- #1

- 39

- 0

## Homework Statement

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)

## Homework Equations

F

_{spring}= -k*(x(t) - x

_{eqb})

ƩF

_{y,mass}= F

_{spring}- m*g = -m*a

_{y}

ƩF

_{y,mass}= -k*(x(t) - x

_{eqb}) - 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.