Problem 22 - A Graphical Approach to Precalculus with Limits

Question

Spring Motion \(\quad\) A weight attached to a spring is pulled down 3 inches below the equilibrium position. (a) Assuming that the frequency is \(\frac{6}{\pi}\) oscillations per second, find a trigonometric model that gives the position of the weight at time \(t\) seconds. (b) What is the period?

Step-by-Step Solution

Answer

The trigonometric model is \( y(t) = 3\cos(12t) \) and the period is \( \frac{\pi}{6} \) seconds.

Step 1: Understand the Problem

We need to find a trigonometric model for the spring's motion and the period of its oscillation. The position of the weight can be modeled by a sine or cosine function.

Step 2: Identify Given Variables

The weight is pulled 3 inches below the equilibrium, making the amplitude 3. The frequency is given as \( \frac{6}{\pi} \) oscillations per second, which is needed to find the angular frequency.

Step 3: Determine Angular Frequency

The angular frequency \( \omega \) is calculated by: \( \omega = 2\pi \times \text{frequency} \). Substitute the given frequency \( \frac{6}{\pi} \):\[ \omega = 2\pi \times \frac{6}{\pi} = 12 \]

Step 4: Construct the Trigonometric Model

Considering the weight is initially displaced and released, assume a cosine function since it generally represents the maximum displacement from equilibrium. Use the formula:\[ y(t) = A \cos(\omega t) \]Substitute \( A = 3 \) and \( \omega = 12 \): \[ y(t) = 3 \cos(12t) \]

Step 5: Find the Period of Oscillation

The period \( T \) of oscillation is given by the formula:\[ T = \frac{2\pi}{\omega} \]Substitute \( \omega = 12 \): \[ T = \frac{2\pi}{12} = \frac{\pi}{6} \] seconds.

Key Concepts

Spring Motion

Spring motion describes how an object connected to a spring moves back and forth from its equilibrium position. This type of motion is often referred to as oscillation due to its repetitive nature. When a weight is attached to a spring and displaced, the energy in the system causes it to move up and down about an equilibrium position. The spring's restoring force attempts to bring the weight back to equilibrium.
Key factors influencing spring motion include:

• Amplitude: The maximum distance from the equilibrium position (3 inches in this exercise).
• Restoring Force: The force exerted by the spring, typically explained by Hooke's law, which states that force is proportional to displacement.
• Energy Exchange: Potential energy in the spring and kinetic energy in the moving mass are exchanged continuously.

In essence, a trigonometric model using sine or cosine functions is ideal to represent the cyclical pattern of spring motion. Here, a cosine function was chosen because it often signifies initial displacement from equilibrium.

Angular Frequency

Angular frequency is a crucial concept in understanding oscillatory motion. It describes how fast the object oscillates and is usually denoted by the Greek letter omega (\(\omega\)). To find the angular frequency, multiply the linear frequency by \(2\pi\).
In mathematical terms, the angular frequency is calculated as follows: \\[\omega = 2\pi \times \text{frequency}\]This conversion is important because it helps relate the oscillatory motion to circular motion, providing a deeper insight into how the trigonometric functions describe motion.
For our exercise:

• The given frequency is \(\frac{6}{\pi}\).
• Substituting it into the equation yields \(\omega = 12\).

This value is essential in accurately modeling the motion using the trigonometric function \(y(t) = A \cos(\omega t)\), allowing you to comprehend how the object moves over time.

Oscillation Period

The oscillation period represents the time it takes for an object to complete one full cycle of motion. It's a key parameter that helps define the characteristics of the oscillatory behavior. The period is inversely related to the frequency and is derived using the angular frequency: \[T = \frac{2\pi}{\omega}\]In practical terms, the period tells you how long it takes for the weight to return to the same position, moving through all possible locations in its range of motion once.
Using our example:

• With \(\omega = 12\), substitute into the formula to find \(T = \frac{\pi}{6}\).
• This means it takes \(\frac{\pi}{6}\) seconds for the weight to complete one full oscillation.

Understanding the period helps in predicting the behavior of the system over time, as repeated cycles continue indefinitely in ideal conditions without damping.