Modeling a Damped Spring System with Differential Equations

joker2014
Messages
21
Reaction score
0

Homework Statement



After a mass weighing 8 pounds is attached to a 5-foot spring, the spring measures 6.6 feet. The entire system is placed in a medium that offers a damping constant of one. Find the equation of motion if the mass is initially released from a point 6 inches below the equilibrium point with a upward velocity of 1 ft/sec.

Homework Equations


F=kx
my''+cy'+ky=0

The Attempt at a Solution


I got k=8/(6.6-5) = 8/1.2
c= 1 as given ?

setting up the eqn I got 1/4 y'' + 1y' + 8/1.2 y = 0
or y'' +4y' + 80/3 y = 0

is this actually right? I tried 3 times and I keep getting this, i don't know i feel suspicious ?!
 
Physics news on Phys.org
"8 pounds" is weight, or force, not mass. What does "damping constant" mean? What are its units and how does it give force?
 
HallsofIvy said:
"8 pounds" is weight, or force, not mass. What does "damping constant" mean? What are its units and how does it give force?
of course 8 lbs is weight, dividing by 32 gives me .25 which is my mass! damping i believe is given in the problem "damping constant of one" .. otherwise if not given i would find it by sqrt of 4*m*k
 
"Damping force" is always opposite to the motion and is, approximately, proportional to the speed or the speed squared. I presume here you are told that it is proportional to speed. But because it is opposite to speed, the force must be -kv.
 
and of course my iniital conditions would be y(0)=-6 y'(0)=1
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top