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Davidllerenav said:And how do I get to that with the constant ##k##?
You have:
Davidllerenav said:##T=2\pi/\omega##
##\omega^2=\frac {U_0a^2}{m}##?
You don't need anything else.
Davidllerenav said:And how do I get to that with the constant ##k##?
Davidllerenav said:##T=2\pi/\omega##
##\omega^2=\frac {U_0a^2}{m}##?
Ok, using that I get ##\omega=\sqrt {\frac {U_0a^2}{m}}## thus ##T=2\pi\sqrt {\frac {m}{U_0a^2}}\Rightarrow T=\frac {2\pi}{a}\sqrt {\frac {m}{U_0}}##, is that right?PeroK said:You have:
You don't need anything else.
Yes.Davidllerenav said:@PeroK I have one question. I can also use the formula ##\omega^2=\frac{k}{m}## right? Also, how can I solve a problem like this if the equilibrium point is not 0?
PeroK said:Yes.
The Taylor series can be expanded about any point. As you posted in post #9 for a point ##a##.
But for example if the equilirbium point is ##y## then would end up with something like ##U_0a^2(x-y)##, what do I do with that y?PeroK said:The Taylor series can be expanded about any point. As you posted in post #9 for a point ##a##.
I understand, thanks. But what i meant was if the equilibrium point was not 0, but anotber constante, say ##c##, such that after using Taylor series I end up with ##U_0a^2(x-c)##, I would have ##U_0a^2x-U_0a^2c##, right? What do I do? Do I just replace that on the equation?timetraveller123 said:at the equilibrium point there would no linear terms in x so
but if your quadratic term is non-zero(positive) then you can do what you did before the coefficient is the "spring constant" if you shift your coordinate system to the y point then then it is similar to what you did before
edit:
just realized you were talking about force so there would be linear force and hence linear term in x so once again you can shift your coordinate system origin to y this makes it similar to what you solved before