Simple Harmonic Motion with Spring

[getty102]

Homework Statement

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)

Homework Equations

$\Sigma$F_y=-k(y_eq-y₀)
$\Sigma$F_y=ma_y

The Attempt at a Solution

The only two forces acting on the mass are the Tension from the spring and weight. So
T_sm-(mg)=ma_y
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?

 

[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.

 

[getty102]

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

 

[rock.freak667]

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

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