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GR9277 #7
Problem
 GREPhysics.NET Official Solution Alternate Solutions
\prob{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$\Rightarrow$}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).

Alternate Solutions
 sharpstones2006-12-01 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.Reply to this comment
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?
bass
2014-07-16 06:11:28
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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 Loops2008-10-05 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.
 spacemanERAU2009-10-22 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.
 keradeek2011-10-03 03:24:37 study the chapter on coupled oscillations in marion and thornton and it will become obvious to you, too.
 speiro2012-09-12 23:56:11 marion 12.18 helped..
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.
kokorets
2006-01-03 15:33:57
kokorets
2006-01-03 15:33:32
 minaesm2009-10-05 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.

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