#$&* course Phy 122 3/18 3 016. `Query 14*********************************************
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Given Solution: `aThe beat frequency is the difference in the frequencies, in this case 277 Hz - 262 Hz = 15 Hz. One ocatave reduces frequency by half, so two octaves lower would give frequencies 1/4 as great. The difference in the frequencies would therefore also be 1/4 as great, resulting in a beat frequency of 1/4 * 15 Hz = 3.75 Hz. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: ok ********************************************* Question: `qquery gen phy problem 12.46 speakers 1.8 meters apart, listener three meters from one and 3.5 m from the other **** gen phy what is the lowest frequency that will permit destructive interference at the location of the listener? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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? ** 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* 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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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? This Query will exit. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!