How Does Mass and Tension Affect Wavelength in a Frictionless System?

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The discussion revolves around a physics problem involving small-amplitude standing waves on a string with one end fixed and the other attached to a mass that can slide on a frictionless rod. The key question is how to determine the wavelength of the standing waves when gravity is neglected. Participants explore various equations related to the mass per unit length, tension, and the mass of the sliding ring, considering limiting cases where the mass approaches zero or infinity. It is clarified that while gravity does not act on the mass, it contributes inertia, affecting the standing wave conditions. The limits lead to specific wavelengths based on boundary conditions, helping to identify the correct answer among the provided options. The conversation emphasizes the importance of understanding these limits to solve the problem effectively.
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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
 
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quantumworld said:
to see the problem with a figure, please click on the link below, it is #85

No, it isn't. I also tried searching for "standing waves", with zero instances found.
 
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.
 
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TOm,
I finally did find that test! page 54 and #85
here it is! :)
http://phys.columbia.edu/~hbar/Physics-GRE.pdf
 
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quantumworld said:
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

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

You're very welcome.

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