ph2_query_16

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course Phy 202

Sunday, July 15, 2012, between 11:20 PM and 11:25 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.

016. `Query 14 ??? labeled ""query 16"" per previous pattern ???

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The pattern occurred as a result of shifting assignments around. In any case, you see the pattern.

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Question: `qquery Principles of Physics and General College Physics 12.40: Beat frequency at 262 and 277 Hz; beat frequency two octaves lower.

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Your solution:

- SOLUTION:

- - a) Neither 266 nor 277 is an integer multiple or integer divisor of the other, and they do not have any common factors -- in fact, 277 is prime -- so their nodes will not concide any more frequently than once per second, and the waves will ""beat"" against each other (277 - 262) = 15 times per second. ???? How does one predict the ""spacing""/timing of the beats within a one-second interval? For example, are they evenly spaced at (1/15) second apart, do they occur in a pattern that features three unevenly-arranged beats but repeats five times per seconds, or ... ? ????

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The relative pattern of peaks and valleys shifts at a constant rate, shifting in and out of phase 15 times every second, resulting in 15 equally-spaced beats every second.

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Interesting idea: Two high-frequency waves could easily produce more than 50 beats per second, so that the beats themselves create a lower-pitched disturbance which we might perceive as sort of a base note. It might or might not be audible in the presence of the original sounds (and might or might not harmonize with the higher frequencies, though unless designed to do so very likely wouldn't...might be an interesting probability to estimate).

If the high-frequency waves were ultrasonic, though, the only thing we could perceive would be the beats.

I imagine this phenomenon has been exploited somewhere, somehow.

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- - b) When the notes are two octaves lower, their frequencies are each 1/4 what they were previously, so their mutual beat frequency is (277 Hz / 4) - (262 Hz / 4) = ([277 Hz - 262 Hz] / 4) = 15 Hz / 4 = 3.75 times per second.

confidence rating #$&*:

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Given Solution:

`aThe beat frequency is the difference in the frequencies, in this case 277 Hz - 262 Hz = 15 Hz.

One ocatave reduces frequency by half, so two octaves lower would give frequencies 1/4 as great. The difference in the frequencies would therefore also be 1/4 as great, resulting in a beat frequency of 1/4 * 15 Hz = 3.75 Hz.

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Self-critique (if necessary): OK

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Self-critique Rating: OK

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Question: `qquery gen phy problem 12.46 speakers 1.8 meters apart, listener three meters from one and 3.5 m from the other **** gen phy what is the lowest frequency that will permit destructive interference at the location of the listener?

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Your solution:

- SOLUTION: .5(wavelength) = (3.5 - 3) = 0.5 m; wavelength = 1 m; when temperature = 20 degC, speed of sound = 343 m s^-1, so frequency = speed of sound / wavelength = 343 Hz (on the low side of F above middle C)

- - Next two lowest frequencies (per request in second part of program): 1) 1.5(wavelength) = 0.5 m; wavelength = (1/3) m; frequency = 3(above) = 1029 Hz. 2) 2.5(wavelength) = 0.5 m; wavelength = 0.2 m; frequency = 5(above) = 1715 Hz

confidence rating #$&*:intellectual), 2.5 (psychological)

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Given Solution:

`aSTUDENT SOLUTION: To solve this problem, I first realize that for destructive interference to occur, the path difference is an odd multiple of half of the wavelength(ex. 1/2, 3/2, 5/2). I used Fig.12-17 in the text to help visualize this problem. Part (A) of the problem asks to calculate the lowest frequency at which destructive interference will occur at the point where two loudspeakers are 2.5m apart and one person stands 3.0m from one speaker and 3.5m from the other. The text states that 'destructive interference occurs at any point whose distance from one speaker is greater than its distance from the other speaker by exactly one-half wavelength.' The path difference in the problem is fixed, therefore the lowest frequency at which destructive interference will occur is directly related to the longest wavelength. To calculate the lowest frequency, I first have to calculate the longest wavelength using the equation

'dL ='lambda/2, where `dL is the path difference.

'lambda=2*'dL

=2(3.5m-3.0m)=1m

Now I can calculate the frequency using

f=v/'lambda. The velocity is 343m/s which is the speed of sound.

f=343m/s/1m=343 Hz.

Thus, the lowest frequency at which destructive interference can occur is at 343Hz.

Keeping in mind that destructive interference occurs if the distance equals an odd multiple of the wavelength, I can calculate (B) part of the problem.

To determine the next wavelength, I use the equation 'dL=3'lambda/2

wavelength=2/3(3.5m-3.0m) =0.33m

Now I calculate the next highest frequency using the equation f=v/wavelength.

f^2=343m/s/0.33m=1030Hz.

