PGRE9677 #85

User avatar
HappyQuark
Posts: 762
Joined: Thu Apr 16, 2009 2:08 am

PGRE9677 #85

Postby HappyQuark » Mon May 03, 2010 3:13 pm

I've been racking my brain on this problem and I've got the possible solutions down to 2 but I can't seem to eliminate either one definitively. Here is the problem

Image

Using the "limiting case" recommendation from the question, I figure that we can eliminate (D) and (E) because for either of these equations to be true, it must be the case that wavelength is not dependent on the mass of the string or the mass of the ring, which couldn't be true.

For (A), (B) and (C), when M \rightarrow 0, \mu/M \rightarrow \infty and the sine component to (C) will just keep it oscillating between its amplitude, so we can eliminate that.

However, I can't for the life of me eliminate (A) or (B). I considered the possibility that only certain wavelengths would be allowed when you have a fixed endpoint, which is technically true, and so I argued that with the fundamental (first harmonic) wave with just two nodes, you would have the \lambda = L, at which point the Cot solution would go to infinity and Tan would go to 0. In this case the Tan solution would be correct. However, if the wavelength = 1/2 L, then the opposite is true. It seems to me the only way that ONLY (B) can be the solution would be if letting the mass of the ring go to infinity would force \lambda = L. This doesn't make sense to me because I know you can create multiple modes for a standing wave even with 1 or both of the endpoints fixed. However, I'm wondering if maybe the problem is trying to suggest, by telling us that the string has a tension T, that the cord is pulled tight. Under this condition I might be willing to accept that the wavelength is restricted to only a wave with 2 nodes.

Any thoughts?

michael
Posts: 50
Joined: Fri Sep 05, 2008 7:21 am

Re: PGRE9677 #85

Postby michael » Mon May 03, 2010 11:49 pm

This is how I would do it:

I agree that the fact that answers D and E dont depend on M and mu make them incorrect - leaving A, B, C.

Take M approaches infinity. Then you know that both ends are nodes (physically its like pinning down the end with the mass cus it is too heavy for the string to move it), and the condition to be satisfied is n \lambda/2 = L look for which equations share this. When M approaches infinity, the solutions of equations for the conditions A, B and C are respectively:
A) cot(2 \pi L/ \lambda) = 0 ~~~~ therefore  ~~~~~~ 2 \pi L / \lambda = \pi (n + 1/2) ~~~~~so ~~~~ n /2 + 1/4 = L
B) tan(2 \pi L/ \lambda) = 0 ~~~~ therefore  ~~~~~~ 2 \pi L / \lambda = \pi n ~~~~~so ~~~~ n \lambda /2 = L
C) sin(2 \pi L/ \lambda) = 0 ~~~~ therefore  ~~~~~~ 2 \pi L / \lambda = \pi n ~~~~~so ~~~~ n \lambda /2 = L

Thus B and C are consistent in this limit with what we know must be the answer - but A is not so kick it out.

To find out between B and C, look at the M goes to zero limit. Since the LHS of the conditions given in all answers will be infinate, the right hand side must be able to be infinite for there to be any valid solutions of lambda - only B can do this, so the answer must be B. I didnt quite get your argument for why C cant be right - I am guessing it was probably the same thing though.

Hope I explained this well (and that i didnt make any mistakes).

User avatar
HappyQuark
Posts: 762
Joined: Thu Apr 16, 2009 2:08 am

Re: PGRE9677 #85

Postby HappyQuark » Tue May 04, 2010 12:59 pm

michael wrote:This is how I would do it:

I agree that the fact that answers D and E dont depend on M and mu make them incorrect - leaving A, B, C.

Take M approaches infinity. Then you know that both ends are nodes (physically its like pinning down the end with the mass cus it is too heavy for the string to move it), and the condition to be satisfied is n \lambda/2 = L look for which equations share this. When M approaches infinity, the solutions of equations for the conditions A, B and C are respectively:
A) cot(2 \pi L/ \lambda) = 0 ~~~~ therefore  ~~~~~~ 2 \pi L / \lambda = \pi (n + 1/2) ~~~~~so ~~~~ n /2 + 1/4 = L
B) tan(2 \pi L/ \lambda) = 0 ~~~~ therefore  ~~~~~~ 2 \pi L / \lambda = \pi n ~~~~~so ~~~~ n \lambda /2 = L
C) sin(2 \pi L/ \lambda) = 0 ~~~~ therefore  ~~~~~~ 2 \pi L / \lambda = \pi n ~~~~~so ~~~~ n \lambda /2 = L

Thus B and C are consistent in this limit with what we know must be the answer - but A is not so kick it out.

To find out between B and C, look at the M goes to zero limit. Since the LHS of the conditions given in all answers will be infinate, the right hand side must be able to be infinite for there to be any valid solutions of lambda - only B can do this, so the answer must be B. I didnt quite get your argument for why C cant be right - I am guessing it was probably the same thing though.

Hope I explained this well (and that i didnt make any mistakes).


Thanks, that makes sense. I'm not sure why I didn't catch that.

I apologize if my reasoning wasn't clear for why I excluded C. Actually my reasoning was identical to yours, I just did a poor job of describing it. I was suggesting that when M\ \rightarrow 0, the LHS (\mu/M\ \rightarrow \infty). Since the RHS needs to be able to equal the LHS for all values of the length and wavelength, the sine term would need to be able to blow up to infinity with some combination of values. It can't so we can ignore it.

Thanks for the help




Return to “Problems, Solutions, and Discussion”

Who is online

Users browsing this forum: No registered users and 1 guest