# CM and Relative Coordinates; Reduced Mass

Midas_Touch
Two particles of mass m1 and m2 are joined by a massless spring of natural length L and force constant k. Initially, m2 is resting on a table and I am holding m2 vertically above m2 at a height L. At time t=0, I project m2 vertically upward with initial velocity v0. Find the positions of the two masses at any subsequent time (before or either mass returns to the table) and describe the motion.

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)