Simple Harmonic Motion equation question: which length and why

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agenttiny200
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Homework Statement


I solved this physics question, but I am unclear about why Amplitude was the amount the spring was stretched by (which should be the new equilibrium point), instead of the amount the person pulled the mass down by (which should be the amplitude). Can anyone help?

On your first trip to Planet X you happen to take along a 300g mass, a 40-cm-long spring, a meter stick, and a stopwatch. You're curious about the acceleration due to gravity on Planet X, where ordinary tasks seem easier than on earth, but you can't find this information in your Visitor's Guide. One night you suspend the spring from the ceiling in your room and hang the mass from it. You find that the mass stretches the spring by 28.2cm. You then pull the mass down 10.4cm and release it. With the stopwatch you find that 10.0 oscillations take 16.7s. What is the gravity of Planet X?

Homework Equations



a=(2πf)2A

The Attempt at a Solution


f (frequency) =(10 oscillations/16.7s)= 0.5988 Hz
m (mass) =0.3 Kg
A (amplitude) = why 28.2 cm (0.282 m) instead of 10.4 cm (0.104 m)?

a=(2π×0.5988Hz)2×0.282m
a=3.99m/s2
 
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You seem to be confusing the maximum acceleration of the mass, (2πf)2A, with the local gravitational acceleration.
Use the resting stretch information to find the spring constant. What differential equation do you get for the oscillation?
 
To find spring constant, there is k = (2πf)2m = 4.25. Not sure what you meant by differential equations, they gave us the number of oscillations (10) per 16.7s in the question. Its supposed to be a simple harmonic question.

Am I supposed to be using the formula g=(k×ΔL)/m? Isn't that for pendulums and not springs?
 
g = (2πf)2 because (g/x) = (k/m) and (k/m) = (2πf)2 ?