Final answer:
The question involves calculating the equation of motion for a mass attached to a spring released from rest in an example of simple harmonic motion. By applying Newton's second law and Hooke's law, the spring constant is determined, and the SHM equations are used to derive position, velocity, and acceleration as functions of time.
Step-by-step explanation:
The question involves a mass attached to a spring that is released from rest and the task is to find the equation of motion for the system. This scenario depicts simple harmonic motion (SHM), which is a common topic in physics, specifically in the study of oscillations. To find the equation of motion, we use Hooke's law and the properties of SHM. When the mass is released from rest 4 inches above the equilibrium position of a spring (stretched 4 inches by the mass), we first need to determine the force constant of the spring, k, using the weight of the mass and the distance of stretch. Then, applying Newton's second law, we can express the motion using the second-order differential equation that characterizes SHM, and solve for the position of the mass as a function of time.
The force constant k is found by converting the weight of the mass to Newtons (since weight is a force) and dividing it by the stretch of the spring converted to meters. The weight is W = mg = 24 lb (using 1 lb = 4.44822 N and 1 inch = 0.0254 m). The spring constant k is W / 4 inches. Next, using the classical SHM equation x(t) = A cos(\(\omega t + \phi\)), where \(\omega = \sqrt{k/m}\) is the angular frequency, we can find the equation of motion. The amplitude A, initial phase \phi, and mass m can be derived from the given conditions (initial displacement and rest condition).
We can then calculate the position, velocity, and acceleration of the mass at any given time t using the derivatives of the equation of motion. For velocity v(t), we take the first derivative of position with respect to time, and for acceleration a(t), we take the second derivative.