91Ó°ÊÓ

To understand how a spring behaves, we need to explore **Hooke's Law**. This fundamental principle of mechanics describes how a spring stretches or compresses in response to applied forces. The law is expressed mathematically as \[ F = kx \], where:

• \( F \) is the force applied to the spring.
• \( k \) is the spring constant, indicating the stiffness of the spring.
• \( x \) is the displacement of the spring from its equilibrium position.

In our exercise, a force of 4 pounds stretches a spring by 2 feet. Applying Hooke's Law, we find the spring constant, \( k = \frac{4}{2} = 2 \) pounds per foot. This means that for every foot the spring is stretched, it resists with 2 pounds of force. Thus, **Hooke's Law** provides the basis for understanding the spring's resistance and contribution to motion equations.

The concept of damping is essential to understanding motion affected by resistive forces. A **Damping Force** acts oppositely to the velocity of an object, reducing its energy over time. This is common in systems where you want to prevent perpetual motion due to friction or other resistive forces.
In our exercise, the damping force is described by the expression \[ rac{7}{8} \] times the instantaneous velocity. This implies that for each unit of velocity, the damping force acts to slow it down by \( \frac{7}{8} \). Hence, the damping coefficient \( c \) is \( \frac{7}{8} \).

• The damping coefficient measures how quickly a system's motion is reduced.
• It's crucial for handling oscillations, ensuring they stabilize over time.

Understanding how damping affects motion helps in forming differential equations, which describe the complete motion of weighted springs in resistive mediums.

The **Spring Constant** is a vital characteristic of any spring, defining its stiffness. Denoted by \( k \), it is determined by how much force is required to stretch or compress the spring by a unit distance.
From Hooke's Law, we know \[ k = \frac{F}{x} \]. In our exercise, a 4-pound force stretches the spring 2 feet, giving \[ k = \frac{4}{2} = 2 \] pounds per foot.

• Higher \( k \) values mean the spring is harder to stretch or compress.
• Lower values indicate a more easily movable spring.

The spring constant not only affects how the spring behaves under load but also how it influences the system's natural frequency and potential energy storage. A firm grasp of \( k \) helps in setting up and interpreting motion equations in dynamic systems.

A **Differential Equation of Motion** formulates how forces acting on a system influence its movement over time. When considering springs and weights, this equation is crucial to predict motion behaviors.
The general form with damping is \[ mx'' + cx' + kx = 0 \], where:

• \( m \) is the mass of the object.
• \( x'' \) represents acceleration.
• \( cx' \) is the damping term aligned with the velocity.
• \( kx \) represents the restoring force of the spring.

In the given problem:

• \( m = \frac{1}{8} \) slug, calculated from the weight \( \frac{4}{32} \) (as gravity \( g = 32 \) ft/s²).
• \( c = \frac{7}{8} \) from the damping force.
• \( k = 2 \) derived from the spring constant calculation.

Plugging these into the differential equation gives \[ \frac{1}{8}x'' + \frac{7}{8}x' + 2x = 0 \]. Solving this equation using the Laplace transform and initial conditions helps to determine \( x(t) \), describing the exact motion of the system.