# Gre Problem # 85

1. Aug 31, 2004

### quantumworld

Another mind boggling problem,
any effort will be relieving...

here it is:
Small-amplitude standing waves of wavelength lambda occur on a string with tension T, mass per unit length mue , and length L. One end of the string is fixed and the other end is attached to a ring of mass M that slides on a frictionless rod. When gravity is neglected, which of the following conditions correctly determines the wavelength? ( you might want to consider the limiting cases M->0 and M->infinity.
(A) mue/M = (2pie/lambda)cot(2pie*L/lambda)
(B) mue/M = (2pie/lambda)tan(2pie*L/lambda)
(C) mue/M = (2pie/lambda)sin(2pie*L/lambda)
(D) Lambda = 2L/n, n=1,2,3...
(E) Lambda = 2L/(n+1/2), n=1,2,3...

my confusion starts with neglecting gravity, what will the added ring do, if no gravity is present?
to see the problem with a figure, please click on the link below, it is #85
http://ftp.ets.org/pub/gre/Physics.pdf [Broken]

Last edited by a moderator: May 1, 2017
2. Sep 1, 2004

### Tom Mattson

Staff Emeritus
No, it isn't. I also tried searching for "standing waves", with zero instances found.

3. Sep 1, 2004

### quantumworld

Sorry Tom,
that question was there couple months ago, it seems that they posted a new one now, I should have double checked before I posted it. but here is my attempted picture:

*fixed end *wavy string *end attached to a ring that slides on a stick

SOrry again, I was trying to draw it, but it was crooked when I posted it.
I will try to search more for that test, to see if it is still somewhere online.
THank you so much for your efforts.

Last edited: Sep 1, 2004
4. Sep 2, 2004

### quantumworld

TOm,
I finally did find that test! page 54 and #85
here it is! :)
http://phys.columbia.edu/~hbar/Physics-GRE.pdf [Broken]

Last edited by a moderator: May 1, 2017
5. Sep 3, 2004

### nrqed

About what confuses you... there is no gravitational force on M but it does provide inertia. Consider the limit $M \rightarrow \infty$, then teh possible wavelengths are $\lambda = 2L, L, 2L/3....$. On the other hand, in the limit $M \rightarrow 0$, we have $\lambda = 4L, 4L/3 ....$ (fixed end/free end boundary conditions). using those limits, it's easy to pick the correct answer.

Pat

Last edited by a moderator: May 1, 2017
6. Sep 3, 2004

### quantumworld

Bingo Pat!
thanks so much!! I just couldn't see it on my own ...

7. Sep 3, 2004

### nrqed

You're very welcome.

It's always nice to get a thank you for answering a question