GR9677 #85



Alternate Solutions 
Herminso 20090921 12:54:05  Consider the motion of the end attached to the ring of mass M:
,
where T is the tension of the right end of the string applied directly on M making a small angle with the horizontal, such that
.
By other hand we know that the string is making a harmonic oscilation of the form , where since the left end of the string is fixed, . Thus,
and ,
with since the wave on the string go at constant velocity given by .
Substituting above,
,
but is restricted to move only at , thus .  

Comments 
Yaroslav_Sh 20170628 10:27:48  Can someone explain me please why do we treat and M as independent parameters?\r\n I would think that if we keep L fixed and increase M by factor of two that would essentially make twice larger as well. Yet if it is indeed the case then consideration of large and small limits of M won\'t change at all.
Yaroslav_Sh 20170628 10:32:20 
Sorry, didn\'t notice that M is mass of the ring, not the string.

  mpdude8 20120420 18:30:42  If ETS gives you a hint, use the damn hint.
Eliminate everything except A and B  those are the only options that can blow up as M > 0.   faith 20101102 00:52:08  by mathematics, (trigonometry for that matter) and taking ETS hints. the only trigonometry ratio that allows infinity is the tan Teta.
only A and B left.
for M=0, becomes tanTeta= infinity: teta are n pi/2 with n odd integer.
say n=1, 2pi L/lamda=pi/2
L=lamda/4 . sketch this wave length which reveal an unclosed wave (antinode at the end). this implies freedom of movement. hence B is the answer.
Dr. D.R. Dopetec 20110925 12:07:49 
What about cot(theta)?

checkyoself 20111012 13:06:22 
cot() blows up when it's argument is integer multiples of pi (incl. zero). In the case where M approaches infinity where we know lambda approaches L, the cot() function doesn't work because it will yield infinity when we want zero.
I don't know about the other limit  you have funny values of lambda which I don't understand (help?). Anyway using just the one limit is enough to identify the answer.

  Herminso 20090921 12:54:05  Consider the motion of the end attached to the ring of mass M:
,
where T is the tension of the right end of the string applied directly on M making a small angle with the horizontal, such that
.
By other hand we know that the string is making a harmonic oscilation of the form , where since the left end of the string is fixed, . Thus,
and ,
with since the wave on the string go at constant velocity given by .
Substituting above,
,
but is restricted to move only at , thus .
walczyk 20110330 17:22:11 
this is awesome. now if there is anyone smart enough to do normal mode oscillation examples, then that would be perfect.

hjq1990 20121014 01:33:52 
Yes, why is ETS bothering contriving 100 questions instead of 10 that require our comprehensive abilities? Is that because in modern time computation is less important than knowing as wide as possible?

  socks 20080917 10:06:09  No explicit knowledge of the modes in the limit is necessary.
From the limit we see that there must be modes for where n is a positive integer. This narrows the choices down to B, C, or D. Choice D is ruled out immediately as we suspect there must be a dependence on the mass of the ring.
In the limit, we see that C allows for no modes since the LHS goes to infinity and the RHS is bounded (except for ). The only physical intuition required to obtain choice B is that there must still exist standing modes even for a massless ring.   FortranMan 20080903 17:07:15  for me, the first key was to realize that rapid movement of the string (short wavelength) would require a light ring mass, hence . Then remember and corresponds to components of the wave.   panos85 20071023 01:32:08  There is a (minor) inaccuracy in this solution. As M goes to we have nodes on both sides so every wavelength of the form with k=1,2,... is acceptable. (Remember the standing waves.) In a similar way, when M=0, every wavelength of the form: with k=0,1,2,... is acceptable. The answer B is the only one that is satisfied when you plug in these values.
QuantumCat 20140923 14:39:10 
Thanks for clearing this up. I was really confused by the = appearance, but this makes perfect sense when I think about it. The analogy I made is that in an openended pipe, you get quarter wavelengths, and this situation appears to be the same.

  dumbguy 20071018 21:52:06  can we still get a better explanation why when M goes to zero lambda=L/4
nontradish 20120419 19:57:40 
I am sure this will not help dumbguy, but maybe it will help someone else. When M > 0 picture it as a quarter of a wave. There is one node on the wall and the first antinode on the rod.

  Mexicana 20071005 14:36:55  I think there's a mistake in the relation for the fundamental harmonic that you are quoting. Instead this should be for (open end) and for (closed end). In this way the limiting conditions are both satisfied by choice (B)
Lego 20090117 11:50:03 
Excellent! This is what i had in mind too!

  cyberdeathreaper 20070204 19:53:47  Also, can someone explain why the wavelength is 4L when M approaches zero?
hamood 20070411 21:36:40 
when m approaches zero
node on the left and antinode on the right
L = lambda/4
wavelength = 4L

  cyberdeathreaper 20070204 19:43:37  It's unclear to me why the wavelengths of the standing waves have to be of any specified value. For example, when M approaches infinity, the wavelength could be 2L, L, 2L/3, etc...
hamood 20070411 21:39:14 
when m is very large you have nodes on both ends

  chri5tina 20061128 06:23:44  answer A) has a cotangent in it, not a cosine.   buddy.epson 20061014 14:18:57  With M>>mu, lambda=2L, as the antinode is in the center at L/2. That gives the term tan(pi)=0, what you want for the limiting case in (b).   alpha 20051107 01:51:52  Good idea.  




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