GREPhysics.NET: GR9677 #85

GR9677 #85

Problem

GREPhysics.NET Official Solution

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Mechanics $\Rightarrow$ }Wave Phenomena

One can solve this problem without knowing anything about mechanics (but with just the barest idea of wave phenomenon theory). Following the hint, one considers the limiting cases:

$M\rightarrow \infty \Rightarrow \mu/M \rightarrow 0$ ... With an infinite M, the string is basically fixed on the rod, and its wavelength is just $\lambda=L$ . One eliminates choice (A) from the fact that $\cos 2\pi = 1 \neq 0$ , as $\mu/M$ demands in this regime.

$M\rightarrow 0 \Rightarrow \mu/M \rightarrow \infty$ ... Without the presence of the mass M, one has $\lambda = 4L$ . Thus, $2\pi L/\lambda=\pi/2$ . Since $\tan x = \sin x/\cos x$ and $\cos \pi/2 = 0$ , one finds that choice (B) is the only one that fits this condition.

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Comments

socks
2008-09-17 10:06:09

No explicit knowledge of the modes in the $M\rightarrow 0$ limit is necessary.

From the $M\rightarrow\infty$ limit we see that there must be modes for $\lambda=2L/n$ 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 $M\rightarrow 0$ limit, we see that C allows for no modes since the LHS goes to infinity and the RHS is bounded (except for $\lambda\rightarrow 0$ ). The only physical intuition required to obtain choice B is that there must still exist standing modes even for a massless ring.

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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 $M \propto \lambda$ . Then remember $tan \theta = \frac{x}{y}$ and $\frac{\lambda \mu}{2 \pi M}$ corresponds to components of the wave.

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panos85
2007-10-23 01:32:08

There is a (minor) inaccuracy in this solution. As M goes to $\infty$ we have nodes on both sides so every wavelength of the form $\lambda=\frac{2L}{k}$ with k=1,2,... is acceptable. (Remember the standing waves.) In a similar way, when M=0, every wavelength of the form: $\lambda=\frac{4L}{2k+1}$ with k=0,1,2,... is acceptable. The answer B is the only one that is satisfied when you plug in these values.

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dumbguy
2007-10-18 21:52:06

can we still get a better explanation why when M goes to zero lambda=L/4

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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 $L=\lambda/4$ for $M\rightarrow 0$ (open end) and $L=\lambda/2$ for $M\rightarrow \infty$ (closed end). In this way the limiting conditions are both satisfied by choice (B)

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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.

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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).

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alpha
2005-11-07 01:51:52

Good idea.

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