# Tricky spring problem. Amplitudes and AccelerationMax

1. Nov 21, 2011

### yigh

1. The problem statement, all variables and given/known data

A block with mass m =6.7 kg is hung from a vertical spring. When the mass hangs in equilibrium, the spring stretches x = 0.26 m. While at this equilibrium position, the mass is then given an initial push downward at v = 4.7 m/s. The block oscillates on the spring without friction.

3)After t = 0.4 s what is the speed of the block?
4)What is the magnitude of the maximum acceleration of the block?

Calculated K = 252.5384615
Oscillation Frequency (omega) = .9771168341

2. Relevant equations

Amplitude = vMax*sqrt(m/k)

3. The attempt at a solution
I got amplitude to be 0.7655.
I used -(omega)*Acos(omega*t) to find V for number 3. I used aMax = omega^2*t

2. Nov 22, 2011

### Rayquesto

I believe there is an equation that you use to find the velocity at any given point given amplitude, stretched distance, k and mass. hold on.... so its the sqrt(k/m) times the sqrt(amplitude^2 - stretched or compressed distance^2) use that

3. Nov 22, 2011

### Rayquesto

oh nevemind. use 9.8m/s^2(t) to find the speed after sometime. isnt the acceleration the same?