course PHY 202 WCČ^lޛ׀assignment #025
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15:59:15 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 --> Beat Frequency is equal to the diffference in the frequencies of the two waves. Thus, 277-262= 15hz. The beat frequency is 15 hz. If the two frequencies are two octaves lower, than they are reduced by a factor of four, so 277/4=69.25, 262/4=65.50. If you subtract 69.25-65.50=3.75hz.
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15:59:26 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 --> ok
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16:13:18 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 --> To be in destructive interference, the two waves must be completely out of phase by one half(.5) wavelength. Given the formula for frequency=velocity/wavelength. Velocity at room temp is 343 m/s. So 343/.05 gives us 6,860 hz.
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16:15:59 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 --> ok, I first needed to calculate the path difference which was 2(3.5m-3.0m)=1m. If I skip this important step, I will yield a much different frequency. I didn't do part B of of the question here, but I did it in homework and understand.
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16:29:26 **** gen phy why is there no highest frequency that will permit destructive interference?
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RESPONSE --> If two waves are traveling in opposite directions, than the frequencies of those waves are determined by hte resonant frequencies of the medium in which they pass through. The cord will have an infiinite number of resonant frequencies, but they will all be a number which is a multiple of the lowest resonant frequency. That's why there is no highest frequency that will permit destructive interference.
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16:29:59 ** You can get any number of half-wavelengths into that .5 meter path difference. **
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RESPONSE --> Please explain further.
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16:31:21 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 amplitudes of the two waves must be equal in order for destructive interference to occur.
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16:32:11 ** The path difference has to be and integer number of wavelengths plus a half wavelength. **
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RESPONSE --> ok
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16:33:23 CRAB NEBULA PROBLEM?
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RESPONSE --> I'm not sure what this question is asking.
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