Kelon McKoy

https://www.dropbox.com/scl/fi/x9isc7x1tun6wul42gcxu/Project-3-Kelon-McKoy.MOV?dl=0

Professor Kate Poirier | OL67 | Fall 2020

Spring Problems I focuses on undamped spring systems. That is, a spring system that lacks any forces counteracting the restorative force of the spring itself. This is a purely theoretical exercise as there are usually energy loses due to heat, drag against a resistive medium like water or air, a dedicated dampening force, or other external forces.

Fig 1 shows three key positions when discussing spring systems. L represents the unloaded spring length. “l” represents the spring elongation, sometimes using “s” for notation. When there is a mass on the spring and the spring system is at rest it is said to be at equilibrium. “y” represents an additional displacement beyond the equilibrium point, in this case either above or below.

Partner Group: Problem 1:

Partner Group: Problem 3:

Partner Group: Problem 4:

Partner Group: Problem 11:

Partner Group: Problem 13:

Partner Group: Problem 19:

Partner Group: Problem 21:

By Sierra Morales and Oscar Pais

In this video we will solve an undamped spring system problem which requires the application of a linear second order equation to solve. This particular problem focuses on the variables associated with the system’s equilibrium, spring elongation and an initial velocity.

Key concepts to understanding this problem are:

- Newton’s Second Law of Motion: Force = mass * acceleration
- Hooke’s Law: spring Force = – spring constant * spring stretch or compression

We will be using the following second order differential equation to solve this problem:

\begin{math} y” + \frac{k}{m} \:y\:=\:0 \end{math}

Where:

- y” represents the an acceleration or deceleration
- k is the spring constant
- m is mass
- y is a displacement value

Note: This problem does not list a value for mass. If you notice the fraction \begin{math} \frac{k}{m} \end{math} has m in the denominator, knowing that that value could not be zero you could either treat \begin{math}\frac{k}{m} \end{math} as a computed value as we did or assign a value of one to mass.

Ultimately our solution will take the form of:

Applying a bit of intuition to this graph shows that over time the frequency of the oscillations begin to decrease, that is to say a lower frequency. However, since this is an undamped spring system notice that the amplitude never decreases. That is the spring system has the ability to restore all of its energy in compression and elongation.

Video Lesson:

https://www.dropbox.com/s/4q5klxo489chygu/RPReplay_Final1606190322.MP4?dl=0

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