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GR9677 #4 |
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Alternate Solutions |
chri5tina
2006-11-27 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. |  |
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Comments |
tinytoon
2008-11-07 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:
^{3/2}})
and
^{3/2}} =kx)
We can clearly see now that:
^{3/2}}) .
We know that:
^{3/2}m}})
In the limit that  >> , this reduces to:
 .
|  | gn0m0n
2008-10-19 17:49:35 | What are we expanding, exactly? | | dcan
2008-04-09 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
2007-10-26 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! | | rmyers
2006-11-29 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
2008-08-18 00:58:43 |
I believe that choice E has the same dimensions as A&B.
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eliasds
2008-08-18 01:43:44 |
I believe that choice E has the same dimensions as A&B.
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wangjj0120
2008-10-11 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.
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| | chri5tina
2006-11-27 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
2006-10-29 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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