Problem 18 - A Graphical Approach to Precalculus with Limits

Question

The formula for the up-and-down motion of a weight on a spring is given by $$s(t)=a \sin \sqrt{\frac{k}{m}} t$$ If the spring constant \(k\) is \(4,\) what mass \(m\) must be used to produce a period of 1 second?

Step-by-Step Solution

Answer

The mass \(m\) needed is approximately 0.101 kg or 101 grams.

Step 1: Understand the relationship between period and angular frequency

The period of a particle in simple harmonic motion is given by the reciprocal of the angular frequency. Given the formula \(s(t)=a \sin \sqrt{\frac{k}{m}} t\), the angular frequency \(\omega\) is \(\sqrt{\frac{k}{m}}\). The period \(T\) is related to \(\omega\) by the formula \[ T = \frac{2\pi}{\omega} \]\.

Step 2: Set up the equation for the period

We want the period \(T\) to be 1 second. Substituting the values into the equation from Step 1, we have \[ 1 = \frac{2\pi}{\sqrt{\frac{k}{m}}} \]Since we're given that \(k = 4\), substitute \(k\) to get:\[ 1 = \frac{2\pi}{\sqrt{\frac{4}{m}}} \]Solve for \(\sqrt{\frac{4}{m}}\) using algebra.

Step 3: Isolate \(\sqrt{\frac{4}{m}}\) and solve for the mass \(m\)

From the equation \(1 = \frac{2\pi}{\sqrt{\frac{4}{m}}}\), we need to solve for \(\sqrt{\frac{4}{m}}\): Multiply both sides by \(\sqrt{\frac{4}{m}}\):\[ \sqrt{\frac{4}{m}} = 2\pi \]Solve for \(\frac{4}{m}\) by squaring both sides:\[ \frac{4}{m} = (2\pi)^2 = 4\pi^2 \]Rearrange to solve for \(m\):\[ m = \frac{4}{4\pi^2} \]Simplify:\[ m = \frac{1}{\pi^2} \]

Step 4: Conclusion

Substitute \(\pi\) as approximately 3.14159, hence \(\pi^2 \approx 9.8696\). Therefore, \[ m \approx \frac{1}{9.8696} \approx 0.10132 \]. The mass \(m\) needed is approximately 0.101 kg or 101 grams.

Key Concepts

Period of Motion

In simple harmonic motion, the "period of motion" is the time it takes to complete one full cycle of the motion. This is crucial for understanding repetitive motions like those seen in springs and pendulums. The formula to find the period \( T \) of a particle undergoing simple harmonic motion is:

• \[ T = \frac{2\pi}{\omega} \]

where \( \omega \) is the angular frequency.In our particular problem, we are given that the period \( T \) is 1 second, meaning the up-and-down motion on the spring repeats every second. Understanding the period helps us calculate other parameters like angular frequency, which tells us how quickly the motion completes a cycle.

Angular Frequency

Angular frequency, represented as \( \omega \), describes how fast something oscillates in simple harmonic motion. It's distinct from the regular frequency because it measures per radians instead of per cycle.
To determine \( \omega \), the equation used is:

• \[ \omega = \sqrt{\frac{k}{m}} \]

where \( k \) is the spring constant, and \( m \) is the mass.This formulation allows you to see how angular frequency is directly impacted by the characteristics of the spring and the attached mass. In our scenario, we find \( \omega \) by rearranging the period formula. Given that the period \( T \) is \( 1 \) second, the relation becomes crucial when isolating the parts in our equation to find the appropriate mass when \( k = 4 \).

Spring Constant

A spring constant, denoted as \( k \), measures a spring's stiffness. It's a vital part of Hooke's Law, which is the foundation of simple harmonic motion. The spring constant tells us how much force is needed to stretch or compress the spring by a unit length.For any spring-based system, such as the one in our exercise, \( k \) can change the period and angular frequency of the motion. In our example, the problem gives \( k = 4 \). This means that the spring is relatively stiff, requiring a noticeable force for displacement and impacting the duration of each oscillation.

Mass Calculation

Finding the correct mass that fits a specified motion is part of solving simple harmonic motion problems. The mass affects both the period and the angular frequency of a system.
To determine the mass, use the equation derived from setting the desired period \( T \):

• \[ m = \frac{4}{4\pi^2} \]

Simplifying gives:

• \[ m = \frac{1}{\pi^2} \]

This mass calculation results in a mass approximately \( 0.10132 \) kg or \( 101 \) grams, assuming \( \pi \approx 3.14159 \). Identifying the right mass satisfies the needs of the formula to achieve the motion's required period.