ph2_query_15

#$&*

course Phy 202

Sunday, July 15, 2012, between 3:50 PM and 3:55 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.

015. `Query 13 ??? Again, I'm going with the filename and ""015"" number. ???

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Question: `qquery experiment to be viewed and read but not performed: transverse and longitudinal waves in aluminum rod

`what is the evidence that the higher-pitched waves are longitudinal while the lower-pitched waves are transverse?

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Your Solution:

- SOLUTION: ???? I can't seem to find the FREQ program, but I happen to have absolute pitch (the colloquial term ""perfect pitch"" is a misnomer because no one's is exact), so I'm able to ""emulate"" the FREQ program to some degree: After I identify the note and whether it's average, ""high,"" or ""low"" for the range of frequencies corresponding to that note in modern times, I can do the math based on A-above-middle-C = 440 Hz and multiplying or dividing by ((the twelfth root of 2)^(number of half-steps / semitones by which the note differs from an A440)). The higher-pitched waves are an A-flat (I hear both what appears to be the fundamental, which appears to be between the first above middle C [so roughly 415 Hz, or about where most Baroque musicians tuned an A-natural], and what appears to be the second harmonic / first overtone, which would be 830 Hz). The lower pitch is the D-natural a tritone / augmented fourth / diminished fifth down from that (i.e., a whole step above middle C), just above 293.66 Hz in equal temperament, with the overtones coming through a lot less for it. ???? We can determine that the higher-pitched waves are longitudinal to begin with and that the lower-pitched waves are generated from transverse waves in the rod by observing two phenomena:

@&

I expect that the longitudinal waves are are two octaves higher, at about 1660 Hz. The transverse waves might well be at 293 Hz.

Naturally since the rod isn't tuned the pitches won't exactly match those of the standard scale.

*@

- - a) 1) i) When the rod is pointed at the camera/microphone, the volume (amplitude) of the A-flat waves ""perceived"" by the microphone increases. This is because the sources of the spherical waves from each end of the rod are collinear with us, with the effect that the compression oscillations that we perceive will be more aligned with each other and therefore generate more constructive interference, increasing the amplitude, as opposed to reaching the microphone at ""angles"" (not really angles because the lines are curved) with each other, which would generate a good deal of destructive interference, as happens when the rod is turned. This lets us know that the movement up and down the rod, as opposed to side to side across it, generates the higher-frequency sound. It also helps that pointing the rod toward the observer increases the total sound intensity by 2k / sqrt([distance from rod center]^2 + [.5(rod length)]^2) to (k/[(distance from rod center - .5[rod length])^2] + k/[distance from rod center + .5(rod length])2]): The observed intensity gains more from one end's moving toward the viewer than it loses from the other end's moving farther away.

- - -- -- ii) When the rod is turned perpendicular to the camera's line of sight, such that the full length rather than the end of the rod is facing the observer, the D-natural wave ""comes through"" a lot more; although some of this perception results from the reduction in the A-flat wave and our ability to hear the D-natural better, much of it also results from the fact that the closer to perpendicular vs. the wavepath the observer's line of perception gets, the more in line the observer is with the longitudinal compression waves that are generated. This suggests that the wave that generates the longitudinal compression waves is oscillating back and forth toward and away from the observer, which means that it is transverse with respect to the rod.

- - b) 1) From the fact that dampening the end of the rod quiets the high frequencies much more than the low ones, we can tell that it is the back-and-forth motion of the particles up and down the rod that generates the high frequencies; this is essentially the ""eliminative""/""negative"" version of the ""additive""/""positive"" proof seen above. The fact that the lower frequencies aren't quieted as much also suggests that there's a pressure node at that end to begin with, further confirming that the lower-frequency waves are transverse.

