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course Phy 202
3/2 10 2.28.11
`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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The chains make about a 90 degree angle with AB because the angle of each is the same and assuming the same length of chain on each side of AB, then they travel the same distance to reach the same point. Change the angles and there would be a difference.
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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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The chains will make a 10 degree angle with AB.
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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 was 4 cm.
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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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The angles are approximately 75 degrees and 15 degrees.
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Can you use trigonometry and/or vectors to figure out the length of the 'leftover section'?
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Using 20 sin 15, we discover the length is approx 5 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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Approximately 4 cm longer
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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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One wave will travel 3 more cycles than the other.
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Answer the same if the wavelength is 18 cm.
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4.5 cycles more
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@& 18 cm is the new wavelength; the 4 cm wavelength was for the preceding question.
12 cm would be only 2/3 of the 18 cm wavelength. *@
For what wavelength would the difference in the distances be half the wavelength?
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24 cm
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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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They will arrive in phase.
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Answer the same for the 18 cm waves.
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They will arrive 180 degrees out of phase
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@& The path difference is 2/3 of a wavelength, so they won't be in phase, but they won't be 180 deg out of phase either.*@
Answer once more if the difference in the distances is half the wavelength.
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Will arrive 180 degrees out of phase
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Give three wavelengths for which the waves would arrive in phase.
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4, 6, 12
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Give three wavelengths for which the waves would arrive 180 degrees out of phase
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8, 10, 14
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@& For 8 cm the 12 cm path difference is 1.5 wavelengths, which will cause them to arrive 180 deg out of phase.
But for a 10 cm wave the 12 cm path difference would be 1.2 wavelengths, so the waves would arrive more in phase than out of phase, though not exactly either.*@
`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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18 degrees and 37 degrees
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@& You don't say how you got these, but since they're right it's likely you did the right thing.*@
Give all possible angles at which 6 cm waves would arrive out of phase.
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All other angles. 15, 17, 20, so on If they arrive in phase at certain degrees then all other angles, they would arrive out of phase.
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@& They wouldn't be 180 deg out of phase, but since they aren't in phase at those lengths the are at least to some extent out of phase.
I intended to ask what wavelengths would be 180 deg out of phase. That would occur around 9 deg and 27 deg.*@
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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5 cm, Assuming the angle adjacent is 40 degrees, then the angle opposite must be 50 degrees.
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@& If the waves meet in phase for three smaller angles, then this in-phase meeting must correspond to a path difference of 4 wavelengths.
The path difference is 20 cm * sin(40 deg) = 13 cm or so. If this is 4 wavelengths then the wavelength must be 3.25 cm.
So the waves will reinforce at angles which result in path differences of 3.25 cm, 7.5 cm and 10.75 cm. It would be possible to use these path lengths to calculate the necessary angles.*@
If two different wavelengths both known to meet in phase at an angle of 40 degrees, what are the two longest possibilities for their wavelengths?
60 cm and 30 cm. Wouldn’t one wavelength need to be half the other in order for them to meet in phase? One would complete one whole cycle while the other completed 2.
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&#This looks good. See my notes. Let me know if you have any questions. &#