Question

A mass weighing 16 pounds is attached to a 5-foot-long spring. At equilibrium the spring measures 8.2 feet (In Hooke's law F = kx, we need to replace x with 8.2-5-3.2 ft.). If the mass is initially released from rest at a point 2 feet above the equilibrium position, (x(0) = -2, x'(0) = 0)

Find the displacement (Equation of the motion) if it is further known that the surrounding medium offers a resistance numerically equal to the instantaneous velocity (Fd=x'(t).

Answer 1

The student's question involves creating an equation of motion for a damped harmonic oscillator using differential equations and understanding of Hooke's Law and resistance proportionate to velocity.

Step-by-step explanation:

understanding of Hooke's Law and resistance proportionate to velocity:

To find the equation of motion for a mass attached to a spring experiencing simple harmonic motion (SHM) and dampening due to resistance proportional to velocity, we would typically use the second-order linear differential equation: m x''(t) + γ x'(t) + k x(t) = 0 ,where m is the mass of the object, γ is the dampening coefficient, k is the spring constant, and x(t) is the displacement as a function of time.

Starting with Hooke's Law, F = -kx, and acknowledging the resistance force Fd = -γ x'(t), we note that the spring constant k can be calculated using the weight of the mass (16 pounds is approximately 71.17 Newtons considering g = 9.81 m/s²) and the displacement (3.2 feet is approximately 0.975 meters) as k = F/x. Given the initial conditions (x(0) = -2 feet and x'(0) = 0), the displacement equation can be solved using methods for linear constant coefficient differential equations, typically resulting in an exponentially decaying sinusoid due to the dampening term γ.