- - -- 2) From the fact that dampening the side of the rod (at least at a non-integer-division point, such that overtones/""harmonics"" aren't disproportionally allowed to persist) quiets the lower frequencies a lot more than it does the higher ones, we can tell that the waves being dampened are the ones that generate the lower-pitched sound waves; specifically, because the transverse oscillation is what is being dampened, we can tell that it is generating the lower-frequency sound waves. From the lesser dampening of the higher-frequency sound waves we can tell that they are being generated by oscillation that is at an angle with, and in the extreme/paradigm case perpendicular to, the oscillations that we are dampening.

confidence rating #$&*: to 2.75 (This all makes sense in my head, but I'm not entirely certain that the causal process that I've proposed is in fact the one that is responsible for and explains the observed phenomena.

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Given Solution: `qSTUDENT RESPONSE: The logitudinal waves had a higher velocity.

That doesn't provide evidence that the high-pitched wave was longitudinal, since we didn't directly measure the velocity of those waves. The higher-pitches waves were damped out much more rapidly by touching the very end of the rod, along its central axis, than by touching the rod at the end but on the side.

The frequency with which pulses arrive at the ear determines the pitch.

The amplitude of the wave affects its intensity, or energy per unit area. For a given pitch the energy falling per unit area is proportional to the square of the amplitude.

Intensity is also proportional to the square of the frequency. **

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Self-critique (if necessary): OK

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Self-critique Rating: OK

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Question: `qquery General College Physics and Principles of Physics 12.08: Compare the intensity of sound at 120 dB with that of a whisper at 20 dB.

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Your solution:

- SOLUTION: Ratio of sound intensities = 10^([difference in dB values / 10]) : 1; 120 dB - 20 dB = 100 dB or 10 B[el] = (10^[100/10]) : 1 or (10^10) : 1 ratio, i.e., 10,000,000,000 (10 billion) times as loud.

confidence rating #$&*:

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Given Solution:

`aThe intensity at 120 dB is found by solving the equation dB = 10 log(I / I_threshold) for I.

We get

log(I / I_threshold) = dB / 10, so that

I / I_threshold = 10^(120 / 10) = 12and

I = I_threshold * 10^12.

Since I_threshold = 10^-12 watts / m^2, we have for dB = 120:

I = 10^-12 watts / m^2 * 10^12 = 1 watt / m^2.

The same process tells us that for dB = 20 watts, I = I_threshold * 10^(20 / 10) = 10^-12 watts / m^2 * 10^2 = 10^-10 watts / m^2.

Dividing 1 watt / m^2 by 10^-10 watts / m^2, we find that the 120 dB sound is 10^10 times as intense, or 10 billion times as intense.

A more elegant solution uses the fact that dB_1 - dB_2 = 10 log(I_1 / I_threshold) - ( 10 log(I_2 / I_threshold) )

= 10 log(I_1 / I_threshold) - ( 10 log(I_2 / I_threshold) )

= 10 {log(I_1) - log( I_threshold) - [ ( log(I_2) - log(I_threshold) ]}

= 10 { log(I_1) - log(I_2)}

= 10 log(I_1 / I_2).

So we have

120 - 20 = 100 = 10 log(I_1 / I_2) and

log(I_1 / I_2) = 100 / 10 = 10 so that

I_1 / I_2 = 10^10.

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Self-critique (if necessary): OK (I shortcut some of the math; for my ability to do it, see ph2_query_14 response 2.)

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Self-critique Rating: OK

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Question: `qquery gen phy 12.30 length of open pipe, 262 Hz at 21 C? **** gen phy What is the length of the pipe?

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Your solution:

- SOLUTION: Velocity = frequency * wavelength; re: longitudinal compression wave, speed of sound in air at this temperature = 343 m s^-1 [i.e., what it is at 20 degC] + (.60 Hz/degC * 1 degC) = 343.6 Hz = frequency * wavelength = 262 cycles second^-1 * wavelength; 343.6 m / 262 cycles = approx. 1.311 meters per cycle. Open pipe length = wavelength / 2 = approx. 0.656 m = approx. 2.15 feet (pipe-organ registers/""ranks"" are measured in feet).