I finally calculate the next highest frequency.

'del L=5/2 'lambda

wavelength=0.20m

f^3=343m/s/0.2m=1715 Hz.

INSTRUCTOR EXPLANATION:

The listener is .5 meters further from one speaker than from the other. If this .5 meter difference results in a half-wavelength lag in the sound from the further speaker, the peaks from the first speaker will meet the troughs from the second. If a half-wavelength corresponds to .5 meters, then the wavelength must be 1 meter. The frequency of a sound with a 1-meter wavelength moving at 343 m/s will be 343 cycles/sec, or 343 Hz.

The next two wavelengths that would result in destructive interference would have 1.5 and 2.5 wavelengths corresponding to the .5 m path difference. The wavelengths would therefore be .5 m / (1.5) = .33 m and .5 m / (2.5) = .2 m, with corresponding frequencies 343 m/s / (.33 m) = 1030 Hz and 343 m/s / (.2 m) = 1720 Hz, approx. ****

**** gen phy why is there no highest frequency that will permit destructive interference?

- RESPONSE: 2(.5 m) can be an integer multiple of an infinite number of increasingly small wavelengths, and 1 / these can be an infinite number of increasingly high frequencies. Confidence = 3.

** You can get any number of half-wavelengths into that .5 meter path difference. **

STUDENT COMMENT:

After reading the solution I understand the formula I am supposed to use a bit better, but I am still kind of confused about the concept of destructive interference.

INSTRUCTOR RESPONSE:

As you change position the relative alignment of 'peaks' and 'valleys' change.

Sometimes peaks from one path arrive at the same time as peaks from the other (in which case valleys will arrive with valleys), and the interference is constructive.

Sometimes peaks from one path arrive at the same time as valleys from the other, and the interference is destructive.

When one path is a whole number of wavelengths longer than the other, peaks meet peaks and the waves reinforce.

When one path is a half a wavelength longer than the other, peaks meet valleys and the waves cancel; the same happens when one path is half a wavelength plus a whole number of wavelengths longer than the other.

STUDENT QUESTION

I got the lowest frequency fine. And I was on the right track with my reasoning.

But which way do you suggest to solve problems like this: your way, or the student’s solution.

INSTRUCTOR RESPONSE

The two solutions are completely equivalent. If you really understand it one way, you'll understand it the other.

However in the interest of time, you should pick one. Whichever way makes more sense to you, that's the way you should think of it.

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Self-critique (if necessary): OK (re: both initial question and follow-up re: why there can be no highest destructively interfering frequency)

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Self-critique Rating: OK (same)

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Question: `qgen phy what must happen in order for the sounds from the two speakers to interfere destructively, assuming that the sources are in phase?

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Your solution:

- SOLUTION: Two further conditions must exist for the speakers to interfere destructively:

- - 1) The speakers must be ""firing""/sounding at the same frequency.

- - 2) The distances between the respective speakers and the observer must differ by an odd integer multiple of half the wavelength of the sound.

confidence rating #$&*:5 (Re: 1], I worry that there may be some special cases, such as when the frequency of one source is a binary multiple or divisor of [i.e., +or- a given number of octaves from] the frequency from the other source, in which destructive interference can occur intermittently [if only instantaneously], but I haven't gone through the formal calculations/diagrams necessary to make myself feel certain whether or not that can happen.)

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That could lead, for example, to every other cycle being at least partially canceled. Might tend to lead to the peception of a sound an octave lower, which would be mixed with the original sound so that you would hear the one-octave interval.

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Given Solution:

`a** The path difference has to be and integer number of wavelengths plus a half wavelength. **

STUDENT QUESTION

The book tells me that for these two speakers to interfere destructively, the distance from one speaker has to be greater than its distance from the other speaker by one-half wavelength. Destructive interference would occur if the distance would equal 1/2, 3/2, 5/2,… wavelengths.

INSTRUCTOR RESPONSE

That is correct.

The given solution to the original problem says this as well.

The book's explanation of course gives you a third option for the most appropriate way to think of the problem.

In any case you need to understand why those path differences result in destructive interference. Once you're clear on that, a wide variety of interference problems become pretty straightforward.

CRAB NEBULA PROBLEM? ???? I recall that there was a Crab Nebula - related problem in the previous Query, but I'm not sure what this means. Are you asking us to solve a problem about the Crab Nebula? If so, which one? ????

This Query will exit. ???? I'm not sure what this means, either. Is it some kind of scripting or other coding artifact? ????

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Self-critique (if necessary): OK

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Self-critique Rating: OK"

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Very good. Check my notes.

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