#$&* course Phy 202 3/6 9:45pm If your solution to stated problem does not match the given solution, you should self-critique per
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Given Solution: `aThe beat frequency is the difference in the frequencies, in this case 277 Hz - 262 Hz = 15 Hz. One ocatave reduces frequency by half, so two octaves lower would give frequencies 1/4 as great. The difference in the frequencies would therefore also be 1/4 as great, resulting in a beat frequency of 1/4 * 15 Hz = 3.75 Hz. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: What frequency is received by a person watching an oncoming ambulance moving at 110 km/h and emitting a steady 800-Hz sound from its siren? The speed of sound on this day is 345 m/s. What frequency does she receive after the ambulance has passed? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: v = 110km/h = 110,000m/3600s = 30.6m/s 800Hz/(1-30.6/345) = 878Hz After: 800Hz / (1+30.6/345) = 735Hz confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The general formula for the Doppler shift is f ' = (c + v_r) / (c - v_s) * f where v_r is the velocity of the receiver, v_s the velocity of the source, and c the speed of sound. f is the frequency of the source, f ' the frequency observed by the receiver. v_r is considered positive if the receiver is moving in the direction of the source, and v_s is considered positive if the source is moving in the direction of the receiver. In this case, the receiver is presumably stationary, while the source (the ambulance) is moving at 110 km/h = 110 * 1000 m / (3600 s) = 30 m/s. The received frequency is thus f ' = (345 m/s + 0) / (345 m/s - 30 m/s) * 800 Hz = 876 Hz. After the ambulance has passed, it is no longer approaching the source and its velocity is negative. In this case we have f ' = (345 m/s + 0) / (345 m/s - (- 30 m/s)) * 800 Hz = 736 Hz. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: 36. Two eagles fly directly toward one another, the first at 15.0 m/s and the second at 20.0 m/s (both relative to the air). Both screech, the first one emitting a frequency of 3200 Hz and the second one emitting a frequency of 3800 Hz. What frequencies do they receive if the speed of sound is 330 m/s? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Bird 1 f'= 3800Hz*(330m/s+20.0m/s)/(330m/s-15.0m/s) = 3556Hz Bird 2 f' = 3800Hz*(330+15)/(330-20) = 4229Hz confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The general formula for the Doppler shift is f ' = (c + v_r) / (c - v_s) * f where v_r is the velocity of the receiver, v_s the velocity of the source, and c the speed of sound. f is the frequency of the source, f ' the frequency observed by the receiver. v_r is considered positive if the receiver is moving in the direction of the source, and v_s is considered positive if the source is moving in the direction of the receiver. Let's assume that the first eagle is the source and the second the receiver. In this case, the receiver is approaching the source at 20 m/s, while the source is approaching the receiver at 15 m/s, and the source is emitting frequency 3200 Hz. The received frequency is thus f ' = (330 m/s + 20 m/s) / (330 m/s - 15 m/s) * 3200 Hz = 3556 Hz, approx. In the second case the first eagle is the receiver, the second the source. The source is approaching the receiver at 20 m/s, the receiver approaching the source at 15 m/s, so f ' = (330 m/s + 15 m/s) / (330 m/s - 20 m/s) * 3200 Hz = 3561 Hz, approx. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qquery gen phy problem 12.46 speakers 1.8 meters apart, listener three meters from one and 3.5 m from the other **** gen phy what is the lowest frequency that will permit destructive interference at the location of the listener? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 1 1/2 'lamda = 3.0m lamda = 6.0m 'd'lamda = 7.0m - 6.0m = 1.0m f1 = 343m/s / 1m = 343Hz confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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: For 2 sounds to be destructive they must differ by 1/2 wavelength so that the peak of one meets with the valley of the other. 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: #*&!