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course Phy 242
March 22 around 4pm. RESUBMITTING this assingment bc it never showed up in my portfolio.
110228`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 you break them into individual triangles, the “hypotenuse” is 20cm, and the extra distance will equal :
20cm sin (theta)
And since the “big/main” triangle has such a tiny angle, the “hypotenuse” and bottom is nearly parallel since it is so long.
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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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10 degrees
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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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The path difference would be 4cm, the base would be 20cm, and the angle is 15 degrees.
A sin 15 degrees = 0.23
0.23(20) = 4.6 approx.
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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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Sin (theta) = (12/20) = 0.6
Theta = sin^(-1)*(0.6) = 37 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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Yes
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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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Path difference would go from 4cm to about 12cm.
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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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3 COMPLETE wavelengths
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Answer the same if the wavelength is 18 cm.
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Close to 1 complete wavelength
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For what wavelength would the difference in the distances be half the wavelength?
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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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In phase
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Answer the same for the 18 cm waves.
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The peaks don’t match
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Answer once more if the difference in the distances is half the wavelength.
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Give three wavelengths for which the waves would arrive in phase.
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If path difference is lambda, then waves reinforce at point of observation.
2 lambda
3 lambda
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Give three wavelengths for which the waves would arrive 180 degrees out of phase
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If path difference is (3/2)lambda, waves arrive out of phase and cancel.
(5/2)lambda
(7/2)lambda
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@& Many were good, but you could have been more specific in some of your solutions.
Compare with the following:
If the difference in the lengths is 12 cm, and if waves of wavelength 4 cm are sent out, then the 12 cm difference corresponds to 3 wavelengths (i.e., 12 cm / (4 cm / wavelength) = 3 wavelengths).
If the difference in lengths is 18 cm, then this distance corresponds to 18 cm / (4 cm / wavelength) = 4.5 wavelengths.
If 18 cm is half the wavelength then the wavelength is 2 * 18 cm = 36 cm.
A cycle of the wave corresponds to 1 wavelength, and a cycle of the wave corresponds to 360 degrees of phase. So a full wavelength represents 360 degrees of phase. The phase can be represented as the angle in a circular model of the sine or cosine function. A full wavelength corresponds to a full cycles around the circle. If one path is 4.5 wavelengths longer than the other then the phase difference corresponds to 4 complete cycles, plus half a cycle. Since the waves in this case are sent out in phase, they will arrive 1/2 cycle (the same as 180 degrees, corresponding to 1/2 wavelength) out of phase.
If the difference in distances is half a wavelength, then the waves are 180 degrees out of phase; peaks of one wave will match up with a valley of the other and the waves will cancel.
For the original 12 cm path difference, a wavelength of 12 cm would give us 1 full wavelength of path difference, so that the waves would arrive in phase. For wavelengths of 6 cm and 4 cm the 12 cm path difference would result, respectively, in path differences of 2 and 3 wavelengths, again giving us in-phase.
To arrive out of phase the path difference would have to be a full wavelength plus a half wavelength. This is the case of the path difference is 1/2, 3/2, 5/2 wavelength, in which case the wavelengths would be 2/1, 2/3, and 2/5 times the path difference, giving us wavelengths:
2/1 * 12 cm = 24 cm
2/3 * 12 cm = 8 cm
2/5 * 12 cm = 4.8 cm.
*@
`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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I’m not sure about these last couple of questions. Is this where I would use tangent, theta, and similar triangles to get the angles?!
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@& All you need is that one triangle, whose hypotenuse is 20 cm, with angle theta opposite the side representing the path difference.*@
Give all possible angles at which 6 cm waves would arrive out of phase.
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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?
****
#$&*
"
@&
If the angle of the chains with the line AB is theta, then the path difference is the side opposite the angle theta in the triangle whose hypotenuse is 20 cm, so that the path difference is
path difference = 20 cm * sin(theta).
