Recent content by Shinaolord

I use the Nspire CAS and I absolutely love it. It's expensive, but in my opinion, since you can't bring a computer with Mathematica on it to exams or quizzes, it's worth it. It also allows you to check things you have to do by hand on tests, but if it's obvious you're not supposed to in a...

E , when considering the total energy, isn't a function of k, it's a function of height (The potential part, The sum is obviously a constant), which in turn can be considered a function of time. ##E(h(t))##EDIT: what you are given is the only necessary info.

How is this relevant?

That's why my original post was as vague as possible, because I didn't understand the true nature of what the question is attempting to ask.

It won't be 2A, but you can get to it with the quantity 2A. Finding the difference between where it would be at equilibrium if released from equilibrium, and you can relate it to this situation. I'm sure I'm not explaining this well, but that's basically what I was getting at. Tomorrow (I'm...

Indeed that was an implicit assumption I made I should've stated that my apologies. I was going to use that to work towards the case where ##L_0 \leq L_n## or ##L_0 > L_n##. Can you see now how we could find x_0 from the information we now have?

And not a very hard problem if you aren't overthinking it. Don't worry, I've been in this same situation, but with a different topic within physics. Once you figure it out you'll have that "aha!" Moment. Anyway, if ##x_{eq}## or ##x_0## is what we're finding, at a point called ##x_0## that is...

My location's forecast;

This week, a high of -5F almost every day, windchills bringing it to -15 to -20F, and the lows at night are as cold as -16F.

I think the 3rd one is exact, but I computed the partial derivatives in my head, so double check before you take my word on it.Actually nvm I don't think it is.

I would like to take this time to point out to The OP, in case he missed my edit, that I made a mistake in my calculations the first time about, and your derivation was correct. I apologize for possibly misleading you.

I'm in the process of doing just that.You are correct. I realized what my mistake was. My apologies.

Ah, that's what I must've been seeing wrong. I knew it was something. But shouldn't your a be x, as your integrating v(t)?

[strike]I don't think your 3rd step is correct. ##a(t)=\frac{d}{dt}v(t)=\frac{d^2}{dt^2}x(t)=\frac{d}{dt}[\frac{d}{dt} x(t)]##.[strike] disregard this comment. I made a mistake originally.

Read my edit. You need the changes, not the absolute in kelvin.