GREPhysics.NET: GR9677 #92

GREPhysics.NET

GR9677 #92

Problem

GREPhysics.NET Official Solution

Alternate Solutions

This problem is still being typed.

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).

See below for user comments and alternate solutions!

Alternate Solutions

walczyk 2011-04-03 02:05:24	Hi all, So I didn't want to do it using taylor expansion so I instead used a change in variables after finding the minimum. $u=x-\sqrt{\frac{a}{2b}}$ After plugging in and rearranging all the terms you get a pretty attractive potential function, obviously the new minimum is at $u=0$ . $V(u)=bu^4+2\sqrt{ab}u^3+2au^2-\frac{a^2}{4b}$ Recall $F=-\frac{dV}{du}$ . So $F=-4bu^3-6\sqrt{ab}u^2-4au$ . Remember for very small oscillations $u \ll 1$ , so higher order terms can be ignored for the ease of this problem. $F=m\ddot{x}$ and you undoubtedly know hooke's law, so $m\ddot{x}=-kx$ and $\omega^2=\frac{k}{m}$ . Unfortunately you cannot drop the quartic term after finding the minimum. If you work through the change of variables you find that the quartic term expands to provide a term $3au^2$ and the quadratic term contributes a $-au^2$ , which is necessary for the right angular frequency. There may be a better answer though, if there is reasoning that there must be a $\sqrt{\frac{a}{m}}$ term. That's all folks! Reply to this comment
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

ETScustomer
2017-10-05 01:52:54

It\\\\\\\'s such a twist that there\\\\\\\'s no $b$ in the answer! Yet, the $b$ matters ( $b=0$ would give a square root of two rather than a square root of four).

Reply to this comment

NEC

Voltsmann
2011-10-12 21:04:53

I think a lot of people have given basically this solution. But I had some trouble understanding it, so maybe my take on it will help someone. We have the functional form of the potential, V(x). We can find the minima via the usual method: taking the derivative and setting it to zero to find x0 s.t. (dV/dx)(x0)=0 (and checking that the second derivative is positive at x0). Then, since we're interested in small oscillations around xo, we can Taylor expand! We have V(dx)=V(x0)+V'(x0)dx+V''(x0)dx^2/2+O(dx^3), where dx = x-x0. Now we can throw out the first term, because it's a constant, so has no effect on the motion. The second term is zero, because x0 was determined by setting V'(x0) = 0. Then we're left with V(dx) = (1/2)V''(x0)dx^2 = 2a*dx^2. Now the force is F = -dV/dx = -4a*dx. Now we have a Hooke's Law force with k = 4a. The frequency of oscillation is then $\sqrt{k/m}$ = 2 $\sqrt{a/m}$ , which is (D).

Reply to this comment

NEC

walczyk
2011-04-03 02:05:24

Hi all, So I didn't want to do it using taylor expansion so I instead used a change in variables after finding the minimum.
$u=x-\sqrt{\frac{a}{2b}}$

After plugging in and rearranging all the terms you get a pretty attractive potential function, obviously the new minimum is at $u=0$ .
$V(u)=bu^4+2\sqrt{ab}u^3+2au^2-\frac{a^2}{4b}$

Recall $F=-\frac{dV}{du}$ .
So $F=-4bu^3-6\sqrt{ab}u^2-4au$ .
Remember for very small oscillations $u \ll 1$ , so higher order terms can be ignored for the ease of this problem.
$F=m\ddot{x}$ and you undoubtedly know hooke's law, so $m\ddot{x}=-kx$ and $\omega^2=\frac{k}{m}$ .

Unfortunately you cannot drop the quartic term after finding the minimum. If you work through the change of variables you find that the quartic term expands to provide a term $3au^2$ and the quadratic term contributes a $-au^2$ , which is necessary for the right angular frequency. There may be a better answer though, if there is reasoning that there must be a $\sqrt{\frac{a}{m}}$ term. That's all folks!

Reply to this comment

Alternate Solution - Unverified

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).

flyboy621
2010-11-05 23:21:29

Well done!

Reply to this comment

Alternate Solution - Unverified

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

Alternate Solution - Unverified

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?

Reply to this comment

NEC

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

Reply to this comment

NEC

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 :(

Reply to this comment

NEC

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 $-4a$

Reply to this comment

NEC

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 $\approx 2ax-12bxx_0^2$ )

Reply to this comment

Answered Question!

naama99
2006-11-15 16:10:27

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

Reply to this comment

NEC

alpha
2005-11-07 14:23:08

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

Reply to this comment

NEC

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:

$F=2ax-4bx^3$

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

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

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

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

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

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

$=-4a*dx$

Reply to this comment

NEC

Post A Comment!

You are replying to:

Hi all, So I didn't want to do it using taylor expansion so I instead used a change in variables after finding the minimum. $u=x-\sqrt{\frac{a}{2b}}$ After plugging in and rearranging all the terms you get a pretty attractive potential function, obviously the new minimum is at $u=0$ . $V(u)=bu^4+2\sqrt{ab}u^3+2au^2-\frac{a^2}{4b}$ Recall $F=-\frac{dV}{du}$ . So $F=-4bu^3-6\sqrt{ab}u^2-4au$ . Remember for very small oscillations $u \ll 1$ , so higher order terms can be ignored for the ease of this problem. $F=m\ddot{x}$ and you undoubtedly know hooke's law, so $m\ddot{x}=-kx$ and $\omega^2=\frac{k}{m}$ . Unfortunately you cannot drop the quartic term after finding the minimum. If you work through the change of variables you find that the quartic term expands to provide a term $3au^2$ and the quadratic term contributes a $-au^2$ , which is necessary for the right angular frequency. There may be a better answer though, if there is reasoning that there must be a $\sqrt{\frac{a}{m}}$ term. That's all folks!

Bare Basic LaTeX Rosetta Stone
LaTeX syntax supported through dollar sign wrappers $, ex., $\alpha^2_0$ produces $\alpha^2_0$ .
type this...	to get...
$\int_0^\infty$	$\int_0^\infty$
$\partial$	$\partial$
$\Rightarrow$	$\Rightarrow$
$\ddot{x},\dot{x}$	$\ddot{x},\dot{x}$
$\sqrt{z}$	$\sqrt{z}$
$\langle my \rangle$	$\langle my \rangle$
$\left( abacadabra \right)_{me}$	$\left( abacadabra \right)_{me}$
$\vec{E}$	$\vec{E}$
$\frac{a}{b}$	$\frac{a}{b}$

The Sidebar Chatbox...
Scroll to see it, or resize your browser to ignore it...

All-Solutions List | Latest Comments | Unanswered Questions (Cries for Help!) |::| About

© 2005-2007 by Yosun Chang, All rights reserved.

This site is not affiliated with ETS, nor is it officially endorsed (yet) by ETS. This site is the work of an individual named Yosun Chang. The solutions, sans user comments and ETS questions, are © 2005 by Yosun Chang except where noted. All rights reserved. / The comments appending particular solutions are due to the respective users. / Educational Testing Service, ETS, the ETS logo, Graduate Record Examinations, and GRE are registered trademarks of Educational Testing Service. The examination questions are © 1997 and © 2001 by ETS.