Waves will arrive in phase if the path difference is 1, 2, 3, ... wavelengths, corresponding for the 6 cm wave to path differences of 6, 12 and 18 cm.
Thus we for the three desired wavelengths we have
6 cm = 20 cm * sin(theta)
12 cm = 20 cm * sin(theta)
18 cm = 20 cm * sin(theta).
Solving we get the respective results
theta = arcsin(6 cm / 20 cm) = 18 deg, approx.
theta = arcsin(12 cm / 20 cm) = 37 deg, approx.
theta = arcsin(18 cm / 20 cm) = 65 deg, approx.
Note that there is no other wavelength that allows the waves to meet in phase. The next candidate would be 24 cm, which is however greater than 20 cm. This won't work because the path difference can't exceed the 20 cm distance between the sources (even at 90 degrees the path difference would be only 20 cm); and algebraically it doesn't work because a 24 cm path difference would require that sin(theta) be greater than 1.
To arrive out of phase the path difference would have to be 1/2, 3/2, 5/2, ... wavelength, corresponding for the 6 cm wave to 3, 9, 15 cm.
3 cm = 20 cm * sin(theta)
9 cm = 20 cm * sin(theta)
15 cm = 20 cm * sin(theta).
Solving we get the respective results
theta = arcsin(3 cm / 20 cm) = 9 deg, approx.
theta = arcsin(9 cm / 20 cm) = 27 deg, approx.
theta = arcsin(15 cm / 20 cm) = 48 deg, approx.
If the waves meet in phase for 40 degrees and for three smaller angles, then in-phase meeting at 40 deg 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.
*@
03-23-2011
03-23-2011
#$&*
course Phy 242
March 22 around 4pm. RESUBMITTING this assingment bc it never showed up in my portfolio.
110228`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.
****
If you break them into individual triangles, the “hypotenuse” is 20cm, and the extra distance will equal :
20cm sin (theta)
And since the “big/main” triangle has such a tiny angle, the “hypotenuse” and bottom is nearly parallel since it is so long.
#$&*
We will measure the (nearly) common direction of these chains not relative to the line AB, but to a 'normal line' perpendicular to AB, and the 'normal direction' is the direction of a normal line. So if the chains make (nearly) right angles with the line AB, they are at (nearly) 0 degrees with the normal direction.
If the chains make an angle of 80 degrees with the line AB, what angle do they make with the normal direction?
****
10 degrees
#$&*
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.
****
The path difference would be 4cm, the base would be 20cm, and the angle is 15 degrees.
A sin 15 degrees = 0.23
0.23(20) = 4.6 approx.
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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?
****
Sin (theta) = (12/20) = 0.6
Theta = sin^(-1)*(0.6) = 37 degrees
#$&*
Can you use trigonometry and/or vectors to figure out the length of the 'leftover section'?
****
Yes
#$&*
What is your best estimate, or your most accurate result, for how much longer one chain must be than the other.
****
Path difference would go from 4cm to about 12cm.
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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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3 COMPLETE wavelengths
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Answer the same if the wavelength is 18 cm.
****
Close to 1 complete wavelength
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For what wavelength would the difference in the distances be half the wavelength?
****
#$&*
If the 4 cm waves are sent out in phase, will they arrive in phase, 180 degrees out of phase, or somewhere in between?
****
In phase
#$&*
Answer the same for the 18 cm waves.
****
The peaks don’t match
#$&*
Answer once more if the difference in the distances is half the wavelength.
****
#$&*
Give three wavelengths for which the waves would arrive in phase.
****
If path difference is lambda, then waves reinforce at point of observation.
2 lambda
3 lambda
#$&*
Give three wavelengths for which the waves would arrive 180 degrees out of phase
****
If path difference is (3/2)lambda, waves arrive out of phase and cancel.
(5/2)lambda
(7/2)lambda
#$&*
@& Many were good, but you could have been more specific in some of your solutions.
