91Ó°ÊÓ

A spring-mass system is a classic physics model where a mass is attached to a spring, and the entire setup is either on a flat surface or hung vertically. In our problem, we deal with a vertical spring supporting a weight. This creates potential energy when the spring is stretched or compressed.
The force exerted by the spring follows Hooke's Law, which states:

• The force exerted is directly proportional to the displacement from its equilibrium position: \( F = kx \), where \( F \) is the force, \( k \) is the spring constant, and \( x \) is the displacement.
• In our example, the spring stretches by \( \frac{1}{3} \) feet under a load of 32 lbs, leading to a spring constant \( k = 96 \text{ lb/ft} \).

A spring-mass system can be damped, meaning it gradually loses energy, preventing indefinite oscillation. The damping is proportional to the velocity of the mass, represented by a damping constant.

Differential equations offer a way to understand how a system evolves over time, describing rates of change. For a damped spring-mass system, the motion is formulated as a second-order linear differential equation: \[ m \frac{d^2y}{dt^2} + c \frac{dy}{dt} + ky = 0. \] Here, each term represents different aspects of the system:

• \( m \frac{d^2y}{dt^2} \) accounts for the mass's acceleration.
• \( c \frac{dy}{dt} \) represents damping, dependent on velocity. In our problem, \( c = 4 \).
• \( ky \) is the restoring force from the spring itself.

Incorporating initial conditions, like initial displacement \( y(0) \) and velocity \( \frac{dy}{dt}(0) \), helps tailor the equation to specific scenarios. In this exercise, the mass is released at equilibrium with a velocity of 12 ft/sec.

To analyze the nature of solutions, we derive a characteristic equation from our differential equation: \[ r^2 + 4r + 96 = 0. \] This is achieved by assuming solutions of the form \( y(t) = e^{rt} \), leading to simplification into an algebraic equation. Solving this equation helps us understand the system's damping and oscillatory behavior.
The characteristic equation is a quadratic with coefficients \( a = 1 \), \( b = 4 \), and \( c = 96 \), and its discriminant \( b^2 - 4ac \) is essential:

• If positive, the roots are real and distinct, indicating overdamping.
• If zero, the roots are real and equal, signifying critical damping.
• If negative, as in our problem \( -368 \), the roots are complex, hinting at underdamped motion.

Underdamped motion occurs when a system oscillates due to a negative discriminant in its characteristic equation. The solution features complex roots, resulting in oscillations that gradually diminish due to the damping effect. In our case, the roots are \( r = -2 \pm i\sqrt{92} \), indicating such behavior.
The general solution to the motion equation is: \[ y(t) = e^{-2t}(C_1\cos(\sqrt{92}t) + C_2\sin(\sqrt{92}t)). \] Here, the \( e^{-2t} \) factor represents an exponential decay, while the trigonometric functions describe oscillations.

• \( C_1 \) and \( C_2 \) are constants determined from initial conditions.
• The oscillation frequency is governed by \( \sqrt{92} \).

In our exercise, using initial conditions \( y(0) = 0 \) and initial velocity \( 12 \) ft/sec, we find \( C_1 = 0 \), leaving the oscillatory component driven by \( C_2 \). The behavior is characterized by diminishing oscillations centred around the equilibrium.