A force of 2 pounds stretches a spring 1 foot. With one end held fixed, a mass weighing 8 pounds is attached to the other end. The system lies on a table that imparts a frictional force numerically equal to (3/2) times the instantaneous velocity.
Initially, the mass is displaced 4 inches above the equilibrium position and released from rest. Find the equation of motion if the motion takes place along a horizontal straight line that is taken at the x-axis.

This is a problem from my Differential Equations class. I'm completely stumped on how to approach this.

Heh, is there a picture? Cuz I don't quite understand what the system looks like from the description.
Like it seems at some points in the description that the spring is aligned/displaced in a horizontal fashion on a tabletop, and in other parts that it is vertically aligned. Maybe I am just being thick headed, but I can't "see" the system...

This post was edited by Azrad on Apr 10 2013 05:20pm

yea, i know what you mean. Unfortunately no picture, I think that's why I'm confused.
I can't seem to understand what it's asking exactly

This post was edited by TritonV8 on Apr 10 2013 05:25pm

i know the generic equation:

mx'' + (Beta)x' + kx = 0 , where m=mass, (Beta)=damping constant, k=spring constant

leaving off units:
m = W/g = 8/32 = 1/4
(Beta) = 3/2 [given]
k = F/spring displacement = 2/1 = 2

(1/4)x'' + (3/2)x' + 2x = 0

this is where i've gotten, but i'm not sure what to do from here. do I solve the DE? Are there initial conditions to find my c's?

screaming horizontal alignment (and differential equation).


screams vertical alignment

I know, right? This question has thrown me for a loop

I think i got it.
I used the equation I found earlier:

(1/4)x'' + (3/2)x' + 2x = 0

with these initial conditions: x(0) = -(1/3), x'(0) = 0
I chose these because it was initally displaced 4 inches above equilibrium, then converted to feet. It's negative because I assigned downward as the positive direction.
For the other initial condition, it stated it was initially at rest, so initial velocity = 0

I then used the Undetermined Coefficients method to solve the homogeneous DE and got:

x = (c1)e^(-2t) + (c2)e^(-4t) , where c1 and c2 are coefficients.

Then plugged in initial conditions to find c1 and c2, solve the system of equations, and got:

x = (-2/3)e^(-2t) + (1/3)e^(-4t)

Does this look like the right method?

Yeah, I eventually went with the spring being vertical and the mass is rubbing against the table leg or something (not very well described ). I did a slightly different way and got the same DE as above^^
And I think your initial conditions are right, so I'll be lazy trust you solved the DE correctly. I mean you are TritonV8, so I don't think that is much of a stretch

This post was edited by Azrad on Apr 10 2013 06:59pm

awesome, thx Azrad