Compare with the following:
If the difference in the lengths is 12 cm, and if waves of wavelength 4 cm are sent out, then the 12 cm difference corresponds to 3 wavelengths (i.e., 12 cm / (4 cm / wavelength) = 3 wavelengths).
If the difference in lengths is 18 cm, then this distance corresponds to 18 cm / (4 cm / wavelength) = 4.5 wavelengths.
If 18 cm is half the wavelength then the wavelength is 2 * 18 cm = 36 cm.
A cycle of the wave corresponds to 1 wavelength, and a cycle of the wave corresponds to 360 degrees of phase. So a full wavelength represents 360 degrees of phase. The phase can be represented as the angle in a circular model of the sine or cosine function. A full wavelength corresponds to a full cycles around the circle. If one path is 4.5 wavelengths longer than the other then the phase difference corresponds to 4 complete cycles, plus half a cycle. Since the waves in this case are sent out in phase, they will arrive 1/2 cycle (the same as 180 degrees, corresponding to 1/2 wavelength) out of phase.
If the difference in distances is half a wavelength, then the waves are 180 degrees out of phase; peaks of one wave will match up with a valley of the other and the waves will cancel.
For the original 12 cm path difference, a wavelength of 12 cm would give us 1 full wavelength of path difference, so that the waves would arrive in phase. For wavelengths of 6 cm and 4 cm the 12 cm path difference would result, respectively, in path differences of 2 and 3 wavelengths, again giving us in-phase.
To arrive out of phase the path difference would have to be a full wavelength plus a half wavelength. This is the case of the path difference is 1/2, 3/2, 5/2 wavelength, in which case the wavelengths would be 2/1, 2/3, and 2/5 times the path difference, giving us wavelengths:
2/1 * 12 cm = 24 cm
2/3 * 12 cm = 8 cm
2/5 * 12 cm = 4.8 cm.
*@
`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.
****
I’m not sure about these last couple of questions. Is this where I would use tangent, theta, and similar triangles to get the angles?!
#$&*
@& All you need is that one triangle, whose hypotenuse is 20 cm, with angle theta opposite the side representing the path difference.*@
Give all possible angles at which 6 cm waves would arrive out of phase.
****
#$&*
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?
****
#$&*
"
@&
If the angle of the chains with the line AB is theta, then the path difference is the side opposite the angle theta in the triangle whose hypotenuse is 20 cm, so that the path difference is
path difference = 20 cm * sin(theta).
Waves will arrive in phase if the path difference is 1, 2, 3, ... wavelengths, corresponding for the 6 cm wave to path differences of 6, 12 and 18 cm.
Thus we for the three desired wavelengths we have
6 cm = 20 cm * sin(theta)
12 cm = 20 cm * sin(theta)
18 cm = 20 cm * sin(theta).
Solving we get the respective results
theta = arcsin(6 cm / 20 cm) = 18 deg, approx.
theta = arcsin(12 cm / 20 cm) = 37 deg, approx.
theta = arcsin(18 cm / 20 cm) = 65 deg, approx.
Note that there is no other wavelength that allows the waves to meet in phase. The next candidate would be 24 cm, which is however greater than 20 cm. This won't work because the path difference can't exceed the 20 cm distance between the sources (even at 90 degrees the path difference would be only 20 cm); and algebraically it doesn't work because a 24 cm path difference would require that sin(theta) be greater than 1.
To arrive out of phase the path difference would have to be 1/2, 3/2, 5/2, ... wavelength, corresponding for the 6 cm wave to 3, 9, 15 cm.
3 cm = 20 cm * sin(theta)
9 cm = 20 cm * sin(theta)
15 cm = 20 cm * sin(theta).
Solving we get the respective results
theta = arcsin(3 cm / 20 cm) = 9 deg, approx.
theta = arcsin(9 cm / 20 cm) = 27 deg, approx.
theta = arcsin(15 cm / 20 cm) = 48 deg, approx.
If the waves meet in phase for 40 degrees and for three smaller angles, then in-phase meeting at 40 deg 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.
*@
03-23-2011
03-23-2011