Question

A mass weighing 4 pounds is attached to a spring whose constant is 2 lbf/ft. The medium offers a damping force that is numerically equal to the instantaneous velocity. The mass is initially at rest at a point 3 feet below the equilibrium position.

Answer 1

The equation of motion for a mass attached to a spring and subject to damping force can be expressed as m * d^2x/dt^2 + b * dx/dt + k * x = 0. Given the mass and spring constant in the question, we need to convert the units to N and m, respectively. This equation represents simple harmonic motion, where the position of the mass oscillates sinusoidally with time.

Step-by-step explanation:

The equation of motion for a mass attached to a spring and subject to damping force can be expressed as:

m * d^2x/dt^2 + b * dx/dt + k * x = 0

where m is the mass, b is the damping constant, k is the spring constant, x is the displacement from equilibrium position, and t is time.

In this case, the mass is 4 pounds (1 pound = 4.448 N) and the spring constant is 2 lbf/ft (1 lbf = 4.448 N), so we need to convert the units:

mass = 4 pounds * 4.448 N/pound = 17.79 N

spring constant = 2 lbf/ft * 4.448 N/lbf / 0.3048 ft/m = 29.11 N/m

This equation represents simple harmonic motion (SHM), where the position of the mass oscillates sinusoidally with time. To solve for the position, velocity, and acceleration at any given time, we need to solve this differential equation.