4
$\begingroup$

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

Simple harmonic motion is a special type of periodic motion or oscillation where the restoring force is directly proportional to the displacement and acts in the direction opposite to that of displacement.

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 or more general? What I mean is, is the first statement always true?

$\endgroup$
  • $\begingroup$ Please tell us the sources of the quotes. $\endgroup$ – Ben Crowell Aug 31 at 23:15
  • 1
    $\begingroup$ If you can approximate $U(x)$ as quadratic you can approximate $F(x)$ as linear, by differentiating. The same works in reverse. The two statements are the same. $\endgroup$ – knzhou Sep 1 at 5:05
  • $\begingroup$ Also, the direction of your question is not clear. You ask if the second statement is "stronger and more general". A stronger requirement holds by definition in fewer situations, i.e. it must be less general. $\endgroup$ – knzhou Sep 1 at 5:08
  • $\begingroup$ @knzhou yes I got it, thanks. And I have edited what I had mistyped earlier to make it 'stronger or more general'. $\endgroup$ – PhysicsMonster_01 Sep 1 at 6:03
  • $\begingroup$ @Ben Crowell the second statement is from wikipedia, and I made the first one myself. $\endgroup$ – PhysicsMonster_01 Sep 1 at 6:06
5
$\begingroup$

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 $U\propto x^2$, then you are also approximating $F\propto -x$. Both of your statements say the same thing. Therefore, they are equally "strong".

$\endgroup$
2
$\begingroup$

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?

$\endgroup$
  • 2
    $\begingroup$ The force you calculated is proportional to the negative displacement. The first statement is not weaker, at least not because of your reasoning. $\endgroup$ – Puk Aug 31 at 13:49
  • 1
    $\begingroup$ Wait, why isn't $\nabla_\nu U=\partial_\nu U$ given that $U$ is a scalar and that the covariant derivative is the same as a partial derivative for a scalar? $\endgroup$ – Dvij Mankad Aug 31 at 16:29
  • 1
    $\begingroup$ The issue with your calculation raised in my previous comment might be related to this but, anyway, your indices in the second term of your final expression are messed up. You have two $\alpha$s upstairs and two $\rho$s downstairs. Might be an independent mistake--I haven't explicitly gone through your calculation. $\endgroup$ – Dvij Mankad Aug 31 at 16:36
  • 1
    $\begingroup$ @DvijMankad , I've followed your lead, recalculated and edited it. This time it needs less restriction. $\endgroup$ – Minotaur Aug 31 at 16:50
  • 1
    $\begingroup$ Finally, since $dx^\rho_{,\nu}=\delta^\rho_\nu$, your first term collapses to $-2Adx^\mu$. And, your second term is a second-order infinitesimal while the first is a first-order one. So, you have to drop the second term leading to no difference between the non-Euclidean case and the Euclidean case. xD $\endgroup$ – Dvij Mankad Aug 31 at 17:03
1
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

The potential energy uniquely determines the acting force on a particle, because in such a case, we have

$$\mathbf{F} = -\nabla U$$

And in the case of one-dimensional motion, this becomes

$$\mathbf{F} = \left<-\frac{dU}{dx}\right>$$

in terms of one-dimensional vectors. Hence with this, the motion is also similarly uniquely-determined. In your case, the statement "directly proportional to the square of the distance from a mean" (paraphrase) defines $U$ up to a constant factor, by the definition of "proportionality". This potential energy function is

$$U_\mathrm{stated}(x) := Kx^2$$

which, disregarding the factor $\frac{1}{2}$ that comes from getting to it by other means, is the potential function for the simple harmonic oscillator. The description isn't really so much anything new or interesting, I'd think, as merely being a restatement of the mathematical equation in words. (Here, we've taken the mean position as $\left<0\right>$ by suitable choice of coordinates in order to simplify things a little.)

That said, your second statement is a bit stronger than the first, because it includes an additional condition of "directed toward the mean" which implies $K > 0$, and the first does not carry this implication. When $K < 0$, you get unbounded "anti-harmonic" motion where the particle has an unstable equilibrium at the mean and compondedly-accelerates away from it under any nonzero perturbation. (Note: a negative proportionality constant is not what is usually meant by "inversely proportional". "Inverse proportionality" means proportionality to the reciprocal.)

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.