OpenStudy (anonymous):

Two in-phase sources of waves are separated by a distance of 4.30 m. These sources produce identical waves that have a wavelength of 4.80 m. On the line between them, there are two places at which the same type of interference occurs. Where are the places located? I have been going crazy with this question. I don't want the answer, just to understand how to find it.

4 years ago
OpenStudy (anonymous):

this is a path length problem |dw:1366675985638:dw| find the amplitudes of the waves A1 and A2 from source S1 and S2 respectively at an arbitrary point P along the line, where P is r1 from source S1 and r2 from source S2 as shown in the drawing. A1 = A2 then at P there is constructive interference A1 = -A2 then at P there is destructive interference does this help?

4 years ago
OpenStudy (anonymous):

Um, I'm not sure we use amplitude for this problem. At least, I'm not sure how to do so with the information given. What I do know is that we are supposed to use the equations \[\left|l _{2}-l _{1} \right|=(m+.5)*\lambda\] Since this is destructive interference. I am just unsure how to utilize the equation.

4 years ago
OpenStudy (anonymous):

how do you know this is destructive interference?

4 years ago
OpenStudy (anonymous):

Well, that is another problem, I'm not too sure I understand why it's destructive, however, that is the correct answer. The answers end up being .95m from the small source and 3.35m from the small source

4 years ago
OpenStudy (anonymous):

well, the amplitude can be expressed as a function of distance 'r' as A = cos(r). so A1 = cos(r1), A2 = cos (r2) try and solve it using this info?

4 years ago
OpenStudy (anonymous):

you got Δr=0,...,nλ for constructive and Δr=nλ/2 for destructive and Δr=r2-r1 ...put the values of λ into the equations and if you have negative r its out of the line,

4 years ago
OpenStudy (anonymous):

for λ=0,4.80,9.6 and so on you have constructive, for λ=2.4,4.8 you have destructive

4 years ago
OpenStudy (anonymous):

ok, that makes sense. The difference between 3.35 and .95 ends up being 2.4. So the difference between l2 and l1 is equal to half a wavelength

4 years ago
OpenStudy (anonymous):

first time i see your equation by the way

4 years ago
OpenStudy (anonymous):

yeah, that's just a weird form our professor gave us of the Δr=nλ/2 which you wrote. I think I see the relationships here. However, when I try to apply them with different numbers, I get the wrong answer For instance, if I say the distance between the sources is 4.2 instead of 4.3, and the wavelength is 5.2 instead of 4.8. With these new numbers, I can't get the new distances where there is destructive interference.

4 years ago
OpenStudy (anonymous):

i think but im not sure that the problem given is wrong because it says two places with the same type of interference and i think that there is one constructive and one destructive ...if you put Δr=0,λ/2 you will find positive values for r2 for example,but when you put Δr=λ οr Δr=3λ/2 and so on you will get negative value for r2 that means its out of the line connecting the two sources

4 years ago
OpenStudy (anonymous):

try for example \[Δr=\left| r2-r1 \right|=4.8\] ..because r2+r1=4.3 you get r2=4.3-r1 and putting it into the previous equation you get negative value for r2

4 years ago
OpenStudy (anonymous):

That's true. However, the example problem I looked had the answers of .95 and 3.35. 3.35-.95=2.4 So, then 4.8/2 = 2.4 or lambda/2 = 2.4 = |r2−r1|

4 years ago
OpenStudy (anonymous):

ah yes that makes sense sorry

4 years ago
OpenStudy (anonymous):

you have 1 point of constructive interference at 2.15,its exactly in the middle so r1=r2=2.15 and 2 points of destructive interference at 3.35 and 0.95 because 3.35+0.95=4.3 and that only because its \[\left| r2-r1 \right|\] and that has two solutions

4 years ago
OpenStudy (anonymous):

r2=r1 for the constructive case

4 years ago
OpenStudy (anonymous):

OH!!! Wow, thank you. You got my brain working again. I just redid the problem using the equations we suggested: lamba/2=deltaR and r1 + r2 = the distance between the sources I solved for r1, input that into the 1st eq. and got r2

4 years ago
OpenStudy (anonymous):

you understood that the destructive interference has 2 solutions because it is an absolute?

4 years ago
OpenStudy (anonymous):

Yeah, I think I have that down. I understood the theory, I just was not sure how to utilize the equations and numbers.

4 years ago
OpenStudy (anonymous):

Thanks a lot for your help

4 years ago
OpenStudy (anonymous):

I had been working on the problem for about 2 hours

4 years ago
OpenStudy (anonymous):

yes its the freaking absolute ...damn maths lol

4 years ago
OpenStudy (anonymous):

lol. I just gotta make sure I keep that in mind on the final haha

4 years ago
OpenStudy (anonymous):

;p

4 years ago
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