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course Phy 242
3/15 6
110228 Physics IINote the additional brief experiments. More will probably be listed prior to the 2/28 class:
• individual_experiments/brief_bottle_experiment_3d.htm
• individual_experiments/brief_bottle_experiment_3d.htm
• individual_experiments/brief_bottle_experiment_3d.htm
• individual_experiments/brief_bottle_experiment_3d.htm
• individual_experiments/brief_rubberband_chain_experiment_1a.htm
• individual_experiments/brief_rubberband_chain_experiment_1b.htm
`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.
Because if they both form 90deg that means they are running parallel with each other to end position
@& The chains will be running parallel if they make any common angle with the line AB, but they only run equal distances if that angle is 90 deg.*@
@& *@
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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?
AB and Normal are perpendicular to each other meaning they form a 90deg angle, so for 80deg angle with AB line then it must form 10deg with Normal
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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.
5 or 6cm just estimate and not finding it specifically.
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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?
90deg, 75deg, and 15deg
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Can you use trigonometry and/or vectors to figure out the length of the 'leftover section'?
Yes knowing all angles and that hypo side = 20cm, leftover section is easily found. Might not have been so obvious if I hadn’t watched you work it out in class.
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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.
sin(15deg) = opposite(leftover sect.)/20cm
opposite = approx. .2588*20cm = approx. 5.18cm
leftover section = approx. 5cm
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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?
12cm is 3 wavelengths assuming that wavelengths are 4cm.
So difference would be 3 cycles
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Answer the same if the wavelength is 18 cm.
12/18 = (2/3), so longer chain will complete 2/3 of a cycle more than the other chain
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For what wavelength would the difference in the distances be half the wavelength?
12/x = .5, where x is wavelength so x = 24cm.
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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?
Would be in phase because at even 3 wavelengths difference by the time the longer chains wave got to point where it was running even with second chain they would be running parallel in phase, I believe.
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Answer the same for the 18 cm waves.
For this one they would some up somewhere in between 180 degrees out of phase.
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Answer once more if the difference in the distances is half the wavelength.
They would be 180 degrees out of phase.
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Give three wavelengths for which the waves would arrive in phase.
1cm, 6cm, 12cm
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Give three wavelengths for which the waves would arrive 180 degrees out of phase
1.5cm, 8cm, 18cm
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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.
0, 90, 180, 270, 360
????I’m a little unsure what the question is about this one?????
@& The condition is that the path difference be a whole number of wavelengths.
Express the path difference in terms of the 20 cm separation and the angle with normal.
Solve for the angle with normal. A number of solutions are possible.*@
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Give all possible angles at which 6 cm waves would arrive out of phase.
again I’m not sure, are you talking about the angle wrt the line AB?
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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?
I think we are talking about wrt to the normal line is that correct???
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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?
I’m lost I can’t spend any more time on it, please let me know what I need to calculate and I’ll resubmit.
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@& Spend up to 10 minutes trying to answer my last question. You'll likely get something. After 10 minutes take a minute to tell me what you've got and what you're thinking.*@