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course PHY 232
2:33 pm 3/16
110228 Physics II
`q001. Two waves propagate along two paperclip chains which are initially separated by 20 cm. The chains meet at a point hundreds of meters away, so they are very nearly parallel. The starting points for the two chains are, as mentioned, 20 cm apart. Let AB be the line connecting these starting points. The chains make nearly equal angles with the line AB.
Explain why, when both chains make about a 90 degree angle with AB, they both lie at a common distance from the point where they meet.
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If the endpoints are the same then the angle the chain makes with the line connecting those ends will be less than or greater than 90. It's easy enough to see in a diagram than to necessarily explain it satisfactorily. An end of a rectangle has 2 90s only if the ends of the lines have a common endpoint, other wise on angle is greater than 90 and one is less than.
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We will measure the (nearly) common direction of these chains not relative to the line AB, but to a 'normal line' perpendicular to AB, and the 'normal direction' is the direction of a normal line. So if the chains make (nearly) right angles with the line AB, they are at (nearly) 0 degrees with the normal direction.
If the chains make an angle of 80 degrees with the line AB, what angle do they make with the normal direction?
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Looking at this as a triangle that forms you can see that you have 2 sides of 80º and 90º so that the 3rd angle has to be 10º.
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Now, if the chains make an angle of 15 degrees with the normal direction, one of the starting points is further from the meeting point than the other. If you imagine the two chains as railroad tracks, with the ties between them making (nearly) right angles with the tracks, then since one 'track' has further to go than the other, there will be one short part of that track that cannot be connected by a 'tie' to the other (i.e., no point in this section can be connected at a right angle to any point of the other). Make a sketch and estimate how long this section will be.
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My estimate is about 1/4 of the 20 cm hyp.
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Now construct the right triangle in your sketch whose hypotenuse is the line AB and one of whose legs is that 'leftover section'. The other leg will correspond to the first possible 'tie' between the 'tracks'. What are the angles of this triangle?
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15, 90 and 75
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Can you use trigonometry and/or vectors to figure out the length of the 'leftover section'?
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The angle that forms this section is congruent to the angle the lines make with the normal line so that makes it 15º and the side is then 20*sin 15 = 5.2 cm.
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What is your best estimate, or your most accurate result, for how much longer one chain must be than the other.
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The most accurate is the one just above which can be given as 5.176 cm
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`q002. At a certain angle one of the chains in the preceding will be 12 cm longer than the other. Suppose waves of wavelength 4 cm are sent out, in phase, along this chain. By how many cycles will the distances traveled by the two waves differ?
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12/4 = 3 cycles
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Answer the same if the wavelength is 18 cm.
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12/18 = 2/3 cycles
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For what wavelength would the difference in the distances be half the wavelength?
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12*2 = 24
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If the 4 cm waves are sent out in phase, will they arrive in phase, 180 degrees out of phase, or somewhere in between?
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For the 12 cm they will be in phase since it's a whole number of wavelengths so that the ends match up after the first wave has travelled the 12 cm.
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Answer the same for the 18 cm waves.
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Here they are 2/3 out of phase which is to say 60º so they interfere destructively but not completely.
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Answer once more if the difference in the distances is half the wavelength.
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For the 24 cm they will be OOP by 180º and will cancel completely since when they meet one wave will be completely opposite the other.
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Give three wavelengths for which the waves would arrive in phase.
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To arrive in phase all of the wavelengths must be ≤12 cm
6, 4, 3 since all of these divide evenly into 12.
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Give three wavelengths for which the waves would arrive 180 degrees out of phase
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180º --> 1/2 wavelength difference
8, 4.8, 3.42857
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`q003. For the preceding situation, give a all possible angles at which 6 cm waves, starting out in phase, would arrive at the meeting point in phase.
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This would occur when the angle generated a side whose length 6 would divide evenly, namely 6, 12, 18, etc. But if the meeting point is fixed then there is only one angle that will generate a side that will allow 6cm waves to arrive in phase, and with d = 20 then 20 sin theta = 6--> theta = 17.5º. If theta changes the point at which the rays meet changes. If that doesn't matter then sin theta can be 0.3, 0.6, 0.9 which gives 17.5, 36.9, and 64.2 respectively.
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Give all possible angles at which 6 cm waves would arrive out of phase.
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Theres all kinds of angles they could arrive OOP, in fact every angle > 0 and < 90 and ≠ one of the angles listed above. Now for only 180º OOP we have: 3, 9, 15 whose corresponding angles are: 8.63, 26.7, and 48.6.
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If two waves generated in phase meet in phase at an angle of 40 degrees, and if they also meet in phase for three different smaller angles, then what is their wavelength?
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Here m*lambda = d sin 40 and in order to be in phase m*lambda/d = a whole number where m*lambda ≤ d. Also sin--> 0 as theta --> 0 so smaller angles means that fewer cycles can occur so here we know 40º is the largest angle and there 3 smaller. This means that m = 4 and then 3, 2, and 1. So lambda= d/4 sin 40. I've worked for about 20 minutes trying to reduce this but without knowing d I don't know if you can find lambda.
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If two different wavelengths both known to meet in phase at an angle of of 40 degrees, what are the two longest possibilities for their wavelengths?
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Following the previous I think that when d* sin 40 =1*lambda is the largest wavelength and when d*sin 40= 2*lambda is the next possible largest.
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@& It isn't stated clearly but this is a continuation of the original situation.
So d = 20 cm, and you can easily find lambda (about 3 cm in this case, if I figure correctly).*@
03-18-2011
@& Very good. See my notes.*@