Question

A mass of 5 kg is attached to a linear spring whose constant is 40 N/m. The damping force applied to the mass is equal to 30 times its velocity, and a force equal to f(t)=20sin2t is applied to the mass. Find and plot the equation of motion if its initial position and velocity are x(0)=0.7 m and x′(0)=−11.8 m/s respectively.

Answer 1

The equation of motion for the mass attached to a linear spring is 5*x'' + 30*x' + 40*x = 20*sin(2*t).

Step-by-step explanation:

To find the equation of motion for a mass attached to a linear spring, we need to consider the forces acting on the mass. The net force on the mass is equal to the force of the spring minus the damping force plus the force applied to the mass. So, the equation of motion is:

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

where x is the position of the mass, m is the mass, b is the damping constant, k is the spring constant, x' is the velocity of the mass, x'' is the acceleration of the mass, and f(t) is the applied force.

In this case, the equation of motion becomes:

5*x'' + 30*x' + 40*x = 20*sin(2*t)