- - CHECK: This corresponds roughly to the ""2-foot open / 1-foot closed"" standard used for middle C [261.63 Hz at ET:12 and A440], with most of the difference attributable to the fact that the open pipe required to produce this pitch at a standard temperature of 0 degC would be shorter (2.07 feet) and the rest attributable to a) the fact that this pitch is slightly above an A440 ET-12 middle C and b) historical differences in tuning standards (late 19th-century European orchestras used As as high as 446, which would make the required pipe length shorter).

@&

Also the length of the pipe isn't exactly equal to the length of the vibrating column. There is some displacement at the open end.

However this would make the effective length of the pipe greater, not less than the length of the vibrating air column. I think.

Ironically I had my Junior Lab under Art Benade, arguably the best acoustical physicist of his time. Neat guy but we didn't do much acoustics that semester.

Also should have paid more attention to the pipe organ I learned to play, but I didn't know much physics at age 12.

*@

confidence rating #$&*:5 intellectually; 2.25 to 2.5 psychologically

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Given Solution:

`aGOOD STUDENT SOLUTION

First we must determine the velocity of the sound waves given the air temperature. We do this using this formula

v = (331 + 0.60 * Temp.) m/s

So v = (331 + 0.60 * 21) m/s

v = 343.6 m/s

The wavelength of the sound is

wavelength = v / f = 343.6 m/s / (262 Hz) = 1.33 meters, approx..

The pipe is open, so it has antinodes at both ends.

* The fundamental frequency occurs when there is a single node between these antinodes. So the length of the pipe corresponds to two node-antinode distances.

* Between a node and an adjacent antinode the distance is 1/4 wavelength. In this case this distance is 1/4 * 1.33 meters = .33 meters, approx..

* The two node-antinode distances between the ends of the pipe therefore correspond to a distance of 2 * .33 meters = .66 meters.

We conclude that the pipe is .64 meters long.

Had the pipe been closed at one end then there would be a node and one end and an antinode at the other and the wavelength of the fundamental would have therefore been 4 times the length; the length of the pipe would then have been 1.33 m / 4 = .33 m.

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Self-critique (if necessary): OK (I got a bit carried away, but it was all in the name of trying to figure out why my answer didn't ""check out"" as corresponding exactly to 2 feet.)

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Self-critique Rating: OK

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Question: `q**** Univ phy 16.79 11th edition 16.72 (10th edition 21.32): Crab nebula 1054 A.D.;, H gas, 4.568 * 10^14 Hz in lab, 4.586 from Crab streamers coming toward Earth. Velocity? Assuming const vel diameter? Ang diameter 5 arc minutes; how far is it?

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Your solution: [Omitted as marked for only University Physics students]

confidence rating #$&*:A]

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Given Solution:

`a** Since fR = fS ( 1 - v/c) we have v = (fR / fS - 1) * c = 3 * 10^8 m/s * ( 4.586 * 10^14 Hz / (4.568 * 10^14 Hz) - 1) = 1.182 * 10^6 m/s, approx.

In the 949 years since the explosion the radius of the nebula would therefore be about 949 years * 365 days / year * 24 hours / day * 3600 seconds / hour * 1.182 * 10^6 m/s = 3.5 * 10^16 meters, the diameter about 7 * 10^16 meters.

5 minutes of arc is 5/60 degrees or 5/60 * pi/180 radians = 1.4 * 10^-3 radians. The diameter is equal to the product of the distance and this angle so the distance is

distance = diameter / angle = 7 * 10^16 m / (1.4 * 10^-3) = 2.4 * 10^19 m.

Dividing by the distance light travels in a year we get the distance in light years, about 6500 light years.

CHECK AGAINST INSTRUCTOR SOLUTION: ** There are about 10^5 seconds in a day, about 3 * 10^7 seconds in a year and about 3 * 10^10 seconds in 1000 years. It's been about 1000 years. So those streamers have had time to move about 1.177 * 10^6 m/s * 3 * 10^10 sec = 3 * 10^16 meters.

