Question

A force of 9 pounds stretches a spring 1 foot. A mass weighing 6.4 pounds is attached to the spring, and the system is then immersed in a medium that offers a damping force numerically equal to 1.2 times the instantaneous velocity. Find the equation of motion If the mass is initially released from rest from a point 1 foot above the equilibrium position.

Answer 1

The equation of motion for a mass-spring-damper system is given by m * d^2x/dt^2 = -kx - bv, where m is the mass, k is the force constant of the spring, b is the damping constant, x is the displacement from the equilibrium position, and v is the instantaneous velocity. This equation describes the forces acting on the mass and can be used to analyze the motion of the system.

Step-by-step explanation:

To find the equation of motion for this system, we need to consider the forces acting on the mass. The net force is equal to the force of the spring minus the damping force. In this case, the force of the spring is given by Hooke's Law as F = kx, where F is the force, k is the force constant of the spring, and x is the displacement from the equilibrium position. The damping force is given by F_d = -bv, where b is the damping constant and v is the instantaneous velocity.

Using Newton's second law, F = ma, we can set up the equation for the motion: m * a = -kx - bv.

Since a = d^2x/dt^2, we have the equation of motion m * d^2x/dt^2 = -kx - bv. This is a second-order linear differential equation that describes the motion of the mass-spring-damper system.

Answer 2

`(d^2x)/(dt^2)+(\beta)/(m)(dx)/(dt)+(k)/(m)x=0`

Step-by-step explanation:

let
`m` be the mass attached, let
`k` be the spring constant and let
`\beta` be the positive damping constant.

-By Newton's second law:

`m(d^2x)/(dt^2)=-kx-\beta (dx)/(dt)`

where
`x(t)` is the displacement from equilibrium position. The equation can be transformed into:

`(d^2x)/(dt^2)+(\beta)/(m)(dx)/(dt)+(k)/(m)x=0` shich is the equation of motion.