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Phy 122
Your 'question form' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
** Question Form_labelMessages **
Piano Question
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Problem Set 6 #14 states:
Problem
Our musical scale is based on the natural harmonics of strings and air columns. We use a scale consisting of a sequence of pitches each of which hass 2^(1/12) the frequency of its predecessor. On the piano keyboard, for example, the frequency ratio of any white or black key to the key just before it (whether white or black) is 2^(1/12). The approximate value of this ratio is 1.0594.
Show that if we start at any key and count up 7 keys we will achieve a ratio of pitches very nearly equal to 3/2 (this ratio is called a 'fifth', corresponding to a span of 5 white keys on the piano).
Show that if we start at any key and count up 12 keys we will achieve a pitch ratio of 2/1 (this ratio is called an 'octave', corresponding to a span of 8 white keys on the piano).
Determine how many keys we must count up from a given key to come as close as possible to the ratios 4/3 (called a 'fourth', corresponding to a span of 4 white keys on the piano), 5/4 (called a 'third', corresponding to a span of 3 white keys on the piano), and 6/5 (called a 'minor third', corresponding to a span of 2 white keys and a black key on the piano). Musicians will note that these ratios are of primary importance in Western music.
How close do the ratios based on the ratio 2^(1/12) come to mimicking the natural ratios of the harmonics of a string?
If 12 divisions of 2 come this close, it seems like a greater number of divisions, for example 26, might do an even better job. Use your calculator to find the ratio that would correspond to 26 subdivisions, and determine how nearly the natural ratios could be mimicked using multiples of this ratio.
It turns out that the next number of divisions that actually improves on 12 is 43. This ratio is used in some Eastern music. How much better can we do with a division of the ratio 2 into 43 equal subdivisions, as compared to 12 subdivisions?
Solution
If we count up 7 keys,
the first will have ratio 2 ^(1/12),
the second a ratio of [2^(1/12)] ^2 = 2^(2/12) = 2^(1/6),
the third a ratio of [2^(1/12)] ^3 = 2^(3/12) = 2^(1/4),
the fourth a ratio of [2^(1/12)] ^4 = 2^(4/12) = 2^(1/3),
....
and the seventh a ratio of [2^(1/12)] ^7 = 2^(7/12).
If you key this into your calculator you will see that it is very close to 3/2, or 1.5.
The other ratios are similarly calculated:
Counting up 12 keys we would obtain a ratio of [2^(1/12)] ^12 = 2^(12/12) = 2^(1) = 2.
The ratio obtained for 5 keys is 2^(5/12), close to the ratio 4/3 = 1.333....
The ratios for 4 and 3 keys are, as indicated above, 2^(1/3) and 2^(1/4). A calculator will verify that these ratios are close to 5/4 = 1.20 and 6/5 = 1.166....
The ratios obtained for the 30-note division of the 2/1 ratio are
2^(1/ 30), 2^(2/ 30), 2^(3/ 30), ..., 2^(n/ 30), where n can be any counting number.
You can figure these ratios out with your calculator. You will not find good matches for most of the naturally occurring ratios 3/2, 4/3, 5/4, .... In any case the matches are not as good as for a division into 12 equal ratios.
The 43-note division is done in a very similar manner, and does result in an improvement over the 12-note scale. However, an instrument with 43 notes to a doubling (and 'octave') is for the most part impractical (imagine a piano with 88 * (43/12) keys). 12 notes does the job well enough that very few people can even learn to distinguish the difference between the musical intervals based on this scale and the natural harmonics.
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I get the progression of the next 7 notes based on the 12 keys.
I don't know how on a test to show that a 26 note octave wouldn't be better. Would I have to show the value for every step and show that nothing came closer to 1.5 or 1.2, etc.( for example, would I need to show 2^(1/30) = 1.023, 2^(2/30) =1.047 all the way up to 2^(29/30)?) If not that, how else would I show it?
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That is what you would need to do.
