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Here is the question:

"A spring stretches 0.4m when a 2kg mass is hung from it. The spring is stretched an additional 0.2m from its equilibrium point and is released. Determine: The Max Acceleration"

The formula for Max Acceleration is: kA/m where: k=spring constant A = Amplitude m = mass

The Spring Constant is 49N which I found and is correct. Now my question is: Why is the Amplitude 0.2m? The video I am watching:Simple Harmonic Motion at point: 46:16, he says the amplitude is 0.2m but wouldn't it be 0.6m since it said additionally in the problem or did we just create a new equilibrium out of nowhere? Thanks! -Guy who has a final tomorrow and is learning an entire semester in 2nights

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closed as off-topic by ZeroTheHero, Buzz, Kyle Kanos, Jon Custer, M. Enns Dec 19 '18 at 3:57

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  • $\begingroup$ When I'm in a sufficiently abstract mood I like to think about it like this: $\text{a 2-nominal} + \text{a 1-nomial}$ is still a 2-nomial and it still has the same quadratic coefficient. $\endgroup$ – dmckee Dec 18 '18 at 3:50
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The amplitude is measured from the equilibrium point. The spring is stretched an additional 0.2m from equilibrium so the amplitude is 0.2m. The position will vary from 0.6m (initial) to 0.2m and back again.

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  • $\begingroup$ Okay. This doesn't help lol, why is the amplitude now 0.2 instead of 0.6? $\endgroup$ – Captain Caboose Dec 18 '18 at 3:19
  • $\begingroup$ I'm sorry, but where is your confusion? The amplitude was never 0.6m. Amplitude is maximum displacement from equilibrium point. The mass starts at equilibrium (x=0.4m) and is displaced by 0.2m (x=0.6m), hence the amplitude is 0.2m. $\endgroup$ – Keefer Rowan Dec 18 '18 at 3:21
  • $\begingroup$ But the problem says it is stretches 0.4m meaning 0.4 from the equilibrium $\endgroup$ – Captain Caboose Dec 18 '18 at 3:22
  • $\begingroup$ @Captain The spring is oscillating around the equilibrium-point it has with the mass, not around it's equilibrium-point without the mass (the reason for this is that it is still remains subject to the weight of the mass the whole time, so you aren't comparing to the case with no mass on it). So you measure the amplitude from the equilibrium-with-weight not the equilibrium-without-weight. $\endgroup$ – dmckee Dec 18 '18 at 3:53

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