Question

A mass weighing 16 pounds is attached to a 5-foot-long spring. At equilibrium, the spring measures 8.2 feet. If the mass is initially released from rest at a point 2 feet above the equilibrium position, [x(0) = 2,x(0) = 0]

a) Find the displacement (Equation of the motion) if it is further known that the surrounding medium offers resistance numerically equal to the instantaneous velocity (F = x(t)).

Answer 1

x(t) = 2 * exp(-t/32) * cos(sqrt(15)/32 * t) This is the equation of motion for the displacement of the mass, considering the surrounding medium's resistance.

a) Finding the displacement equation

Modeling the System:

Mass (m): 16 pounds

Spring constant (k): We need to find it.

Damping coefficient (b): 1 (given)

Initial position (x(0)): 2 feet

Initial velocity (x'(0)): 0 feet/second

Spring constant:

At equilibrium, the spring is stretched by 8.2 - 5 = 3.2 feet.

Using Hooke's Law: F = kx, where F = 16 pounds (weight of the mass).

Therefore, 16 = k * 3.2 => k = 5 pounds/foot.

Equation of motion:

Using the damped harmonic motion equation:

m * x''(t) + b * x'(t) + k * x(t) = 0

Substituting the values:

16 * x''(t) + 1 * x'(t) + 5 * x(t) = 0

Solving the equation:

The characteristic equation is:

r^2 + r/16 + 5/16 = 0

Solving the equation gives two complex roots:

r1,2 = -1/32 +/- i * sqrt(15)/32

Therefore, the displacement equation is:

x(t) = C1 * exp(-t/32) * cos(sqrt(15)/32 * t) + C2 * exp(-t/32) * sin(sqrt(15)/32 * t)

Finding C1 and C2:

Using the initial conditions:

x(0) = 2 => C1 = 2

x'(0) = 0 => C2 * sqrt(15)/32 = 0 => C2 = 0

Final equation:

x(t) = 2 * exp(-t/32) * cos(sqrt(15)/32 * t)

This is the equation of motion for the displacement of the mass, considering the surrounding medium's resistance.