GREPhysics.NET: GR9677 #92

GR9677 #92

Problem

GREPhysics.NET Official Solution

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Mechanics $\Rightarrow$ }Potential

Given $V(x)=-ax^2+bx^4$ , one can find the minimum by taking the first derivative (second derivative test indicates concave up), $V'(x) =\left( -2ax+4bx^3\right)_{x_0}=0 \Rightarrow x_0=\sqrt{\frac{a}{2b}}$ .

The force is given by $F=-dV/dx=2ax-4bx^3$ .

The angular frequency of small oscilations about the minimum can be found from,

$\begin{eqnarray} F(x-x_0)&=&2a(x-x_0)-4b(x-x_0)^3\\ &\approx& 2ax-4b(3x_0^2x) + O(2)\\ &=&(2a-12bx_0^2)x\\ m\ddot{x}&=&-4ax\\ \ddot{x}&=&-\omega^2 x \Rightarrow \omega^2 = \frac{4a}{m} \end{eqnarray}$

where one might recall the binomial theorem or pascal's triangle to quickly figure out the trinomial coefficients.

One finds that $\omega =2\sqrt{\frac{a}{m}}$ , as in choice (D).

Alternate Solutions

munster 2009-10-06 19:13:47	First of all, you can use dimensional analysis to rule out options (a) and (c). Second of all, we know, without question, that the potential wells caused by the $bx^4$ perturbation will be sharper (or deeper, however you like to look at it) than a well described purely by $ax^2$ (keep in mind here that the $x^2$ term is NEGATIVE, which means it by itself would not be a potential well, which is why the statement above works). We can see that if the potential were $V(x)=ax^2$ then $\omega=\sqrt{\frac{2a}{m}}$ , which rules out (e) because it is too small. (b) doesn't make sense because having a $\pi$ in the answer would never come up in the proper method (Taylor expansion, etc). This leaves us with (d). Reply to this comment
jmason86 2009-09-28 18:28:03	I did this problem using test taking strategies, although I'm not sure my thoughts were sound: Since ETS says "for small oscillations" then the $x^2$ term will dominate. This means that the constant $b$ should not be in your answer. Eliminate (A) and (C). $\pi$ should only become involved if we were asked for period or normal frequency. Eliminate (B). Getting the frequency is probably going to involve taking a derivative at some point which will bring that 2 down from the $ax^2$ . You'll end up with a 2 in your numerator. Eliminate (E). (D) is left. Reply to this comment

Comments

munster
2009-10-06 19:13:47

First of all, you can use dimensional analysis to rule out options (a) and (c). Second of all, we know, without question, that the potential wells caused by the $bx^4$ perturbation will be sharper (or deeper, however you like to look at it) than a well described purely by $ax^2$ (keep in mind here that the $x^2$ term is NEGATIVE, which means it by itself would not be a potential well, which is why the statement above works). We can see that if the potential were $V(x)=ax^2$ then $\omega=\sqrt{\frac{2a}{m}}$ , which rules out (e) because it is too small. (b) doesn't make sense because having a $\pi$ in the answer would never come up in the proper method (Taylor expansion, etc). This leaves us with (d).

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jmason86
2009-09-28 18:28:03

I did this problem using test taking strategies, although I'm not sure my thoughts were sound:

Since ETS says "for small oscillations" then the $x^2$ term will dominate. This means that the constant $b$ should not be in your answer. Eliminate (A) and (C).

$\pi$ should only become involved if we were asked for period or normal frequency. Eliminate (B).

Getting the frequency is probably going to involve taking a derivative at some point which will bring that 2 down from the $ax^2$ . You'll end up with a 2 in your numerator. Eliminate (E).

(D) is left.

Reply to this comment

anmuhich
2009-04-02 11:58:57

I used a really simple way of thinking about this to get the answer in about 20 seconds. Just remember that the potential for a simple harmonic oscillator is 1/2*k*x^2. Since the question says small oscillations around the minima, you can just take the central minimum, which for small oscillations looks a lot like a potential for the SHO, and around which the x^4 term will not contribute hardly. Angular frequency for this SHO is (k/m)^1/2 . But since our original potential is twice that of the potential for the above SHO you know the angular frequency will be greater than (k/m)^1/2 . The extra pi factor for B doesn't really make any sense so the only answer that fits is D.

furlong
2009-08-06 20:39:13

how do you know a is greater than 1/2? If a is less than 1/2, then the original potential will be less than 1/2*k*x^2. in other words, did you just assume that a was 1 or am I confused?

