Need help with this second order DE

[Craw]

Its for physics dealing with a second order homogeneous DE. I've only done first order problems before.

I neglected gravity to just look at the restoring/retarding forces, so in a damped oscillation of a spring I have so far...

?F[sub:3qj2y7qg]x[/sub:3qj2y7qg]: -kx - bv = ma[sub:3qj2y7qg]x[/sub:3qj2y7qg] where b is the damping coefficient and k is the spring constant

-kx - b(dx/dt) = m(d[sup:3qj2y7qg]2[/sup:3qj2y7qg]x/dt[sup:3qj2y7qg]2[/sup:3qj2y7qg])

And then I don't know what to do from there. Can anyone help me out with the easiest way to do this problem?

 

[galactus]

\(\displaystyle m\frac{d^{2}x}{dt^{2}}=-b\frac{dx}{dt}-kx\)

b is the damping constant and the negative sign is a consequence of the fact that the damping force acts in a direction opposite the motion.

Divide by m and we get the equation for free-damped motion.

\(\displaystyle \frac{d^{2}x}{dt^{2}}+(\frac{b}{m})\frac{dx}{dt}+(\frac{k}{m})x=0\)

\(\displaystyle \frac{d^{2}x}{dt^{2}}+2{\lambda}\frac{dx}{dt}+{\omega}^{2}x=0\)

Where \(\displaystyle \frac{b}{m}=2{\lambda}, \;\ {\omega}^{2}=\frac{k}{m}\)

The \(\displaystyle 2{\lambda}\) is done out of convenience since the auxilliary equation is \(\displaystyle m^{2}+2m+{\omega}^{2}=0\)

The roots are then:

\(\displaystyle m_{1}=-{\lambda}+\sqrt{{\lambda}^{2}-{\omega}^{2}}, \;\ m_{2}=-{\lambda}-\sqrt{{\lambda}^{2}-{\omega}^{2}}\)

There are 3 cases: overdamped, critically damped, underdamped.

For critically damped:

\(\displaystyle {\lambda}^{2}-{\omega}^{2}=0\)

Any tiny increase in the damping force results in oscillatory motion, the general solution is

\(\displaystyle x(t)=c_{1}e^{m_{1}t}+c_{2}te^{m_{1}t}\)

or

\(\displaystyle x(t)=e^{-{\lambda}t}(c_{1}+c_{2}t)\)

Does that help any?.

Incidentally, the cases for overdamped is when the discriminat >0 and for underdamped when it is <0

 

[Craw]

Ok yeah I understand what you did there.

Oh nevermind, I got it, thanks for your help.