91Ó°ÊÓ

Hooke's Law is a fundamental principle in physics that describes how springs behave. When a spring is either stretched or compressed, it exerts a restoring force that tries to return it to its equilibrium position. The magnitude of this force is directly proportional to the displacement of the spring from its natural length. Mathematically, this is expressed as: \[ F = kx \]Here,

• \( F \) represents the force exerted by the spring,
• \( k \) is the spring constant, which measures the stiffness of the spring,
• \( x \) is the displacement from the equilibrium position.

In our exercise, a 16-pound weight stretches the spring 6 inches (0.5 feet), which we use to calculate the spring constant: \[ k = \frac{F}{x} = \frac{16}{0.5} = 32 \frac{\text{lb}}{\text{ft}} \]This spring constant \( k \) tells us how much force per unit length is needed to stretch or compress the spring.

The natural frequency of a spring-mass system is crucial because it defines how the system oscillates when disturbed from its equilibrium position, without external forces acting on it. This frequency is determined by the mass of the object attached to the spring and the spring constant. The formula to find the natural frequency \( \omega_n \) is: \[ \omega_n = \sqrt{\frac{k}{m}} \]Where,

• \( k \) is the spring constant,
• \( m \) is the mass of the object.

In our problem, the mass is calculated using weight \( W \), where \( W = mg \) and rearranging gives \( m = \frac{W}{32 \frac{\text{lb}}{\text{ft}}} = 0.5 \text{slug} \). Plugging this into the formula: \[ \omega_n = \sqrt{\frac{32}{0.5}} = \sqrt{64} = 8 \frac{1}{s} \]So, the natural frequency of our spring-mass system is 8 radians per second. This value helps us understand the system’s undisturbed oscillation properties.

A particular solution is one part of solving a non-homogeneous differential equation, which includes both the homogeneous solution and the particular solution. The homogeneous solution represents natural oscillations of the system, while the particular solution accounts for external forces. For the differential equation \[ \frac{d^2 y}{dt^2} + 64 y = 4 \cos t \]we assume a form for the particular solution that fits the type of external force. For our given external force \( f(t) = 4 \cos t \), we assume a particular solution of the form: \[ y_p = A \cos t \]Substituting this guess into the differential equation, we find: \[-A \cos t + 64A \cos t = 4 \cos t \]Solving, we get: \[A = \frac{4}{63} \]So the particular solution becomes: \[ y_p = \frac{4}{63} \cos t \]Combined with the homogeneous solution, the general form for the system's displacement is: \[ y(t) = C \cos(8t) + D \sin(8t) + \frac{4}{63} \cos t \]