# Need some help with differential equations for mechanics.

1. Jan 30, 2013

### uber_kim

1. The problem statement, all variables and given/known data

I'm having problems with some differential equations, just need to know where I'm going wrong.

2. Relevant equations

3. The attempt at a solution

a) mv$\stackrel{dv}{dx}$=F(x)
mvdv=F(x)dx
m∫vdv=∫F(x)dx
v^2=v$v^{2}_{0}$+$\stackrel{2}{m}$∫F(x)dx

Setting F(x)=-kx
v^2=$v^{2}_{0}$-$\stackrel{k}{m}$(x^2-$x^{2}_{0}$)

I then have to find the position as a function of time..
$\stackrel{dx}{dt}$=$\sqrt{v^{2}_{0}-\stackrel{k}{m}(x^2-x^{2}_{0}}$
dx/$\sqrt{v^{2}_{0}-\stackrel{k}{m}(x^2-x^{2}_{0}}$=dt

I'm not sure how to do that integral, though, or if that's even right.

b) This problem involves a disc moving along a rough surface, so it has friction (F) and linear air resistance (-bv) acting on it.

ma=-bv+F
mdv/dt=-bv+F
-$\stackrel{m}{b}$∫dv/v=∫Fdt
-$\stackrel{m}{b}$ln($\stackrel{v}{v_0}$=Ft
e^(-$\stackrel{m}{b}$)v/v_o=e^(Ft)
v=v_0e^(-Fbt/m)

Thanks for any help!

2. Jan 30, 2013

### voko

For a), transform the integrand to this form: $\displaystyle \frac {a} {\sqrt {1 - (cx)^2 } }$, then use the substitution $u = cx$.

For b), you are not doing it correctly. You should have gotten $\displaystyle \int \frac {dv} {F - bv}$.