GREPhysics.NET: GR9677 #85

GREPhysics.NET

GR9677 #85

Problem

GREPhysics.NET Official Solution

Alternate Solutions

This problem is still being typed.

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.

See below for user comments and alternate solutions!

Alternate Solutions

Herminso
2009-09-21 12:54:05

Consider the motion of the end attached to the ring of mass M:

$-T\sin {\theta}=M\ddot{y}$ ,

where T is the tension of the right end of the string applied directly on M making a small $\theta$ angle with the horizontal, such that

$\sin {\theta}\simeq\sin {\theta}=\frac{\triangle y}{\triangle x}=\frac{\partial y}{\partial x}$ .

By other hand we know that the string is making a harmonic oscilation of the form $y=A\sin {kx}+B\cos {kx}$ , where $B=0$ since the left end of the string is fixed, $y(x=0)=0$ . Thus,

$\frac{\partial y}{\partial x}=Ak\cos {kx}$ and $\ddot{y}=-Ak^2\dot{x}^2\sin {kx}+A\ddot{x}k \cos {kx}$ ,

with $\ddot{x}=0$ since the wave on the string go at constant velocity given by $v^2=\dot{x}^2=T/\mu$ .

Substituting above,

$-T(Ak\cos {kx})=M(-Ak^2\frac{T}{\mu}\sin {kx})$ $\Rightarrow$ $\mu/M=k\tan {kx}$ ,

but $M$ is restricted to move only at $x=L$ , thus $\mu/M=k\tan {kL}$ .

Reply to this comment

Alternate Solution - Unverified

Comments

Yaroslav_Sh
2017-06-28 10:27:48

Can someone explain me please why do we treat $\\mu$ 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 $\\mu$ twice larger as well. Yet if it is indeed the case then consideration of large and small limits of M won\'t change $\\frac{\\mu}{M}$ at all.

Yaroslav_Sh
2017-06-28 10:32:20

Sorry, didn\'t notice that M is mass of the ring, not the string.

Reply to this comment

NEC

mpdude8
2012-04-20 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.

Reply to this comment

NEC

faith
2010-11-02 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
2011-09-25 12:07:49

What about cot(theta)?

checkyoself
2011-10-12 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.

Reply to this comment

NEC

Herminso
2009-09-21 12:54:05

Consider the motion of the end attached to the ring of mass M:

$-T\sin {\theta}=M\ddot{y}$ ,

where T is the tension of the right end of the string applied directly on M making a small $\theta$ angle with the horizontal, such that

$\sin {\theta}\simeq\sin {\theta}=\frac{\triangle y}{\triangle x}=\frac{\partial y}{\partial x}$ .

By other hand we know that the string is making a harmonic oscilation of the form $y=A\sin {kx}+B\cos {kx}$ , where $B=0$ since the left end of the string is fixed, $y(x=0)=0$ . Thus,

$\frac{\partial y}{\partial x}=Ak\cos {kx}$ and $\ddot{y}=-Ak^2\dot{x}^2\sin {kx}+A\ddot{x}k \cos {kx}$ ,

with $\ddot{x}=0$ since the wave on the string go at constant velocity given by $v^2=\dot{x}^2=T/\mu$ .

Substituting above,

$-T(Ak\cos {kx})=M(-Ak^2\frac{T}{\mu}\sin {kx})$ $\Rightarrow$ $\mu/M=k\tan {kx}$ ,

but $M$ is restricted to move only at $x=L$ , thus $\mu/M=k\tan {kL}$ .

walczyk
2011-03-30 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
2012-10-14 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?

Reply to this comment

Alternate Solution - Unverified

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.

Reply to this comment

NEC

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.

Reply to this comment

NEC

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.

QuantumCat
2014-09-23 14:39:10

Thanks for clearing this up. I was really confused by the $\lambda$ = $4L$ appearance, but this makes perfect sense when I think about it. The analogy I made is that in an open-ended pipe, you get quarter wavelengths, and this situation appears to be the same.

Reply to this comment

NEC

dumbguy
2007-10-18 21:52:06

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

nontradish
2012-04-19 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.

Reply to this comment

NEC

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)

Lego
2009-01-17 11:50:03

Excellent! This is what i had in mind too!

Reply to this comment

NEC

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

Reply to this comment

Answered Question!

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

Reply to this comment

Answered Question!

chri5tina
2006-11-28 06:23:44

answer A) has a cotangent in it, not a cosine.

Reply to this comment

Typo Alert!

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).

Reply to this comment

Typo Alert!

alpha
2005-11-07 01:51:52

Good idea.

Reply to this comment

NEC

Post A Comment!

Bare Basic LaTeX Rosetta Stone
LaTeX syntax supported through dollar sign wrappers $, ex., $\alpha^2_0$ produces $\alpha^2_0$ .
type this...	to get...
$\int_0^\infty$	$\int_0^\infty$
$\partial$	$\partial$
$\Rightarrow$	$\Rightarrow$
$\ddot{x},\dot{x}$	$\ddot{x},\dot{x}$
$\sqrt{z}$	$\sqrt{z}$
$\langle my \rangle$	$\langle my \rangle$
$\left( abacadabra \right)_{me}$	$\left( abacadabra \right)_{me}$
$\vec{E}$	$\vec{E}$
$\frac{a}{b}$	$\frac{a}{b}$

The Sidebar Chatbox...
Scroll to see it, or resize your browser to ignore it...

All-Solutions List | Latest Comments | Unanswered Questions (Cries for Help!) |::| About

© 2005-2007 by Yosun Chang, All rights reserved.

This site is not affiliated with ETS, nor is it officially endorsed (yet) by ETS. This site is the work of an individual named Yosun Chang. The solutions, sans user comments and ETS questions, are © 2005 by Yosun Chang except where noted. All rights reserved. / The comments appending particular solutions are due to the respective users. / Educational Testing Service, ETS, the ETS logo, Graduate Record Examinations, and GRE are registered trademarks of Educational Testing Service. The examination questions are © 1997 and © 2001 by ETS.