#$&*
course Phy 202
Saturday, July 14, 2012, at 6:55 PM
If your solution to stated problem does not match the given solution, you should self-critique per instructions at http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm.Your solution, attempt at solution. If you are unable to attempt a solution, give a phrase-by-phrase interpretation of the problem along with a statement of what you do or do not understand about it. This response should be given, based on the work you did in completing the assignment, before you look at the given solution.
012. `Query 10 ??? I don't see Queries 10 or 11 on the assignment list, but it's numbered 12 in the online file name, so I'm going with that. ???
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Question: `q**** Query introductory set six, problems 11-13 **** given the length of a string how do we determine the wavelengths of the first few harmonics?
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Your Solution:
- SOLUTION: Wavelengths of the first few harmonics = 2(integer divisions of the string length); frequencies of ' ' ' = 1 / wavelengths of ' ' ' = 1 / (integer divisions of string) = integer multiples of fundamental:
- - 0) Fundamental's wavelength = 2(length of string): When string is pulled and released, ([nut/finger] to other end of string) = equilibrium point to next equilibrium point, which is only half of a wave rather than a full wave; therefore, wavelength = 2(string length).
- - 1) First harmonic's wavelength = 2(1/2 of fundamental's) = length of string; frequency is doubled, i.e., up an octave from fundamental
- - 2) Second harmonic's wavelength = 2(1/3 of fundamental's) = 2/3 (length of string) = 2(2/3 of the 1/2 harmonic); frequency: octave + perfect fifth up from fundamental or perfect fifth up from first harmonic
- - 3) Third harmonic's wavelength = 2(1/4 of fundamental's) = 1/2 (length of string) = 2(3/4 of the [1/3 = 2/3 of the 1/2] harmonic); frequency = 2 octaves (also = octave + perfect fifth + perfect fourth) up from fundamental or perfect fourth up from second harmonic
- - 4) Fourth harmonic's wavelength = 2(1/5 of fundamental's) = 2/5 (length of string) = 2(4/5 of the [1/4 = 3/4 of the (1/3 = 2/3 of the 1/2)] harmonic); frequency = 2 octaves + major third (also = octave + perfect fifth + perfect fourth + major third) up from fundamental or major third up from third harmonic
- - 5) Fifth harmonic's wavelength = 2(1/6 of fundamental's) = 1/3 (length of string) = 2(5/6 of the [1/5 = 4/5 of the (1/4 = 3/4 of the [1/3 = 2/3 of the 1/2])] harmonic); frequency = 2 octaves + perfect fifth (also = octave + perfect fifth + perfect fourth + major third + minor third) up from fundamental or minor third up from fourth harmonic
- - 6) Sixth harmonic's wavelength = 2(1/7 of fundamental's) = 2/7 (length of string), etc., etc., but when expressing in terms of intervals, note that this ""septimal"" interval will not be equal to adding on another minor third as this interval would be approximated in twelve-tone equal temperament (powers of twelfth root of two). Instead, the seventh that is created will be slightly flatter and less dissonant than a minor seventh as usually defined; for details, look up ""harmonic seventh"" or ""septimal minor seventh.""
confidence rating #$&*:5 (the notion of the string's length being a half-wavelength is intellectually understandable but psychologically counterintuitive; also, I hope that I got the multiplication by two of all half-wavelengths right)
Psychologically it's not difficult to see that the second harmonic consists of a complete cycle contained within the string. The ends are nodes, as is the middle.
Since the string can also form a standing wave with nodes at the ends and an antinode in the middle, it's easy to see that this fundamental mode has double the wavelength of the second harmonic, so has wavelength equal to double the length of the string.************
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Given Solution: `q** As wavelength decreases you can fit more half-waves onto the string. You can fit one half-wave, or 2 half-waves, or 3, etc..
So you get
1 half-wavelength = string length, or wavelength = 2 * string length; using `lambda to stand for wavelength and L for string length this would be
1 * 1/2 `lambda = L so `lambda = 2 L.
For 2 wavelengths fit into the string you get
2 * 1/2 `lambda = L so `lambda = L.
For 3 wavelengths you get
3 * 1/2 `lambda = L so `lambda = 2/3 L; etc.
Your wavelengths are therefore 2L, L, 2/3 L, 1/2 L, etc..
FOR A STRING FREE AT ONE END:
The wavelengths of the first few harmonics are found by the node - antinode distance between the ends. The first node corresponds to 1/4 wavelength. The second harmonic is from node to antinode to node to antinode, or 4/3. the third and fourth harmonics would therefore be 5/4 and 7/4 respectively. **
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Self-critique (if necessary): OK, except that my numbering terminology was off: I thought that the fundamental was the ""0th""/""zero[e]th"" harmonic rather than the first, with the effect that my numberings of ""harmonics"" corresponded to the conventional numberings of ""overtones"" (per convention, #:Overtone = #:Harmonic - 1).
There's no zeroth harmonic, and the second harmonic is the first overtone. There's no zeroth overtone.
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Self-critique Rating: OK
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Question: `q**** Given the wavelengths of the first few harmonics and the velocity of a wave disturbance in the string, how do we determine the frequencies of the first few harmonics?
