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

Answer 2

This question involves the concepts of differential equations and a damped spring-mass system.

The differential equation to describe this system is "x'' + 0.67x' + 2.33x = 0".

The general differential equation used to represent a spring-mass system that is damped is given as follows:

$m\frac{d^2x}{dt^2}+c\frac{dx}{dt}+kx=0\\\\mx''+cx'+kx=0$

where,

m = mass = 3 kg

x'' = second derivative of displacement with respect to time

c = damping coefficient = 2 N.s/m

x' = first derivative of displacement with respect to time

k = spring constant = 7 N/m

x = displacement

Therefore,

$3x''+2x'+7x=0\\dividing\ whole\ equation\ by\ 3:\\x''+\frac{2}{3}x'+\frac{7}{3}x=0$

x'' + 0.67x' + 2.33x = 0

Learn more about the damped spring-mass system here:

https://brainly.com/question/16750036?referrer=searchResults

The attached picture shows a damped spring-mass system.