Question

(1 point) Suppose a spring with spring constant 7 N/m is horizontal and has one end attached to a wall and the other end attached to a 3 kg mass. Suppose that the friction of the mass with the floor (i.e., the damping constant) is 2 N⋅s/m. Set up a differential equation that describes this system. Let x to denote the displacement, in meters, of the mass from its equilibrium position, and give your answer in terms of x,x′,x′′. Assume that positive displacement means the mass is farther from the wall than when the system is at equilibrium.

Answer 1

Answer:

x'' + (2/3)x' +$\frac{7}{3}$x=0

Explanation:

Given:

spring constant, k= 7 N/m ; mass = 3kg  ;and damping constant, c=2 N-s/m

a)

Differential Equation:

m($d^{2} x/dt^{2})= -kx-cdx/dt$   ........... (1)

Substitute the values of m,k, and c in (1).

3($d^{2} x/dt^{2})= -7x-(2dx/dt$)

3($d^{2} x/dt^{2}) + 7x + (2dx/dt$)=0

i.e , 3x'' +2x' + 7x =0

By dividing the equation with ;

   x'' + (2/3)x' +$\frac{7}{3} x$=0

The differential equation for the system is

x'' + (2/3)x' +$\frac{7}{3}$x=0