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Your solution:
- SOLUTION: From the wavelength of a harmonic and the knowledge of which harmonic it is, or from a set of harmonics whose ratios we can calculate and whose numbers and shared fundamental we can thereby determine, we can calculate the string length as described in the Solution above. Wave velocity (i.e., velocity of transverse wave going up and down string, not of longitudinal-compression ""sound wave"" generated when transverse wave strikes air) = wave frequency * wavelength, so frequency = (speed of wave / wavelength) and therefore = (speed of wave / [2(string length) / (#:Harmonic or [#:Overtone - 1])]), i.e., ([#:Harmonic or (#:Overtone - 1)][wave velocity]) / (2[string length]). See above Response for more details re: computing frequencies; note also that the velocity of the transverse wave is determined by other factors (sqrt[tension/(xsec. area * density)], if I recall correctly). CHECK of this caveat: Book gives velocity formula as sqrt(tension/[mass/length]), but that should be the same; density * xsec. area = (mass / volume) * xsec. area = (mass / length).
confidence rating #$&*:at least now that I have the nomenclature straight
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Given Solution:
`a** The frequency is the number of crests passing per unit of time.
We can imagine a 1-second chunk of the wave divided into segments each equal to the wavelength. The number of peaks is equal to the length of the entire chunk divided by the length of a 1-wavelength segment. This is the number of peaks passing per second.
So frequency is equal to the wave velocity divided by the wavelength. **
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Self-critique (if necessary): OK; the ""overkill"" came from my expressing this ratio in terms of the properties of the harmonics and their shared fundamental and generating string.
That wasn't really overkill. The ratios of the harmonics are the basis of music.
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Self-critique Rating: OK
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Question: `q **** Given the tension and mass density of a string how do we determine the velocity of the wave in the string?
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Your solution:
- SOLUTION: In previous Solution, see ""note also that the velocity of the transverse wave"" and following. Taking ""mass density"" to mean ""mass per unit length,"" i.e., ""density [as normally construed] * cross-sectional area,"" velocity = sqrt(tension / [mass density]).
confidence rating #$&*:
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Given Solution:
`a** We divide tension by mass per unit length:
v = sqrt ( tension / (mass/length) ). **
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Self-critique (if necessary): OK
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Self-critique Rating: OK
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Question: `q**** gen phy explain in your own words the meaning of the principal of superposition
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Your solution:
- SOLUTION: When two waves are superimposed on each other, they form a new wave whose amplitude at any given point is equal to the sum of their amplitudes (including those amplitudes' respective positive/negative signs) at that point.
- ALTERNATIVE SOLUTION*: Yakka foob mog. Grug pubbawup zink wattoom gazork. Chumble spuzz. (I love loopholes.)
- - * Source: Calvin and Hobbes, 1995 January 9: See, e.g., http://www.cooperativeindividualism.org/calvin-on-scientific-law.gif (http://www.gocomics.com/calvinandhobbes/1995/01/09)
confidence rating #$&*:5 (I think that the above covers it, but I'm not 100% sure.)
Both my 43-year-old computer scientist / former semiprofessional bouncer of a son-in-law and, independently, my 11-year-old budding engineer of a grandson have both been seriously warped by Calvin and Hobbs, the former before I knew him and the latter with my active encouragement.
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Given Solution:
`a** the principle of superposition tells us that when two different waveforms meet, or are present in a medium, the displacements of the two waveforms are added at each point to create the waveform that will be seen. **
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Self-critique (if necessary): OK/3 (possibly not ""OK"" because depending on how strictly one defines the term ""amplitude,"" e.g., as signifying only absval[maximum displacement from equilibrium position], my use of the term ""amplitude at any given point"" in place of ""displacement"" could be incorrect.)
"displacement" or "displacement from equilibrium" is the appropriate term, but your meaning was clear.
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Self-critique Rating: OK
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Question: `q **** gen phy what does it mean to say that the angle of reflection is equal to the angle of incidence?
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Your solution: When matter or energy traveling in a straight line bounces off a surface, the angle between its path of travel to the surface and the line coplanar with both the surface and that path will have the same magnitude as will the angle between its path away from that surface and the line coplanar with both that surface and its path away.
confidence rating #$&*:5 (not entirely sure about matter; I've specified ""traveling in a straight line"" to account for that, and it seems like it would have to be that way [e.g., an air hockey puck bouncing off one of the 'rink' walls], but there may be some kind of exception of which I'm unaware.)
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Given Solution:
`a** angle of incidence with a surface is the angle with the perpendicular to that surface; when a ray comes in at a given angle of incidence it reflects at an equal angle on the other side of that perpendicular **
Your solution ??? assumed meaning: ""Self-critique (if necessary)"" ???: OK (I was defining it with respect to the surface, such that the angles would have measures of [90 degrees - the angle from the perpendicular mentioned in the Given Solution], but their magnitudes would still be equal.)
Confidence Rating ??? assumed unintentionally copied/pasted here ???: [N/A: nothing to correspond to]
The result for a wave is actually derived using Huygens' Principle, which states that every point on a traveling wave serves as a source for the subsequent wave.
The wave in this case would be a plane wave, advancing in parallel planes. The direction perpendicular to these planes would be the direction of propagation. Applying Huygens' Principle it's fairly easy to see how this results in equal angles of incidence and reflection.
Refraction, which results from a change in the velocity of propagation at the interface between two media, is also modeled in this manner.
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Self-critique Rating: OK"
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#*&!
&#Your work looks good. See my notes. Let me know if you have any questions. &#