GR9677 #92



Alternate Solutions 
walczyk 20110403 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.
After plugging in and rearranging all the terms you get a pretty attractive potential function, obviously the new minimum is at .
Recall .
So .
Remember for very small oscillations , so higher order terms can be ignored for the ease of this problem.
and you undoubtedly know hooke's law, so and .
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 and the quadratic term contributes a , which is necessary for the right angular frequency. There may be a better answer though, if there is reasoning that there must be a term. That's all folks!
  munster 20091006 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 perturbation will be sharper (or deeper, however you like to look at it) than a well described purely by (keep in mind here that the 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 then , which rules out (e) because it is too small. (b) doesn't make sense because having a in the answer would never come up in the proper method (Taylor expansion, etc). This leaves us with (d).   jmason86 20090928 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 term will dominate. This means that the constant should not be in your answer. Eliminate (A) and (C).
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 . You'll end up with a 2 in your numerator. Eliminate (E).
(D) is left.  

Comments 
ETScustomer 20171005 01:52:54  It\\\\\\\'s such a twist that there\\\\\\\'s no in the answer! Yet, the matters ( would give a square root of two rather than a square root of four).   Voltsmann 20111012 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 = xx0. 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 = 2, which is (D).   walczyk 20110403 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.
After plugging in and rearranging all the terms you get a pretty attractive potential function, obviously the new minimum is at .
Recall .
So .
Remember for very small oscillations , so higher order terms can be ignored for the ease of this problem.
and you undoubtedly know hooke's law, so and .
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 and the quadratic term contributes a , which is necessary for the right angular frequency. There may be a better answer though, if there is reasoning that there must be a term. That's all folks!
  munster 20091006 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 perturbation will be sharper (or deeper, however you like to look at it) than a well described purely by (keep in mind here that the 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 then , which rules out (e) because it is too small. (b) doesn't make sense because having a in the answer would never come up in the proper method (Taylor expansion, etc). This leaves us with (d).
flyboy621 20101105 23:21:29 
Well done!

  jmason86 20090928 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 term will dominate. This means that the constant should not be in your answer. Eliminate (A) and (C).
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 . You'll end up with a 2 in your numerator. Eliminate (E).
(D) is left.   anmuhich 20090402 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 20090806 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?

  plapas 20090401 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 secondderivative 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 20090928 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

  theodiggers 20071019 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 20071019 20:10:42 
But that triangle expansion method is some sick shit

sawtooth 20071030 07:47:35 
I agree, I prefer finding the minimum , Taylor expanding around :
dropping everything except the coefficient of creating something of the form:
and we remember that usually .

sawtooth 20071030 07:52:22 
Ah, and ofcourse, since is a minimum 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
since this is the oscillating quantity... I think!

carlosoctavius 20080828 07:48:24 
How are you getting ? From the Taylor expansion of V(x) we should have Which leads to the wrong answer :(

  TheXDestroyer 20070929 22:42:05  Can anyone explain how did we move from F(xx0) to mx''=4ax
Thanks
antithesis 20071005 07:19:50 
Plug in the value you found for into the line with , you get

  cyberdeathreaper 20070218 19:50:59  I agree  I don't see how this approximation is done. Can someone expand on the steps?
alpha 20070330 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 ... in the approximation, higher powers of x are dropped out, so we end up with )

  naama99 20061115 16:10:27  Can someone please elaborate a little on how to do the approcimation? Thanks.   alpha 20051107 14:23:08  kolndom, isn't that what's already there?   kolndom 20051107 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=2ax4bx^3
=2a(L+dx)4b(L+dx)^3
=2aL(1+dx/L)4bL^3(1+dx/L)^3
=2aL4bL^3+2a*dx4bL^3*3dx/L
=2a*dx12b*L^2*dx
=2a*dx12b*(a/2b)*dx
=2a*dx6a*dx
=4a*dx
:)
Anastomosis 20080411 13:12:36 
Mmmm LaTeX:

 

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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 = xx0. 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 = 2 , which is (D).

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