating of the spring-mass system. In the given problem, damping is crucial, and here's why:
  • Damping Constant (\(c\)): The damping force is often proportional to the velocity. In the exercise, the damping constant was given as 4, implying that the resistance force was four times the instantaneous velocity.
  • Types of Damping: Systems can exhibit different damping behaviors based on the discriminant of their characteristic equation:
    • Overdamped: The system returns to rest without oscillating (\(\Delta > 0\)).
    • Critically Damped: The system returns to equilibrium faster without oscillating (\(\Delta = 0\)).
    • Underdamped: The system oscillates with decreasing amplitude (\(\Delta < 0\)), which is the case in the exercise.
Damping in engineering ensures that mechanical systems can efficiently come to rest, maintaining stability and safety.
Characteristic Equation
The characteristic equation is a cornerstone in modeling systems like the spring-mass system. It helps us determine how the system behaves over time.Here鈥檚 how it works:
  • The standard form for a second-order differential equation with constant coefficients is: \( m\frac{d^2y}{dt^2} + c\frac{dy}{dt} + ky = 0 \).
  • The characteristic equation derived from this is: \( mr^2 + cr + k = 0 \), where \( r \) represents the roots.
  • The roots are found using the quadratic formula: \( r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), providing insight into the system's response.
  • In the exercise, the discriminant \( \Delta = c^2 - 4mk \) was calculated as \(-368\), showing that the roots are complex, indicating an underdamped system.
Understanding the characteristic equation gives a mathematical way to predict the motion's behavior, whether it stabilizes quickly, oscillates, or gradually returns to rest.

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