GREPhysics.NET
GR | #
<< Previous | Next >>
Login | Register
   
  GR9277 #7
Problem
GREPhysics.NET Official Solution    Alternate Solutions
\prob{7}
9277_7

A cylindrical tube of mass M can slide on a horizontal wire. Two identical pendulums, each of mass m and length l, hang from the ends of the tube, as shown above. For small oscillations of the pendulums in the planet of the paper, the eigenfrequencies of the normal modes of oscillation of this system are 0, $\sqrt{\frac{g(M+2m}{lM}}$, and


  1. $\sqrt{g/l}$
  2. $\sqrt{\frac{g(M+m)}{lM}}$
  3. $\sqrt{\frac{gm}{lM}}$
  4. $\sqrt{\frac{gm}{l(M+m)}}$
  5. $\sqrt{\frac{gm}{l(M+2m)}}$
Mechanics}Normal Modes

For normal mode oscillations, there is always a symmetric mode where the masses move together as if just one mass.

There are three degrees of freedom in this system, and ETS is nice enough to supply the test-taker with two of them. Since the symmetric mode frequency is not listed, choose choice it!---as in (A).
See below for user comments and alternate solutions! See below for user comments and alternate solutions!
Alternate Solutions
sharpstones
2006-12-01 09:33:27		At the least you should be able to figure out the modes. In one case all are moving together as a single mass so there is no frequency (they give you this one). In another case the two pendulums move in the same direction and the tube M moves in the opposite direction. In the last case the two pendulums move opposite directions and the tube stands still. -- Clearly the last case which you are trying to find will be independent of M so (A) is the only answer the works.Alternate Solution - Unverified
Comments
Yaroslav Shustrov
2017-09-04 10:47:25		I do not understand the following point. When we are speaking about certain normal mode we imply that every degree of freedom of the system oscillates with the same frequency. We can see that for 2 eigenfrequences proposed in the problem. \r\n However when I imagine that M stands still, while two pendulums oscillate with the same frequency in the opposite directions I arrive to the contradiction. In this case frequency of M is different, it equals zero. How come?NEC
bass
2014-07-16 06:11:28		???????????????????????????????????????????????????????????????????????????????????????????
BASS http://www.livingscriptures.com/jpshoes.asp?cheap=products-c181.html		NEC
explainthatagain
2006-12-12 10:44:45		Ok, can someone explain that to me again. For some reason saying A is an obvious choice doesn't work for me. How would someone know the formula to find the eigenfrequencies from. Does it follow the logic that you can go with neither pendulum is moving, they are moving in oposite directions, or they are moving in the same direction?
Poop Loops
2008-10-05 13:00:23		 That is exactly it. 0 is when there is no oscillation, the big formula is when both masses are moving in the same direction, but since the tube can move, it has to move opposite to conserve momentum, and you are looking for the one where both masses are moving in opposite directions.
spacemanERAU
2009-10-22 09:10:12		 well...like EVERY ETS question...this question is not designed for the test taker to "solve". I mean if you want you can set up the differential equations and solve them but good luck finishing the PGRE in 180 minutes.
keradeek
2011-10-03 03:24:37		 study the chapter on coupled oscillations in marion and thornton and it will become obvious to you, too.
speiro
2012-09-12 23:56:11		 marion 12.18 helped..
NEC
sharpstones
2006-12-01 09:33:27		At the least you should be able to figure out the modes. In one case all are moving together as a single mass so there is no frequency (they give you this one). In another case the two pendulums move in the same direction and the tube M moves in the opposite direction. In the last case the two pendulums move opposite directions and the tube stands still. -- Clearly the last case which you are trying to find will be independent of M so (A) is the only answer the works.Alternate Solution - Unverified
kokorets
2006-01-03 15:33:57		NEC
kokorets
2006-01-03 15:33:32
minaesm
2009-10-05 04:29:43		 What I try and work till now is that I use limits: in the frequency which is given in the question think of this : What if the M is really bigger than two smaller m s? it's the answer . It goes to answer A.
It also works in similar questions.
NEC

Post A Comment!
You are replying to:
I do not understand the following point. When we are speaking about certain normal mode we imply that every degree of freedom of the system oscillates with the same frequency. We can see that for 2 eigenfrequences proposed in the problem. \r\n However when I imagine that M stands still, while two pendulums oscillate with the same frequency in the opposite directions I arrive to the contradiction. In this case frequency of M is different, it equals zero. How come?

Username:
Password:
Click here to register.
This comment is best classified as a: (mouseover)
 
Mouseover the respective type above for an explanation of each type.

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