course Phy 232 6/18 4pmThis is the last assignment that I will be turning in before I go to Europe. I will be in Denmark July 3rd and will begin working on the class again. Dr. Goldstein is the American professor from Virginia Tech who will be the grader for the class that I am taking in Denmark. He has agreed to proctor tests while I am in Denmark. I will submit a form with his contact information for proctoring. 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): Okay. Self-critique Rating:3 ********************************************* 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: In order for there to be destructive interference, the difference between the two speakers must be an odd multiple of half of the wave length. For example (2n-1)*wavelength/2 for all n’s from 1 to infinity. Since it is the lowest frequency, it has to be the longest possible wavelength, so n = 1. The listener is 0.5 meters apart. So the wavelength is 1 meter long. We know that the speed of sound is 343 m/s, so the frequency is 343 Hz.
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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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Okay. 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: The path difference must be an integer of half of a wavelength.
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Given Solution: `a** The path difference has to be and integer number of wavelengths plus a half wavelength. ** CRAB NEBULA PROBLEM? This Query will exit. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Okay. Self-critique Rating:3 "