A massless spring is hanging vertically. With no load on the spring, it has a length of 0.29 m. When a mass of 0.33 kg is hung on it, the equilibrium length is 0.92 m. At t=0, the mass (which is at the equilibrium point) is given a velocity of 4.80 m/s downward.
At t=0.70 s, what is the acceleration of the mass? (Positive for upward acceleration, negative for downward)
Fspring = -k*(x(t) - xeqb)
ƩFy,mass = Fspring - m*g = -m*ay
ƩFy,mass = -k*(x(t) - xeqb) - m*g = -m*a
x(t) = A*cos(ω*t)
The Attempt at a Solution
Not sure how to go about this. I attempted to find the acceleration using the sum of the forces on the mass (which is accelerating downward) but I ran into a roadblock with finding k. I thought the equation for the position of a mass undergoing SHM might help out but I am not sure how to solve for A, because A is not equal to the equilibrium position.