- #1

Midas_Touch

okay, so i set up the equations and motions and used langrangian equations to yield y = c0+c1t-1/2gt^2

where co and c1 are constants. so for this equation, i have to take the first derivative and plug in t=0. so does that mean, y = c0 and y' = c1?

further, i let the centrifugal force equal to the spring force

F_r = mu*(r'') = -k(r-L) where mu is m1m2/M where M = m1+m2

where L is the unstretched length

anyway since the equation is a simple harmonic equation, i could find two constants b1, and b2 and i got b1 = r - L and b2 = r'/w ( where w =omega) when i plugged in t=0 for the derivatives.

however, how do these initial conditions fit in with the answer, with the position equations y1 = Y + (m2/M)*r and y2= Y - (m1/M)*r

and ultimately, the answer to the problem

which is y1 = (L + (m1/M)*v0*t - 0.5gt^2) + (m2*v0/M*w)sin(wt)

and y2 = ( (m1/M)*v0*t - 0.5gt^2) - (m1*v0/M*w)sin(wt)