# How to find the frequency of small oscillation of a particle in a given potential? [closed]

A particle of mass $m$ is in a potential $$V(x)=-\frac12ax^2+\frac14bx^4$$ where $a$ and $b$ are positive constants. The equilibrium points occur when the potential $V$ is either minimum or maximum, i.e., $$\frac{d}{dx}V = 0 \Rightarrow x(bx^2-a) = 0$$ The roots of the above solution is, $x=0,\pm\sqrt\frac{a}{b}$. Now for the nature of stability, $$\bigg(\frac{d}{dx}\bigg)^2V = 3bx^2-a$$Substituting the equilibrium point we into the above equation, it's clear that $x=0$ is not a stable equilibrium point but $\pm\sqrt\frac{a}{b}$ are stable. The plot of $V(x)$ v. $x$ makes it even more clear.

If we perturb the red ball a little, it will not return to it's equilibrium position, but the green balls will stand a little perturbation and will return to it's equilibrium position, $\pm\sqrt\frac{a}{b}$. Now, to find the frequency, I expanded the potential $V(x)$ in Taylor series at $x=\pm\sqrt\frac{a}{b}$, $$V(x) = -\frac{a}{4b}+a\bigg(x-\sqrt\frac{a}{b}\bigg)^2+b\sqrt\frac{a}{b}\bigg(x-\sqrt\frac{a}{b}\bigg)^3+\cdots$$ I don't care about any other term except for that one term, $ax^2$ term. So, I get the force $F$ as, $$F=-\textbf{grad } V\Rightarrow -kx = -ax \Rightarrow k = a$$ Or finally the frequency of the small oscillation will be, $$\omega = \sqrt\frac{k}{m} = \sqrt\frac{a}{m}$$

But the frequency incorrect. According to the book I am studying, the frequency of the small oscillation is supposed to be $\omega = \sqrt\frac{2a}{m}$. I have been trying this since morning and I am very very tired now and I don't know what mistake I making. I wrote down all my thought process. Please help me figure out, I am literally screaming at my notebook.

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• @Farcher Thank you. I can't believe that I found the derivative of $x^2$ wrong! I was checking and re-checking my Taylor series expansion which was correct. – Ayatana May 19 '17 at 15:47
• Your Taylor series looks suspect. It will have terms in "x" when expanding around a minimum. The first non-constant terms should be second order. It is a general property of a minimum. The Taylor formula can be found here, for example mathworld.wolfram.com/TaylorSeries.html Note that f' will be zero at the minimum and the first term of interest is the one with the second derivative. – nasu May 19 '17 at 16:02
• @Ayatana is your problem solved now or should I write up an answer? – noah May 19 '17 at 16:03
• @noah Please write an answer. I still don't understand. Maybe I am too dumb for this. – Ayatana May 19 '17 at 16:06

Using the Taylor expansion of the potential around $x_0=\sqrt{\frac{a}{b}}$ you get $U(x_0+\epsilon)=U(x_0) +U'(x_0)\epsilon +1/2 U''(x_0)\epsilon^2+...$ The first term is just a constant that can be dropped, the second term is zero because $U'(x_0)=0$ and the third term will be $\frac{1}{2} (-a+3bx)|_{x=x_0} \epsilon^2$ or

$U= \frac{1}{2} (2a) \epsilon^2$

It is then clear that the elastic constant for this potential is 2a. (Just by comparison wit $U=\frac{1}{2}kx^2$)

I think you are forgetting that the second term, expanded, yields another $-3x^2$:

$$\left(x - \sqrt{\frac{a}{b}}\right)^3 =\\ x^3 - 3\sqrt{\frac{a}{b}}x^2+...$$

Somewhere you need to take care of your signs.

• Yeah, I realized that, $V=\cdots+ax^2-3ax^2+\cdots = -2ax^2$ but the force is now $F=-\textbf{grad }V = -2a \times 2x = -4ax$ and subsequently, $\omega = \sqrt{4a / m}$. – Ayatana May 19 '17 at 15:45
• If you want to know the frequency of an oscillation around a point $A$, you should only care about terms $(x-A)^2$ and not $x^2$, no? – noah May 19 '17 at 15:47
• @noah yeah, thats what I did, I expanded Taylor series around $A$. I hadn't noticed the $-3ax^3$ term, and with correct differentiation, I get yet another wrong answer. – Ayatana May 19 '17 at 15:55

If you expand the potential around the minimum correctly you get:

$V(x)=-\frac{a^2}{4 b}+b x^3 \sqrt{\frac{a}{b}}+a x \sqrt{\frac{a}{b}}-2 a x^2+\ldots$

This means that the real term of the second order is $2ax^2$ and not $ax^2$. This is how you get the real oscillation frequency.

• The expansion around minimum should have no term linear in x. The first derivative is zero at the minimum. – nasu May 19 '17 at 15:58
• That contribution does not come from the derivative of the potential but from the expansion of the second order contribution and the third one. When you expand $(x-\sqrt{\frac{a}{b}})^2$, for instance, you get a first order in x due to the double product – Alessandro Mininno May 19 '17 at 16:02
• Oh, I see what you did. You need an expansion in terms of the displacement from equilibrium and not from origin. Then the potential look like the one for a harmonic oscillator. And you don't need to take the gradient of it to find the elastic constant. – nasu May 19 '17 at 16:13

$$V(x) = -\frac{1}{2} a x^2 + \frac{1}{4}bx^4\\ V'(x) = -ax + bx^3\\ V''(x) = -a + 3bx^2$$

Now expand your potential around $\pm\sqrt{\frac{a}{b}}$:

$$V(x) = V\left(\pm\sqrt{\frac{a}{b}}\right) + \left(x \pm\sqrt{\frac{a}{b}}\right) V'\left(\pm\sqrt{\frac{a}{b}}\right) + \left(x\pm\sqrt{\frac{a}{b}}\right)^2 V''\left(\pm\sqrt{\frac{a}{b}}\right) + \mathcal{O}\left(\left(\pm\sqrt{\frac{a}{b}}\right)^3\right)\\[6ex] \approx-\frac{a^2}{2b} + \frac{a^2}{4b} + \left(x\pm\sqrt{\frac{a}{b}}\right)^2 (-a + 3a) = -\frac{a^2}{2b} + \frac{a^2}{4b} + 2a\left(x\pm\sqrt{\frac{a}{b}}\right)^2$$ from which you can already see that the restoring force at the stable points will be $2a$.

Another approach is, rather than expanding the potential about a minimum, start with the force on the particle:

$$F(x) = -\frac{\mathrm{d}V}{\mathrm{d}x} = ax - bx^3$$

Now change variables:

$$x' = x - \sqrt{\frac{a}{b}} \rightarrow x = x' + \sqrt{\frac{a}{b}}$$

and then

$$F(x') = a\left(x' + \sqrt{\frac{a}{b}}\right) - b\left(x' + \sqrt{\frac{a}{b}}\right)^3$$

which is zero for $x' = 0$.

For small $x' = \epsilon$, we're interested in the terms linear in $\epsilon$

$$F(\epsilon) = a\epsilon - 3b\frac{a}{b}\epsilon + \mathcal{O}(\epsilon^2) \approx -2a\epsilon \Rightarrow \omega = \sqrt{\frac{2a}{m}}$$