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However, no question on your test would ask you to do such an extensive investigation. Any related question on the test would test your understanding by asking a more limited question.
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Likewise how do I show that 43 is better? I did 2^(25/43) and got 1.496295739 and 2^(26/43) = 1.520411032. Neither one is closer to 1.5 than 2^(7/12) = 1.498307077. So I'm not sure how to show that the 43 note octave is better or how it is better.
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2^(25/43) is very close to 1.5, though not closer than 2^(7/12).
However other intervals are also important. A scale has to approximate not only the ratio 3/2, but also the ratios 4/3, 5/4 and 6/5 (perfect fourth, perfect third and minor third). The 43-tone scale is the first one that's better at approximating all these intervals than the 12-tone scale.
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Thanks for your help.
Sorry if this one sounds whiny.
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A question like that on the test would be worth whining about.
However you certainly understand what you need to understand.
Check my notes.
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critiqued_student work modified 130317__________
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Phy 122
Your 'question form' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
** Question Form_labelMessages **
Ray Tracing with Cylinders
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I'm working on the Lab and I've come to this point:
Using a book, a CD case or any flat container measure the distance behind the cylinder at which the sharpest image forms. Measure also the radius of the cylinder.
I did not have two lamps of the same size so I used two identical flashlights 1 foot apart and 10 feet from the front of the two liter bottle.
Sharpest image forms at 3.4 cm behind the cylinder
Radius of the cylinder was 5.6cm. (Circumference of 35cm/2`pi)
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As explained in Index of Refraction using a Liquid and also in Class Notes #18, find the index of refraction of water.
D / R = (2 - n) / (2n - 2)
D/R = 3.4cm/5.6cm = .607 = x
n = (2x + 2) / (2x + 1)
n = (2* .607 +2)/(2* .607 +1)
n = 1.45 approx.
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Then using a ray-tracing analysis, as describe in Class Notes, answer the following:
1. If a ray of light parallel to the central ray strikes the cylinder at a distance equal to 1/4 of the cylinder's radius then what is its angle of incidence on the cylinder?
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I've watched the clips from Class Notes #17 and #18 but I can't really piece together the fundamental principal for figuring out the angle of incidence. I'm sure it involves a tangent but beyond that I'm stumped. Circular geometry is not intuitive for me beyond area, radius, circumference, but if I get a formula I can usually figure out the principals by back-learning.
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Thanks.
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You're right about a tangent being the key.
A radial line from the center of the cylinder to the point of incidence will be perpendicular to the surface at that point, so you need to find the angle between the incident ray and the radial line.
There is a ray through the center of the cylinder (called the central ray) which is parallel to the incident ray. The radial line is a transversal of the two parallel rays. The angle made by that ray and the radial line and the angle between the incident ray and the radial line are therefore equal.
So you need only solve the right triangle whose hypotenuse is the segment of the radian line from the center to the point of incidence, one of whose legs lies along the central ray. (Sketch the triangle by sketching the radial line from center to point of incidence, then from that point at a perpendicular to the central ray, then from the point of intersection with the central ray back to the center).
The length of the hypotenuse is the radius, the leg from the point of intersection to the central ray is 1/4 the radius. By the Pythagorean Theorem the remaining leg is sqrt( r^2 - (1/4 r)^2 ) = sqrt(15/16) * r = sqrt(15) / 4 * r. The angle between the radial line and the central ray is therefore the arcTangent of (1/4 r) / (sqrt(15) / 4 * r). This angle is equal to the angle of incidence: theta_incidence = arcTan(1/4 r) / (sqrt(15) / 4 * r) = arcTan(1 / sqrt(15) ), which is about .25 radians or about 14.5 degrees.
If you want to visualize this in terms of vectors, as presented in Phy 121, the vector from the center to the point of incidence has magnitude r, with x and y components of magnitudes 1/4 r and sqrt(15) / 4 * r. Using theta = arcTan ( y comp / x com) you get the result.
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