Problem 16 - Calculus: An Integrated Approach to Functions and Their Rates of Change

Question

A spring with a \(2-\mathrm{kg}\) mass has a natural length of \(0.6 \mathrm{~m} .\) A \(10 \mathrm{~N}\) force is required to compress it to a length of \(0.5 \mathrm{~m}\). If the spring is compressed to \(0.4 \mathrm{~m}\) and released, find the position of the mass at time \(t\). Assume a frictionless system.

Step-by-Step Solution

Answer

The position of the mass at time \( t \) is given by \( x(t) = 0.2 \cos(10t) \) m.

Step 1: Determine the spring constant k

From Hooke's law, the force required to change the spring's length away from its natural length is given by \( F = k \Delta x \), where \( k \) is the spring constant and \( \Delta x \) is the change in length. Given that a force of 10 N is required to compress the spring by 0.1 m (from 0.6 m to 0.5 m), we can rearrange the equation and solve for \( k \) as \( k = F / \Delta x = 10 / 0.1 = 100 \) N/m.

Step 2: Set up the differential equation for the spring motion

Assuming no damping (frictionless system), the motion of the spring follows the simple harmonic oscillator model and is governed by the following ordinary differential equation \( m \frac{d^2 x}{dt^2} + k x = 0 \), where \( m = 2 \) kg is the spring's mass, \( x \) is the position of the spring at time \( t \), and \( \frac{d^2 x}{dt^2} \) is the acceleration. Substituting values gives \( 2 \frac{d^2 x}{dt^2} + 100 x = 0 \).

Step 3: Solve the differential equation

The differential equation is a 2nd order constant coefficient homogeneous differential equation with solution of the form \( x = A \cos(\omega t + \phi) \), where \( \omega = \sqrt{k / m} = \sqrt{100 / 2} = 10 \) rad/s, \( A \) is the amplitude, \( t \) is the time, and \( \phi \) is the phase angle. To find \( A \) and \( \phi \), we need initial conditions. Here, we're given that the spring is compressed from 0.6 m (natural length) to 0.4 m. So \( x(0) = \delta = 0.6 - 0.4 = 0.2 \) m initially and \( x'(0) = 0 \) since the velocity is 0 at maximum compression. These imply \( A = \delta \) and \( \phi = 0 \). So the solution is \( x(t) = \delta \cos(\omega t) = 0.2 \cos(10t) \) m.

Key Concepts

Hooke's Law

Hooke's Law helps us understand how springs behave when they are stretched or compressed. It states that the force required to stretch or compress a spring is proportional to the distance it is stretched or compressed from its natural length. This relationship is expressed mathematically as:

• \( F = k \Delta x \)
• Where:
  • \( F \) is the force applied to the spring (in Newtons, N)
  • \( k \) is the spring constant (in N/m)
  • \( \Delta x \) is the change in spring length (in meters, m)

To understand this better, let's connect it to our exercise example. A force of 10 N compresses the spring from its natural length of 0.6 m to 0.5 m. This means the spring's length changes by 0.1 m. Using Hooke's Law, we calculated the spring constant \( k \) as 100 N/m. This constant tells us how stiff the spring is.

The higher the value of \( k \), the more force is needed to compress or stretch the spring. Understanding Hooke's Law is crucial in studying oscillations and vibrations in various systems, such as car suspensions and earthquake-prone building designs.

Differential Equation

To understand how a spring moves over time, we use a mathematical tool called a differential equation. In simple harmonic motion, the system can be described by Newton's second law of motion. For our exercise, since we assume no friction, the equation simplifies to:

• \( m \frac{d^2 x}{dt^2} + k x = 0 \)
• Where:
  • \( m \) is the mass attached to the spring (in kilograms, kg)
  • \( \frac{d^2 x}{dt^2} \) represents the acceleration (in m/s²)
  • \( x \) is the displacement from the spring's natural length (in meters, m)

When we plug in our values, with mass \( m = 2 \) kg and spring constant \( k = 100 \) N/m, we get the differential equation:

\( 2 \frac{d^2 x}{dt^2} + 100 x = 0 \).

This specific form of the equation describes how the mass moves back and forth over time in simple harmonic motion. Solving this equation helps us find the spring's position at any given time, crucial for understanding oscillating systems such as pendulums or heartbeats.

Spring Constant

The spring constant, noted as \( k \), is an essential component in understanding springs and their behavior. It indicates how stiff a spring is.

• A larger \( k \) means the spring is stiffer, requiring more force to change its length.
• A smaller \( k \) indicates a more flexible spring that stretches or compresses more readily.

In our exercise, we determined \( k \) to be 100 N/m. We derived this value from observing how much force was needed to compress the spring by a specific amount. Specifically, applying a force of 10 N reduced its length by 0.1 m.

Understanding the spring constant helps in analyzing systems that involve springs, such as designing shock absorbers in vehicles or providing cushioning in athletic shoes. It is also a fundamental part of building models for complex engineering systems where precise motion is necessary.