How to find the period of small oscillations given the potential?

[Davidllerenav]

You posted the relevant equation in your OP.

$x''+\omega^{2}x=0$?

 

[PeroK]

$x''+\omega^{2}x=0$?

Yes, and the other one.

 

[Davidllerenav]

Yes, and the other one.

$T=2\pi/\omega$? One question, can't I just write $F$ as $mx''$ such that $mx''=-U_0a^2x$? Thus $x''=-\frac {U_0a^2}{m}x$ and since $x''+\omega^{2}x=0\Rightarrow x''=-\omega^2x$ then $\omega^2=\frac {U_0a^2}{m}$?

 

[PeroK]

$T=2\pi/\omega$? One question, can't I just write $F$ as $mx''$ such that $mx''=-U_0a^2x$? Thus $x''=-\frac {U_0a^2}{m}x$ and since $x''+\omega^{2}x=0\Rightarrow x''=-\omega^2x$ then $\omega^2=\frac {U_0a^2}{m}$?

Yes, that's the idea.

 

[Davidllerenav]

Yes, that's the idea.

And how do I get to that with the constant $k$?

 

[PeroK]

And how do I get to that with the constant $k$?

You have:

$T=2\pi/\omega$

$\omega^2=\frac {U_0a^2}{m}$?

You don't need anything else.

 

[Davidllerenav]

You have:
You don't need anything else.

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?

 

[Davidllerenav]

@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]

@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?

Yes.

The Taylor series can be expanded about any point. As you posted in post #9 for a point $a$.

 

[Davidllerenav]

Yes.

The Taylor series can be expanded about any point. As you posted in post #9 for a point $a$.

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?

 

[timetraveller123]

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

 

[Davidllerenav]

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

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]

no what i am saying is even if the equilibrium point is not at zero you can make it zero by shifting your coordinate system to that point or you don't even have to shift . The new oscillation is just same as the one about the equilibrium about 0 but just that now it is at c. you still use the coefficient in front of the linear force term . in this case the oscillation frequency would be same because of the sine nature of the potential so in this case there is another stable equilibrium point at $\frac{2 \pi}{a}$ the ball oscillates about that point the same way as in the point at origin as the ball doesn't which equilibrium point it is in too all the stable points are the same