GR9677 #4



Alternate Solutions 
djh101 20140919 21:38:40  To expand a little bit on the quick and dirty method (point 2, specifically):
1. Small oscillations aren't going to depend on displacement. Eliminate BDE.
2. Experience with small oscillations should tell you that from the potential you are going to get a Hooke'sLawesgue force, which will give you k. To get the frequency, take the square root of k and divide by the square root of m. Only A has root m in the denominator.   archard 20100514 10:06:55  The way I did it is to remember from freshman physics that the period of an oscillation (and thus the frequency) is independent of displacement. That eliminates B, D and E. A and C are the same save for a square root, so one of them has incorrect dimensions. Choose the one with dimensions of frequency, A.   chri5tina 20061127 05:11:24  It is possible to do this problem w/o the binomial expansion.
Take F = k*x, plug in dV/dx for F and solve for k.
Plug k into w = (k/m)^1/2 and you'll have an expression with (R^2 + x^2)^3/2 in the denominator.
At this point you can remember that R >> x and simply cancel out x as negligible, which leaves R^3 in the denominator and the correct solution.  

Comments 
djh101 20140919 21:38:40  To expand a little bit on the quick and dirty method (point 2, specifically):
1. Small oscillations aren't going to depend on displacement. Eliminate BDE.
2. Experience with small oscillations should tell you that from the potential you are going to get a Hooke'sLawesgue force, which will give you k. To get the frequency, take the square root of k and divide by the square root of m. Only A has root m in the denominator.   maxdp 20130924 13:55:12  It seems stiner905's comment hasn't been noted since s/he didn't mark it correctly, so I'll repost it.
"After the binomial expansion, the potential should have a "" instead of a "+" in the expansion term. However, a minus exists in the force equation because the test charge has charge q."
In other words, there are two sign errors in the current solution that cancel out.   shak 20100813 15:02:04  The best and easiest way to solve this problem is using Lagrangian equation of motion.. it is one dimensional system.. so small charge q is moving only along the xaxis.
Lagrangian is
L=E V (1)
and
E=m*x(dot)^2/2 (2)
V= same potential in problem 3.
  archard 20100514 10:06:55  The way I did it is to remember from freshman physics that the period of an oscillation (and thus the frequency) is independent of displacement. That eliminates B, D and E. A and C are the same save for a square root, so one of them has incorrect dimensions. Choose the one with dimensions of frequency, A.
nkqed 20111024 17:01:10 
I'm sorry but this is the wrong way to think about it. x is not the displacement. We are displacing the particle perpendicular to x. And if you consider the situation where R is not >> x then we will have some dependence on x.

ian 20121005 09:56:53 
nkqed:
x is the displacement  look at the diagram. Also, the problem only asks us to consider the case where R >> x. I don't see anything wrong with archard's approach.

Prufrock 20130916 14:44:53 
This is a good way to approach it. Barring possible pathological cases, the frequency in a SHO "obviously" doesn't depend on the displacement.

  Muphrid 20091008 14:06:53  It may be faster to keep working with electrical potential to find the spring constant. Simple harmonic oscillators satisfy , as has been said, but they also satisfy
and have angular frequency
It's important to remember that what you know from problem (3) is , potential energy per unit charge, not , which is real potential energy. This is what alleviates the sign confusion later on.
As in the base solution, you can binomial expand and approximate it as
Note that the 1 can be dropped; this just gives a constant overall potential that we're not interested in. This leaves us with
And now multiply through by , the charge of the test charge, to get real potential energy:
Recognize this as a simple harmonic potential, where
And thus
  tinytoon 20081107 14:48:10  Also, you could derive the electric field from first principles to get the answer (although very inefficient). This is, nonetheless, a valid alternative:
We only care about the horizontal component because the force in the vertical component is zero:
.
Using and , we get:
and
We can clearly see now that:
.
We know that:
In the limit that >> , this reduces to:
.
  gn0m0n 20081019 17:49:35  What are we expanding, exactly?   dcan 20080409 16:32:33  Can someone tell me why it doesn't work to use conservation of energy? Supposedly the potential energy is qV, but when I solve for I'm off by a factor of 1/.   Gaffer 20071026 14:12:55  This problem has some particularly generous writers.
For small oscillations, I believe it is safe to say must be a constant, that is not dependent upon x. This knocks off BDE. Then all you need to remember is that will most likely involve a square root due to the diffEQ yosun mentioned and you have choice A, no dimensional analysis or calculation necessary.
Though of course it is always better to know why a particular formula is correct, sometimes quick and dirty is the way to go.
Oh that they were all this easy!
jmason86 20090929 16:40:35 
This is exactly how I did the problem too. Definitely the right strategy under time pressure.

  rmyers 20061129 16:16:57  This can also be done just by dimensional analysis. Qq/(4pi(e0)R^2) has units of force = kg * m/s^2 and angular frequency is just 1/s. Only A & B are possible answers after this. From there I suppose you can just use the idea that the angular frequency usually doesn't depend on initial displacement in cases like these.
eliasds 20080818 00:58:43 
I believe that choice E has the same dimensions as A&B.

eliasds 20080818 01:43:44 
I believe that choice E has the same dimensions as A&B.

wangjj0120 20081011 01:12:02 
rmyers:
your comment "you can just use the idea that the angular frequency usually doesn't depend on initial displacement" is not always true. You can find the x dependence from the exact solution.

  chri5tina 20061127 05:11:24  It is possible to do this problem w/o the binomial expansion.
Take F = k*x, plug in dV/dx for F and solve for k.
Plug k into w = (k/m)^1/2 and you'll have an expression with (R^2 + x^2)^3/2 in the denominator.
At this point you can remember that R >> x and simply cancel out x as negligible, which leaves R^3 in the denominator and the correct solution.   stiner905 20061029 12:26:06  After the binomial expansion, the potential should have a "" instead of a "+" in the expansion term. However, a minus exists in the force equation because the test charge has charge q.  

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