# Simple Harmonic Motion with Spring

1. Mar 6, 2012

### getty102

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

A massless spring is hanging vertically. With no load on the spring, it has a length of 0.22 m. When a mass of 0.68 kg is hung on it, the equilibrium length is 0.75 m. At t=0, the mass (which is at the equilibrium point) is given a velocity of 4.88 m/s downward.
At t=046s, what is the acceleration of the mass? (Positive for upward acceleration, negative for downward)

2. Relevant equations

$\Sigma$Fy=-k(yeq-y0)
$\Sigma$Fy=may

3. The attempt at a solution

The only two forces acting on the mass are the Tension from the spring and weight. So
Tsm-(mg)=may
I am now stuck wondering if that was a good place to start. Is the traditional way to solve this problem to find the spring constant first?

2. Mar 6, 2012

### rock.freak667

Yes you should find the spring constant first as you will need it later on to get the acceleration at t= 0.46.

3. Mar 6, 2012

### getty102

I have k=(mg)/(yeq-y0). I don't think this is right.

4. Mar 7, 2012

### rock.freak667

Right, well you are given m, y0 and yeq and you know g. So you can get k.