91Ó°ÊÓ

Hooke's Law is a fundamental principle in physics that relates the force exerted by a spring to its displacement from equilibrium. It is given by the equation: \[ F = kx \] Where:

• \textbf{F} is the force applied to the spring (in pounds or Newtons).
• \textbf{k} is the spring constant (a measure of the stiffness of the spring, in pounds per foot or Newtons per meter).
• \textbf{x} is the displacement of the spring from its equilibrium position (in feet or meters).

In the given exercise, a 6 lb weight stretches a spring 6 inches (0.5 feet). Using Hooke's Law, we find the spring constant: \[ k = \frac{F}{x} = \frac{6 \text{ lb}}{0.5 \text{ ft}} = 12 \text{ lb/ft} \] This spring constant will be used in the differential equation describing the motion of the spring-mass system.

A non-homogeneous differential equation includes an external force or function on one side. These equations model many physical systems where external influences are considered. The general form is: \[ m \frac{d^2u}{dt^2} + ku = f(t) \] In our exercise, the spring-mass system is influenced by an external force given by \[ f(t) = 2 \text{ cos}(5t) \]. Combining with Hooke's Law, the equation becomes: \[ 0.1875 \frac{d^2u}{dt^2} + 12u = 2 \text{ cos}(5t) \] Simplifying: \[ \frac{d^2u}{dt^2} + 64u = 8 \text{ cos}(5t) \] This equation must be solved by finding both the homogeneous solution (solution to the related homogeneous equation without the external force) and a particular solution corresponding to the external force.

The spring-mass system is a classic problem in physics and engineering, modeling oscillatory motion. It considers a mass attached to a spring that obeys Hooke's Law. The motion can be described by a second-order linear differential equation: \[ m \frac{d^2u}{dt^2} + ku = 0 \] The mass (\textbf{m}) can be found using the weight equation \textbf{W} = \textbf{mg}, where g is the gravitational acceleration (32 ft/s\(^2\) in feet). For our problem: \[ m = \frac{W}{g} = \frac{6 \text{ lb}}{32 \text{ ft/s}^2} = 0.1875 \text{ slugs} \] Ignoring damping (frictional forces), we solve the homogeneous part of the equation first: \[ \frac{d^2u}{dt^2} + 64u = 0 \] The characteristic equation of this homogeneous differential equation is: \[ r^2 + 64 = 0 \] Solving it: \[ r = \pm 8i \] Thus, the solution of the homogeneous equation is: \[ u_h(t) = c_1 \text{ cos}(8t) + c_2 \text{ sin}(8t) \] Where \textbf{c\textsubscript{1}} and \textbf{c\textsubscript{2}} are constants determined by initial conditions.

External forces add complexity to the differential equations by making them non-homogeneous. To find the complete solution in such cases, we need to find a particular solution, which in this exercise is due to the external force \[ f(t) = 2 \text{ cos}(5t) \]. Assume a particular solution in the same form as the external force: \[ u_p(t) = A \text{ cos}(5t) + B \text{ sin}(5t) \] Substituting this into the differential equation and equating coefficients, we find:\[ A = \frac{8}{39}, \text{ and } B = 0 \] Thus, the particular solution is: \[ u_p(t) = \frac{8}{39} \text{ cos}(5t) \] Combining the homogeneous and particular solutions gives the general solution: \[ u(t) = c_1 \text{ cos}(8t) + c_2 \text{ sin}(8t) + \frac{8}{39} \text{ cos}(5t) \] Using initial conditions, we solve for \textbf{c\textsubscript{1}} and \textbf{c\textsubscript{2}}. Here, the object is raised 3 inches (0.25 feet) above equilibrium and released: \[ u(0) = -0.25 \text{ ft} \text{ and } u'(0) = 0 \] After applying these conditions: \[ c_1 = \frac{-47}{39}, \text{ and } c_2 = 0 \] The final solution for the displacement is: \[ u(t) = \frac{-47}{39} \text{ cos}(8t) + \frac{8}{39} \text{ cos}(5t) \]