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Sound wave interfernce on a large circle

by jwxie
Tags: circle, interfernce, sound, wave
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jwxie
#1
Feb19-11, 12:27 PM
P: 283
1. The problem statement, all variables and given/known data

Given two isotropic point sources of sound [itex]\[s_{1}\][/itex] and [itex]\[s_{2}\][/itex]. The sources emit waves in phase at wavelength 0.50m; they are separated by D = 1.75m. If we move a sound detector along a large circle centered at the midpoint between the sources, at how many points do waves arrive at the detector (a) exactly in phase, and (b) exactly out of phase?


2. Relevant equations

When the ratio of [itex]\[\frac{\Delta L}{\lambda }\][/itex] is 0, 1, 2, or integral multiple of
[itex]\lambda[/itex], we have a fully constructive interference. When the ratio is odd multiple of [itex]\lambda[/itex] (thus, ratio is 1/2, 3/2, 5/2...etc), we have fully destructive interference.

3. The attempt at a solution

The answers to both are 14.

This is what I had initially.... assuming r >>> D


It should be positive to say that [itex]\[s_{1}\][/itex] and [itex]\[s_{2}\][/itex] is constructive at point a, because the both will travel at identical path length.
At point b, the difference should always be 1.75m, because [itex]\[s_{2}\][/itex] has to travel an additional 1.75m (the distance of which the two sources are separated).
--edited---
but 7/2 is neither fully constructive nor fully destructive right??? am i correct???

Then I am lost with how to get 14 of them. I know both have 14 because of the symmetry. But what points should I label on the circle???
I was looking up on the Internet, and we have to get the ratio, 1.75/0.5 = 3.5, and this means 7/2. But I am confused.

I know that [itex]\[L_{1} - L_{2} = n \lambda \][/itex] (constructive). What does this ratio, n, means at all???

Similarly, for destructive, we have [itex]\[L_{1} - L_{2} = (n+\frac{1}{2} )\lambda \][/itex], and n (the ratio) is 3.

How do we use the ratio, n, to solve this problem (or any interference of sound wave problems?) What is the game plan for solving any interference problems?

I appreciate any helps! Thanks
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Redbelly98
#2
Feb19-11, 08:44 PM
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Quote Quote by jwxie View Post
2. Relevant equations

When the ratio of [itex]\[\frac{\Delta L}{\lambda }\][/itex] is 0, 1, 2, or integral multiple of
[itex]\lambda[/itex], we have a fully constructive interference. When the ratio is odd multiple of [itex]\lambda[/itex] (thus, ratio is 1/2, 3/2, 5/2...etc), we have fully destructive interference.
Okay, what is the next number in that sequence, 1/2, 3/2, 5/2, ___?

3. The attempt at a solution

The answers to both are 14.

This is what I had initially.... assuming r >>> D


It should be positive to say that [itex]\[s_{1}\][/itex] and [itex]\[s_{2}\][/itex] is constructive at point a, because the both will travel at identical path length.
At point b, the difference should always be 1.75m, because [itex]\[s_{2}\][/itex] has to travel an additional 1.75m (the distance of which the two sources are separated).
--edited---
but 7/2 is neither fully constructive nor fully destructive right??? am i correct???
Well, what was your answer to my earlier question???

Then I am lost with how to get 14 of them. I know both have 14 because of the symmetry. But what points should I label on the circle???
I was looking up on the Internet, and we have to get the ratio, 1.75/0.5 = 3.5, and this means 7/2. But I am confused.

I know that [itex]\[L_{1} - L_{2} = n \lambda \][/itex] (constructive). What does this ratio, n, means at all???

Similarly, for destructive, we have [itex]\[L_{1} - L_{2} = (n+\frac{1}{2} )\lambda \][/itex], and n (the ratio) is 3.

How do we use the ratio, n, to solve this problem (or any interference of sound wave problems?) What is the game plan for solving any interference problems?

I appreciate any helps! Thanks
For starters, just think about points from A to B. You calculated that ΔL/λ goes from 0 to 3.5. How many integers (0, 1, 2, ...) are included in that range? That would correspond to points where constructive interference occurs.
jwxie
#3
Feb20-11, 10:26 AM
P: 283
hi, Redbelly98
Thanks for the help.
Right. the next number would be 7/2.

I have 0,1,2,3 and 3 integer numbers in that range.
I know that 2pi is one wavelength.
Two problems arise:

[1] what should I label on point B? I know point a would be 0 anyway because the path difference is 0.
So point c (on the other side of point a) should also be 0? Which means I should label point b as 3.5? Doing so I did get 14 points of constructive and 14 points of destructive.

[2] The second question is, again, how do I choose the label (knowing the ratio)?? What does that tells us? As I said earlier, one wavelength corresponds to 2pi right? Moreover, earlier we said for constructive we have 3.5, or 7/2, and for destructive the ratio is 3. So what does ur last point help determing the points?

I really appreicate your guide here! Something is emerging....
:] Thanks

Redbelly98
#4
Feb20-11, 05:29 PM
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P: 12,071
Sound wave interfernce on a large circle

Quote Quote by jwxie View Post
hi, Redbelly98
Thanks for the help.
Right. the next number would be 7/2.

I have 0,1,2,3 and 3 integer numbers in that range.
I know that 2pi is one wavelength.
While 2pi radians corresponds to a wavelength, we are not talking about radians here. 1, 2, 3, etc. means ΔL is λ, 2λ, 3λ.

Two problems arise:

[1] what should I label on point B? I know point a would be 0 anyway because the path difference is 0. So point c (on the other side of point a) should also be 0? Which means I should label point b as 3.5? Doing so I did get 14 points of constructive and 14 points of destructive.
Yes, that's correct.

[2] The second question is, again, how do I choose the label (knowing the ratio)?? What does that tells us? As I said earlier, one wavelength corresponds to 2pi right? Moreover, earlier we said for constructive we have 3.5, or 7/2, and for destructive the ratio is 3. So what does ur last point help determing the points?
As I said earlier in this post, one wavelength corresponds to 1, 2 wavelengths corresponds to 2, etc.
jwxie
#5
Feb20-11, 07:38 PM
P: 283
Great. Thanks for the calcification.


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