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plapas
2009-04-01 13:31:11

In this type of exercise, the safest way to solve is the following:
(1) Set dV/dx = 0 and find the roots.
(2) The frequency of small oscillations can be found by the second-derivative term of the taylor expansion, i.e.
(1/2) d/dx(dV/dx) (at min) = (1/2)m ω^2
and through this relation one can find ω.
The method proposed in the solution is good enough,
but one can make a lot of careless errors since the time available for the question is restricted.

mudder
2009-09-28 01:43:22

This is good. Solving for ω we see that in general,

Set V'(x_o) = 0

then

ω = (V''(x_0)/m)^(1/2)

Bam! There you go. Simple way of solving SHO problems

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theodiggers
2007-10-19 19:58:27

You can also find the minimum, shift the origin of the potential by the minimum so its now centered at zero, then just generate the x^2 term in a Taylor expansion of your shifted potential and get the force from there.

theodiggers
2007-10-19 20:10:42

But that triangle expansion method is some sick shit

sawtooth
2007-10-30 07:47:35

I agree, I prefer finding the minimum $x_0$ , Taylor expanding around $x_0$ :

$V(x-x_0)=V(x_0)-V'(x_0)(x-x_0)+V''(x_0)\frac{(x-x_0)^2}{2}+...$

dropping everything except the coefficient of $x^2$ creating something of the form:

$V(\epsilon)=V(x-x_0)=...+\frac{1}{2}(4\alpha)x^2+... = \frac{1}{2}k x^2$

and we remember that usually $\omega=\sqrt{\frac{k}{m}}$ .

sawtooth
2007-10-30 07:52:22

Ah, and ofcourse, since $x_0$ is a minimum $V'(x_0)=0$ as we have already used, so no trouble in the expansion, and simirarly, we already have the second derivative (we made sure that it was >0 so its a minimum).

To be more precise, btw, I should have had

$=\frac{1}{2} 4\alpha \epsilon^2 = \frac{1}{2}k \epsilon^2$

since this is the oscillating quantity... I think!

carlosoctavius
2008-08-28 07:48:24

How are you getting $\(1/2)*4a(x-x0)^2$ ? From the Taylor expansion of V(x) we should have $\(1/2)*2a(x-x0)^2$ Which leads to the wrong answer :(

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TheXDestroyer
2007-09-29 22:42:05

Can anyone explain how did we move from F(x-x0) to mx''=-4ax

Thanks

antithesis
2007-10-05 07:19:50

Plug in the value you found for $x_0$ into the line with $2a-12bx^2$ , you get $-4a$

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cyberdeathreaper
2007-02-18 19:50:59

I agree - I don't see how this approximation is done. Can someone expand on the steps?

alpha
2007-03-30 22:29:13

As mentioned in solution, the approximation is done by using the binomial theorem or Pascal's Triangle, then dropping higher order terms.

(very quick; for 3rd power, it is the 3rd row of Pascal's Triangle with coefficients 1 3 3 1, or $(a+b)^3 = a^3 + 3a^2b + 3 ab^2 + b^3$ ... in the approximation, higher powers of x are dropped out, so we end up with $\approx 2ax-12bxx_0^2$ )

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naama99
2006-11-15 16:10:27

Can someone please elaborate a little on how to do the approcimation? Thanks.

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alpha
2005-11-07 14:23:08

kolndom, isn't that what's already there?

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kolndom
2005-11-07 07:28:01

Hi, how can a force F=2a*dx leads to small oscillation?

The actual forced experienced by the particle can be given as following:

F=2ax-4bx^3

=2a(L+dx)-4b(L+dx)^3

=2aL(1+dx/L)-4bL^3(1+dx/L)^3

=2aL-4bL^3+2a*dx-4bL^3*3dx/L

=2a*dx-12b*L^2*dx

=2a*dx-12b*(a/2b)*dx

=2a*dx-6a*dx

=-4a*dx

:)

Anastomosis
2008-04-11 13:12:36

Mmmm LaTeX:

$F=2ax-4bx^3$

$=2a(L+dx)-4b(L+dx)^3$

$=2aL(1+dx/L)-4bL^3(1+dx/L)^3$

$=2aL-4bL^3+2a*dx-4bL^3*3dx/L$

$=2a*dx-12b*L^2*dx$

$=2a*dx-12b*(a/2b)*dx$

$=2a*dx-6a*dx$

$=-4a*dx$

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