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.

 

[asteriatic]

I would say that you have started in the right direction by identifying the forces acting on the mass and using the equations for Newton's second law. However, to fully solve the problem, you will need to find the spring constant first. This can be done by using the given information about the equilibrium length and the mass hanging on the spring. Once you have the spring constant, you can plug it into your equation and solve for the acceleration of the mass at t=0.46s. It is always important to start by identifying the relevant variables and equations, and then proceed with finding the missing values.