GREPhysics.NET: GR9677 #4

GR9677 #4

Problem

GREPhysics.NET Official Solution

This problem is still being typed.

Electromagnetism $\Rightarrow$ }Small Oscillations

The potential is determined in the previous problem to be $V=\frac{Q}{4\pi \epsilon_0 \sqrt{R^2+x^2}}$ . The field is given by $\vec{E}=-\nabla V$ . Before taking derivatives, one can simplify the potential since it is given that $R >> x$ .

Binomial expand it ( $(1+y)^n\approx 1+ny$ , for $y$ small) to get

$V\approx \frac{Q}{4\pi \epsilon_0 R}(1+\frac{x^2}{2R^2})<br />$

Taking the derivative, using the equations $\vec{E}=-\nabla V$ , and $\vec{F}=q\vec{E}$ , one gets,

$-q\frac{Q}{4\pi \epsilon_0 R}\frac{x}{R^2}=F=m\ddot{x}.<br />$

Small oscillations have the same form as simple harmonic oscillations, i.e., $\ddot{x} = -\omega^2 x$ . The angular frequency is $\omega=\sqrt{q\frac{Q}{4\pi \epsilon_0 R^3 m}}$ , as in choice (A).

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.

Reply to this comment

Comments

Muphrid
2009-10-08 14:06:53

It may be faster to keep working with electrical potential to find the spring constant. Simple harmonic oscillators satisfy $F(x) = -kx$ , as has been said, but they also satisfy

$U(x) \approx U_0 + \frac{1}{2} k x^2$

and have angular frequency

$\omega = \sqrt{\frac{k}{m}}$

It's important to remember that what you know from problem (3) is $V(x)$ , potential energy per unit charge, not $U(x)$ , which is real potential energy. This is what alleviates the sign confusion later on.

As in the base solution, you can binomial expand $V(x)$ and approximate it as

$V(x) = \frac{Q}{4\pi \epsilon_0} \frac{1}{R} \left( 1 - \frac{1}{2} \frac{x^2}{R^2} \right)$

Note that the 1 can be dropped; this just gives a constant overall potential that we're not interested in. This leaves us with

$V(x) = V_0 - \frac{1}{2} \frac{Q x^2}{4 \pi \epsilon_0 R^3}$

And now multiply through by $-q$ , the charge of the test charge, to get real potential energy:

$U(x) = -q V_0 + \frac{1}{2} \frac{qQ x^2}{4 \pi \epsilon_0 R^3}$

Recognize this as a simple harmonic potential, where

$\frac{1}{2} k x^2 = \frac{1}{2} \frac{qQ x^2}{4 \pi \epsilon_0 R^3} \rightarrow k = \frac{qQ}{4 \pi \epsilon_0 R^3}$

And thus

$\omega = \sqrt{\frac{k}{m}} = \sqrt{\frac{qQ}{4 \pi \epsilon_0 m R^3}}$

Reply to this comment

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:

$E = \int \frac{dq}{4\pi\epsilon_0 r^2}$

We only care about the horizontal component because the force in the vertical component is zero:

$E = \int \frac{dqcos\theta}{4\pi\epsilon_0 r^2}$ .

Using $cos\theta = \frac{x}{r} = \frac{x}{\sqrt{x^2+R^2}}$ and $r^2 = x^2+R^2$ , we get:

$E = \frac{qx}{4\pi\epsilon_0 (x^2+R^2)^{3/2}}$

and

$F = \frac{qQx}{4\pi\epsilon_0 (x^2+R^2)^{3/2}} =kx$

We can clearly see now that:

$k = \frac{qQ}{4\pi\epsilon_0 (x^2+R^2)^{3/2}}$ .

We know that:

$\omega = \sqrt{k/m} = \sqrt{\frac{qQ}{4\pi\epsilon_0 (x^2+R^2)^{3/2}m}}$

In the limit that $R$ >> $x$ , this reduces to:

$\omega = \sqrt{k/m} = \sqrt{\frac{qQ}{4\pi\epsilon_0 R^3m}}$ .

Reply to this comment

gn0m0n
2008-10-19 17:49:35

What are we expanding, exactly?

Reply to this comment

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 $\omega$ I'm off by a factor of 1/ $\sqrt{2}$ .

Reply to this comment

Gaffer
2007-10-26 14:12:55

This problem has some particularly generous writers.

For small oscillations, I believe it is safe to say $\omega$ must be a constant, that is not dependent upon x. This knocks off BDE. Then all you need to remember is that $\omega$ 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
2009-09-29 16:40:35

This is exactly how I did the problem too. Definitely the right strategy under time pressure.

Reply to this comment

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.

eliasds
2008-08-18 01:43:44

I believe that choice E has the same dimensions as A&B.

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.

Reply to this comment

chri5tina
2006-11-27 05:11:24

Reply to this comment

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.

Reply to this comment

Post A Comment!

You are replying to:

This problem has some particularly generous writers. For small oscillations, I believe it is safe to say $\omega$ must be a constant, that is not dependent upon x. This knocks off BDE. Then all you need to remember is that $\omega$ 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!

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...

<a href="http://www6.shoutmix.com/?grephysics">View shoutbox</a>
ShoutMix chat widget

Free 3D Virtual World. Largest User-Created Fantasy World. Don't Play, Experience. Join Now!

Bored, got time to kill, and want to earn micro-cash by mindlessly clicking on links? Join CrownGPT. Click Offers. Click the Paid to Click button... then click to your heart's delight.

FREE 3D Virtual World Chat - Join Second Life Today!

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

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.

Bare Basic LaTeX Rosetta Stone