course Phy 202 assignment #014ñCèËêBy×ßïçb¾„˜‚´zþ˜ž”ª
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21:16:21 Query 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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RESPONSE --> hte beat frequancy is f2-f1 277Hz-262Hz=15Hz 2 octaves lower 277Hz/4=69.25Hz 262Hz04=65.5Hz 69.25Hz--65.5Hz=3.75Hz
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21:16:32 The 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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RESPONSE --> I got the same answer
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21:34:46 Query 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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RESPONSE --> In order for it to be destructive interference, one has to be half a wavelength behind or infront of the other. The differaance in the two is .5m. THe wavelength therefore has to be 1m. By putting itinto the equation, you get: lambda=v/f or 1=343m/s/ f f=343Hz
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21:36:12 STUDENT 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. ****
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RESPONSE --> i did not determine the next 2 frequancies but i know how they are found and have gone back and wrot it in my notes.
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22:12:33 **** gen phy why is there no highest frequency that will permit destructive interference?
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RESPONSE --> Cause there is no possibilty to escape the sound.
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22:19:45 ** You can get any number of half-wavelengths into that .5 meter path difference. **
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RESPONSE --> I wasn't real sure how to answer the question but i DO understand that.
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22:22:57 gen 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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RESPONSE --> The waves have to be off. It happens when the distance from one point is differant by half a wavelength from the otehr.
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22:23:12 ** The path difference has to be and integer number of wavelengths plus a half wavelength. **
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RESPONSE --> Basically what I said.
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22:23:35 CRAB NEBULA PROBLEM?
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RESPONSE --> ??
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22:23:44 This Query will exit.
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RESPONSE -->
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