Vertical Springs Concept Clarifier

A spring with k=50 Nmk=50\ \frac{N}{m}and length 10 cm10\ cmis hanging from the ceiling (the spring is massless).

In this problem, to keep the math a bit simpler, feel free to use g=10m/s2g=10 m/s^2. (No penalty for using 9.81 if you want, though.)

a) If we attach a 50 g50\ gmass to the spring and then let it fall down, by how much does the spring stretch?
b) What is the new equilibrium position of the oscillation? (Measured from the ceiling, in centimeters.)

Let's say that, after hanging the mass, we now pull it down further by an amount equal to your answer in part (a). That is, if your answer in part (a) was 5 cm, we now pull it down by an additional 5 cm. Your answer in part (a) is now the amplitude of the simple harmonic motion for the questions below.
c) What is the lowest position of the mass now? (Measured from the ceiling, in centimeters.)
d) What is the period of oscillation?
e) If you model the displacement of the mass in the form x(t)=Acos(ωt+φ)x(t)=A\cos(\omega t+\varphi), what is the phase constant?
f) When the mass passes through the center of the oscillation range, what is its speed?
g) What is the acceleration when the magnitude of the acceleration when the mass is at the top or bottom of its oscillation?
h) When the mass passes through the center of the oscillation range, what is the magnitude of its acceleration?

More Vertical Springs Questions: