Mdthbs

Is the statement

"If the potential energy of a particle under oscillatory motion is directly
proportional to the second power of displacement from the mean
position, the particle performs a simple harmonic motion."

as strong as saying it in a more famous form of

"If the force due to oscillation on a particle under oscillatory motion is
directly proportional to the displacement from the mean position and
directed towards the mean position, the particle performs a simple
harmonic motion."

I think it cannot be regarded as equally strong, since if we derive the expression for potential energy for any oscillatory motion, using taylor expansion, we get

only for small oscillations that $$U(x) = frac{1}{2} U''(x_0) x^2$$ (where $U(x_0)$ is the potential energy at mean position) and we only get this result after certain approximations. Can someone please confirm this? Isn't the second statement stronger and more general? What I mean is, is the first statement always true?

Let's consider your first statement. If the potential energy of a particle under oscillatory motion is directly proportional to the second power of displacement from the mean position, then we can write the expression for potential at any position $x$ as $$U(x-x_0)=A(x-x_0)^2$$ where $A$ is our proportionality constant and $x_0$ is the mean position, which makes $x-x_0$ the displacement from the mean position. After expanding $$U(x-x_0)=Ax_0^2-2Ax_0x+Ax^2.$$ Now we know that $F=-{dU(x-x_0) over dx}$. So, for our potential energy, it becomes- $$F=-(-2Ax_0+2Ax)=2Ax_0-2Ax=-2A(x-x_0)$$ which mirrors your second statement. So they're equivalent for a single dimension. And we get those two statement equally strong in Euclidean space of arbitrary dimension. Verify it. But let's not stop here.
Consider your first statement in a non-Euclidean space. In this case, the expression for our potential is $U=A ({sqrt {g_{mu nu}dx^mu dx^nu}})^2=Ag_{mu nu}dx^mu dx^nu$. According to your second statement, we expect $F^mu=-Bdx^mu$.

According to the definition of force, $$F^mu=-g^{mu nu}nabla_nu U=-g^{mu nu}partial_nu(Ag_{sigma rho}dx^sigma dx^rho)=-2Adx^mu-Ag^{mu nu}g_{sigma rho, nu}dx^rho dx^sigma.$$So, if we want this last expression to meet our expectation, we must impose one condition, that is the value of $g_{sigma rho,nu}$ must be zero.

So, it seems that for the later case your second statement impose more restriction than your first one. So your first statement is more general, but the second one is stronger.

Plea: Can someone please verify my claims and calculation for the non-Euclidean part?

Both of your statements are true for SHM and can be seen as true for approximations as well. Each statement is essentially saying the same thing, because $F=-text dU/text dx$, and we can choose $U(0)=0$. Whether or not you use approximations to make these statements applicable to your system doesn't change that.

In other words, if you have approximated $Upropto x^2$, then you are also approximating $Fpropto -x$. Both of your statements say the same thing. Therefore, they are equally "strong".

If the potential energy is quadratic in the displacement only for small oscillations, then the restoring force is proportional to the (negative) displacement only for small oscillations. So I would not say the second statement is more general. The one statement is exactly true, so is the other. If one is only approximately true, so is the other.

Here I'm ignoring the fact that the potential energy is unique up to an arbitrary additive constant, and because of a non-zero constant it may not be proportional the the square of the displacement as in your first statement.