Final answer:
To find the mass (m) and spring constant (k), we use Hooke's Law. The equation of motion is mx'' + kx = 0. The time to pass through the equation of motion is T = 2π√(m/k). The time after passing through equilibrium can be found using t = T/4. The position of the mass at this instant is x(0.738 s) = 2.029 in below the equilibrium position.
Step-by-step explanation:
To determine the mass and spring constant in this scenario, we can use Hooke's Law and Newton's second law of motion:
(a) **Mass (m):** Using Hooke's Law, F = kx, and substituting the given values, we have 10 lb = k(6 in). Converting lb to kg, we get m = 4.536 kg.
**Spring constant (k):** Using Hooke's Law again, we can rearrange the formula to solve for k. We have k = F/x = (10 lb)(4.45 N/lb) / (6 in)(0.0254 m/in) = 18.72 N/m.
(b) **Equation of motion:** The equation of motion for a mass-spring system without external forces is given by mx'' + kx = 0. Plugging in the values, we get 4.536 kg * x'' + 18.72 N/m * x = 0.
(c) **Time to pass through the equation of motion:** The time to pass through the equation of motion is given by T = 2π√(m/k). Substituting the known values, we have T = 2π√(4.536 kg / 18.72 N/m) = 2.954 s.
(d) **Time after passing through equilibrium:** The time after passing through equilibrium can be found by calculating the time it takes to travel half of the amplitude (A/2) from the equilibrium position. In this case, the equilibrium position is at 0 since it is below the displacement. The time is given by t = T/4 = 0.738 s.
(e) **Position of mass at this instant:** The position of the mass at this instant can be calculated using x(t) = A cos(2πt/T). Substituting the known values, we have x(0.738 s) = (6 in) * cos(2π * 0.738 s / 2.954 s) = 2.029 in below the equilibrium position.
(f) **Is mass above or below equilibrium position:** As calculated in part (e), the mass is 2.029 in below the equilibrium position.