91Ó°ÊÓ

A differential equation is a key mathematical tool that relates a function with its derivatives. In this spring mass system exercise, we deal with a second-order linear homogeneous differential equation. Such equations are crucial in modeling physical systems that undergo oscillatory motion, like a weight on a spring.
The differential equation for our spring-mass system is expressed as: \[ m \frac{d^2x}{dt^2} + kx = 0 \] where:

• \( m \) is the mass of the object, which we converted to slugs for compatibility with our force units.
• \( k \) is the spring constant, expressing how stiff the spring is.
• \( x \) is the displacement from the equilibrium position.

Understanding this equation means recognizing how the mass's acceleration (\( \frac{d^2x}{dt^2} \)) interacts with the spring's restorative force (\( kx \)). This results in a system where the mass oscillates due to these forces acting in opposition to displacement.

Hooke's Law is the foundation of understanding how springs behave under load.
The law stipulates that the force \( F \) exerted by a spring is directly proportional to the displacement \( x \) from its equilibrium position, represented as:\[ F = -k \times x \] Here:

• \( k \) is the spring constant, which dictates how resistant the spring is to being stretched or compressed.
• The negative sign depicts the direction of the force, which is always opposite to the direction of displacement.
For this specific exercise, Hooke's law helps us set up the differential equation by knowing the spring constant \( k = 48 \text{ lb/ft} \). As the spring returns to equilibrium, it does so while providing a force proportional to how much it's stretched beyond that point. Since the mass is in motion, Hooke's Law assures that the restoring force leads to a consistent oscillatory pattern when there is no external damping or forcing.

Initial conditions are essential in determining the specific behavior of a differential equation solution. In this spring mass system, we need them to find the particular values for our displacement formula. The initial conditions predict how the system will behave at a starting point.
In our case:

• The initial position is \( x(0) = -\frac{5}{12} \) feet (converted from 5 inches).
• The initial velocity is \( \frac{dx}{dt}(0) = 4 \text{ ft/sec} \).

With these, we can solve for constants \( A \) and \( B \) in the general solution form: \[ x(t) = A \cos(\omega t) + B \sin(\omega t) \]These values come directly from plugging in our initial position and velocity into the formula, ensuring our solution respects these starting conditions and thus accurately depicts the system's motion over time.