# Mass hanging on a spring - find acceleration

## 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

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.

ehild
Homework Helper
First find the spring constant.

ehild

The spring constant, k, is equal to the weight force divided by the change in position.

k = (m*g)/Δx

which we are all given.

Once I've found the spring constant, now what do I do? I am confused on how to find x(t=0.7s) when I have an unknown amplitude.

ehild
Homework Helper
The mass will oscillate about the new equilibrium position. It starts SHM from equilibrium point with maximum velocity. Write up the equations for displacement, velocity, acceleration of this SHM.

ehild