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Mechanics}Wave Phenomena


There's a long way to solve this problem and then a short. One looks at the choices to find the one that first the physical deduction: when \mu_l=\mu_r, the whole incident wave should be transmitted, with 0 reflection. Moreover, in the limit of \mu_r>>\mu_l there should be 0 transmission. Choice (C) is the only one that fits this condition, leading to a ratio of 1 for \mu_l=\mu_r.

One can also calculate the exact form of the transmission coefficient for this multi-density string. Take the following,


At the boundary between different density parts, one applies continuity \psi_i(x=0)+\psi_r(x=0)=\psi_t(x=0) to get 1+R=T.

One applies m \ddot{\psi} = 0 = \frac{\partial \psi}{\partial x}(x<0) - \frac{\partial \psi}{\partial x}(x>0), where m=0 since there is no point particle situated at the origin, to obtain ik_l(1-R)=ik_r T.

Recalling the nifty relation \omega = ck and c=\sqrt{F/\mu}, one solves for T to get T=\frac{2k_l}{k_r+k+l}=\frac{\sqrt{\mu_l/\mu_r}}{1+\sqrt{\mu_l/\mu_r}}, as in choice (C).

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Comments
apr2010
2010-04-07 04:10:06
Limits will do the trick:

\mu_l / \mu_r \rightarrow 0 \Leftrightarrow A \rightarrow 0

only C fullfills this.

\mu_l / \mu_r \rightarrow 1 \Leftrightarrow A \rightarrow 1

only B and C fullfill this


NEC
BerkeleyEric
2010-04-06 23:35:36
A) and E) can be eliminated immediately from physical intuition. B) and D) can then be eliminated since they do not tend to zero as \frac{\mu_l}{\mu_r} goes to zero.NEC
schadenfraude
2008-11-02 09:28:50
You may also notice that the correct answer should be similar to the form 2m1/(m1+m2) because this problem is analogous to an elastic collision with one particle initially at rest: the transmitted wave is analogous to the motion of the particle that was initially at rest after being hit by an incident particle. This formula has helped me about four times on this test alone, so it is probably a good one to memorize.NEC
yosun
2005-11-11 22:54:35
keflavich: good luck to u too, and to all a good night. i'm out for the night. everyone: have fun on saturday!NEC
keflavich
2005-11-11 21:52:25
Good call, thanks for the revision. Looking at limiting cases seems like a good idea in general. I hope I remember that when I'm taking the test.

Good luck to all taking it tomorrow, and Yosun, thanks for making the site.
NEC
yosun
2005-11-11 21:22:51
keflavich: one can also avoid the formalism by looking at the limit of high density on the right wave. should be no transmission if so. (see revised sol'n above)NEC
keflavich
2005-11-11 19:05:29
Actually, A, B, and C all fit that 'easy' deduction, though it's pretty easy to eliminate A. I ended up having to flip a coin between b an c, since they only differe by which \mu is on top.
physicsisgod
2008-10-30 15:36:04
B doesn't go to zero when \mu _{l} << \mu_{r}

A doesn't depend on either densities at all.

Only C fits the limiting cases.
NEC

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