What is the relationship between 1D potential and simple harmonic motion?

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    1d Potential Shm
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The discussion revolves around understanding the relationship between a one-dimensional potential function and simple harmonic motion (SHM). The potential function U(x) has a minimum at x = 0, which is established by differentiating U(x) and finding that the derivative equals zero at this point. The participant expresses confusion about determining the maximum energy for a particle to remain in the potential and how energy relates to potential, questioning the problem's wording. To demonstrate SHM, it is suggested to expand the potential function using a Taylor series around the minimum, focusing on the second derivative to find the oscillation characteristics. Overall, the analysis emphasizes the connection between potential minima and the conditions for SHM.
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1D Potential
1. Homework Statement
Given:

U(x) = U_0 - U_n*e^-(kx^2), U_0, U_n and k are all constant.

i) What is the maximum E for the particle to remain in potential?
ii) Show that the potential has a minimum at x = 0.
iii) If a particle of mass m is placed near the minimum at x = 0, and displaced slightly, show that it does shm. find the period of this motion.2. Homework Equations

\frac{Df}{Dx} = -U(x)

3. The Attempt at a Solution

i) First, when I received this problem, it was in very bad handwriting. My immediate thought for this part of the problem was to differentiate U(x) with respect to x since I know the relevant equation from above, make the function negative, and set it equal to zero. However, I don't think Energy and potential are related like that, since as far as I know, the force has that relationship, not the energy - and I know very well that force and energy have different units =).

So my question for part i): Do you think I miswrote this problem? Does finding E-Max make sense given a potential?

ii) I'm not very comfortable with this problem. To find minima, I differentiate and set equal to zero. Differentiating U(x), we get

U(x) = -2xkU_n * e^(-kx^2)

And to find minima, we set U(x) to zero and solve for x.

0 = 2xkU_n, x = 0.

Then, I plug in values of -1 and 1 into the original equation, as well as zero, and zero clearly gives the lowest value (the x^2 ensures that!)

Is that sufficient to prove that there is a minimum at x = 0?

Finally,

iii) If a particle of mass m is placed near the minimum at x = 0, and displaced slightly, show that it does simple harmonic motion. Find the period of this motion.

Now this one, I'm a bit stumped on. While I don't know the graph of this motion, I do know that at x = 0 there is a minima, so it looks kind of like the minima in a parabola or something. (I believe this particle, at x = 0, is at 'stable' equilibrium, but I might be confusing terms...) treating this minima as a kind of 'ditch' that the particle cannot escape from without sufficient potential to overcome this 'ditch', the particle, starting at, say, -1, will start racing down, pass the ditch and, ignoring friction, make it to +1 before changing direction, passing x = 0 and moving back to -1. (oscillating!)

Clearly, this is SHM, and I know that for a SHM, F = -kx. I don't really know how to show this mathematically though. I was planning on rearranging the potential formula so that it looked something like this:

U(x) = U_0 - U_n*e^-kx^2,

U_0 - U(x) = U_n*e^-kx^2, Let [U_0 - U(x)]/(U_n) = Q,<br /> <br /> Q = e^-kx^2,<br /> <br /> LN(Q) = -kx^2, ...

But this feels definitely wrong.

If anyone can help me out on this, I'd be very appreciative. Thanks for taking your time to help me =).
 
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For the last part you need to expand the expression of potential (Taylor series) around the minimum and keep only the first term in the expansion.
Keep in mind that the first order derivative is zero at the minimum so the first non-zero terms will be the one with the second derivative.
 
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