Homework Help: 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.