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GR9677 #81
Problem
 GREPhysics.NET Official Solution Alternate Solutions
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Wave Phenomena}Beats

One remembers the relation for beats, i.e., . Beat phenomenon occurs when two waves occur at nearly the same frequency.

Taking as the fundamental frequency, one deduces the harmonic to be . , and thus the answer is choice (B).

One can derive the relation for beats by recalling the fact that one gets beat phenomenon when one superposes two sound waves of similar frequency , say, of the form ,

where to get beats phenomenon one must have , and since there are two beats per period, one has .  Alternate Solutions
 ramparts2009-10-01 12:43:38 Limits, limits, limits! No need to know about beat frequencies and all that fancy *physics* stuff to get this right (I didn't! :P). Just play the GRE game. One can figure pretty easily that if the string tensions are the same, the amplitude stays at 1, because then they're the same string and there's absolutely no reason for the wave to change. That eliminates D and E. Meanwhile, if the right side is EXTREMELY heavy, you expect the amplitude to be pretty darn small - so as goes to , the amplitude should go to zero. This eliminates A and C, leaving only B. There you go :)Reply to this comment jmason862009-09-22 19:30:44 Basically the same solution but using some test taking strategy: ETS gave two answers with Harmonic = 6. The correct answer is PROBABLY one of these (since all the others are different). Eliminate (C) (D) and (E). 73*6 = 438. 440-438=2 but since the actual frequency is just over 73, the difference should be less than 2. Eliminate (A)... (B) remains. Reply to this comment calvin_physics
2014-03-27 14:24:00
Why does it ask "number" of beats?
Why not say what is the beat frequency?

number - seems integer to me. ramparts
2009-10-01 12:43:38
Limits, limits, limits! No need to know about beat frequencies and all that fancy *physics* stuff to get this right (I didn't! :P). Just play the GRE game. One can figure pretty easily that if the string tensions are the same, the amplitude stays at 1, because then they're the same string and there's absolutely no reason for the wave to change. That eliminates D and E. Meanwhile, if the right side is EXTREMELY heavy, you expect the amplitude to be pretty darn small - so as goes to , the amplitude should go to zero. This eliminates A and C, leaving only B. There you go :)
 ramparts2009-10-01 12:44:32 Ugh! Delete this. I was looking at problem 80. I was wondering why all the solutions talked about beat frequencies ;)
 memorial2010-07-08 08:19:18 wrong problem, buddy. we're on #81, not #80. jmason86
2009-09-22 19:30:44
Basically the same solution but using some test taking strategy:

ETS gave two answers with Harmonic = 6. The correct answer is PROBABLY one of these (since all the others are different). Eliminate (C) (D) and (E).

73*6 = 438. 440-438=2 but since the actual frequency is just over 73, the difference should be less than 2. Eliminate (A)... (B) remains.
 shak2010-08-19 07:02:42 Can u please explain me, why are u dividing frequency of A4 to the frequency of D2 to find harmonics? even , if then answer is close to 6 , but it is smaller than 6, and therefore answer should be 5,i.e lowest? thank you
 alemsalem2010-09-15 03:23:42 what matters is the difference not who is bigger, if you use the fifth harmonic the difference would be bigger, therefore more beats per second, the two waves add to give (approximately): rnsin(w1t)+sin(w2t) = sin(wt)*cos( w1-w2)/2 trnhere w is (w1+w2)/2 since they're almost equal you can take it to be anyone of them, |w2-w1| is the beat (angular) frequency, you're required to minimize the absolute value (cosine is even)
 flyboy6212010-11-05 21:25:05 The nth harmonic of D2 is D2*n. You can quickly estimate 440/73 is closer to 6 than any of the other answers, so the 6th harmonic would minimize the beat frequency. Then multiply 73.4 by 6 to get 440.4, which would be .4 Hz, very close to .5.      LaTeX syntax supported through dollar sign wrappers $, ex.,$\alpha^2_0$produces . type this... to get...$\int_0^\infty\partial\Rightarrow\ddot{x},\dot{x}\sqrt{z}\langle my \rangle\left( abacadabra \right)_{me}\vec{E}\frac{a}{b}\$