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GR9677 #85
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Alternate Solutions |
Herminso
2009-09-21 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, =0) . Thus,
 and  ,
with  since the wave on the string go at constant velocity given by  .
Substituting above,
=M(-Ak^2\frac{T}{\mu}\sin {kx}))   ,
but  is restricted to move only at  , thus  . |  |
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Comments |
Herminso
2009-09-21 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 . | | socks
2008-09-17 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
2008-09-03 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
2007-10-23 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. | | dumbguy
2007-10-18 21:52:06 | can we still get a better explanation why when M goes to zero lambda=L/4 | | Mexicana
2007-10-05 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
2009-01-17 11:50:03 |
Excellent! This is what i had in mind too!
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| | cyberdeathreaper
2007-02-04 19:53:47 | Also, can someone explain why the wavelength is 4L when M approaches zero?
hamood
2007-04-11 21:36:40 |
when m approaches zero
node on the left and antinode on the right
L = lambda/4
wavelength = 4L
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|  | cyberdeathreaper
2007-02-04 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
2007-04-11 21:39:14 |
when m is very large you have nodes on both ends
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| | chri5tina
2006-11-28 06:23:44 | answer A) has a cotangent in it, not a cosine. |  | buddy.epson
2006-10-14 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
2005-11-07 01:51:52 | Good idea. | |
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