# Homework Help: SHM involving a spring

1. Apr 21, 2012

### Mdhiggenz

1. The problem statement, all variables and given/known data
A 1.6-kg block on a horizontal frictionless surface is attached to a spring whose force constant is 190 N/m. The block is pulled from its equilibrium position at x = 0 m to a displacement x = +0.080 m and is released from rest. The block then executes simple harmonic motion along the x-axis (horizontal). The velocity of the block at time t = 0.40 s is closest to:

2. Relevant equations

3. The attempt at a solution

v=-Aωcos(ωt)

=-0.080*(10.90)cos(10.90*.40)=.30m/s

Am I missing a step?

Thank you

2. Apr 21, 2012

### Staff: Mentor

You are using x = Asin(ωt). But that assumes that at t = 0 the mass is at x = 0.

(Pick a form that starts at maximum displacement.)

3. Apr 21, 2012

### Mdhiggenz

I dont quite understand what you are saying,

isnt the maximum displacement .080 which is the amplitude, im not sure how to incorporate it.

4. Apr 21, 2012

### Staff: Mentor

Where did this come from?

When they say "at time t = 0.40 s" they are measuring the time from the moment it was released, which is the point of maximum displacement.

5. Apr 21, 2012

### Mdhiggenz

Which means I use -waSin(wt)

since the graph of sin starts at zero?

6. Apr 21, 2012

Yes.