That would be the distance of the closest streamers from the center of the nebula. The other side of the nebula would be an equal distance on the other side of the center. So the diameter would be about 6 * 10^16 meters.

A light year is about 300,000 km/sec * 3 * 10^7 sec/year = 9 * 10^12 km = 9 * 10^15 meters. So the nebula is about 3 * 10^16 meters / (9 * 10^15 m / light yr) = 3 light years in diameter, approx.

5 seconds of arc is 5/60 of a degree or 5 / (60 * 360) = 1 / 4300 of the circumference of a full circle, approx.

If 1/4300 of the circumference is 6 * 10^16 meters then the circumference is about 4300 times this distance or about 2.6 * 10^20 meters.

The circumference is 1 / (2 pi) times the radius. We're at the center of this circle since it is from here than the angular diameter is observed, so the distance is about 1 / (2 pi) * 2.6 * 10^20 meters = 4 * 10^19 meters.

This is about 4 * 10^19 meters / (9 * 10^15 meters / light year) = 4400 light years distant.

Check my arithmetic. **

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Self-critique (if necessary): [N/A]

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Self-critique Rating: [N/A]

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Question: `q **** query univ phy 16.66 (21.26 10th edition). 200 mHz refl from fetal heart wall moving toward sound; refl sound mixed with transmitted sound, 85 beats / sec. Speed of sound 1500 m/s.

What is the speed of the fetal heart at the instant the measurement is made?

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Your solution: [Omitted as marked for only University Physics students]

confidence rating #$&*:A]

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Given Solution:

`a. ** 200 MHz is 200 * 10^6 Hz = 2 * 10^8 Hz or 200,000,000 Hz.

The frequency of the wave reflected from the heart will be greater, according to the Doppler shift.

The number of beats is equal to the difference in the frequencies of the two sounds. So the frequency of the reflected sound is 200,000,085 Hz.

The frequency of the sound as experienced by the heart (which is in effect a moving 'listener') is fL = (1 + vL / v) * fs = (1 + vHeart / v) * 2.00 MHz, where v is 1500 m/s.

This sound is then 'bounced back', with the heart now in the role of the source emitting sounds at frequency fs = (1 + vHeart / v) * 2.00 MHz, the 'old' fL. The 'new' fL is

fL = v / (v - vs) * fs = v / (v - vHeart) * (1 + vHeart / v) * 2.00 MHz.

This fL is the 200,000,085 Hz frequency. So we have

200,000,085 Hz = 1500 m/s / (v - vHeart) * (1 + vHeart / v) * 2.00 MHz and

v / (v - vHeart) * (1 + vHeart / v) = 200,000,085 Hz / (200,000,000 Hz) = 1.000000475.

A slight rearrangement gives us

(v + vHeart) / (v - vHeart) = 1.000000475 so that

v + vHeart = 1.000000475 v - 1.000000475 vHeart and

2.000000475 vHeart = .000000475 v, with solution

vHeart = .000000475 v / (2.000000475), very close to

vHeart = .000000475 v / 2 = .000000475 * 1500 m/s / 2 = .00032 m/s,

about .3 millimeters / sec. **

STUDENT COMMENT

My final answer was twice the answer in the given solution. I thought that I used the Doppler effect equation correctly; however, I may have solved for the unknown incorrectly.

INSTRUCTOR RESPONSE

The equations tell you the frequency that would be perceived by a hypothetical detector on the heart.

Suppose that each time the detector records a 'peak', it sends out a pulse. The pulses are sent out at the frequency of the detected wave. The source of these pulses is the detector, which is moving toward the 'listener', and as a result they are detected at an even higher frequency.

Thus the doubled number of beats.

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Self-critique (if necessary): [N/A]

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Self-critique Rating: [N/A]"

@&

Good. Check my notes.

*@