Calculating Spring Displacement with Added Mass in Simple Harmonic Motion

[thesandalman]

Homework Statement

A 0.24 kg mass is suspended on a spring which stretches a distance of 5.9 cm. The mass is then pulled down an additional distance of 13.5 cm and released. What is the displacement from the equilibrium position with the mass attached (in cm) after 0.46 s? Take up to be positive and use g = 9.81 m/s2.

Homework Equations

Hint:
Get k from the displacement as it was done in this example. The equation will have
ω = (k/m)1/2,

and the phase will be such that it will be a cosine with a negative amplitude, because it starts at a negative displacement.

Be careful about the sign!
You have to get
y=-x[0]*cos(sqrt(t^2*g/x))
where from the eq-position
it follows that omega=sqrt(g/x)

The Attempt at a Solution

I tried doing the equations stated above however, I am still confused. I was not sure if x[0] means the distance the spring stretches at first or what? Also if X would then be the initial spring stretch + the extra distance pulled or what? Any help would be must appreciated!

 

[arkofnoah]

x[0] is simply the amplitude, i.e. maximum displacement from the equilibrium position.

The equilibrium position is defined as the position of the mass when there is no oscillation, so go figure out if you need to add up 13.5 cm and 5.9 cm or not.

After that find the spring constant k and then just use that cosine equation to solve for the displacement.