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Constructive Interference of Sound

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lafalfa
#1
Dec13-09, 07:50 PM
P: 15
Hi,

Two speakers, A and B, are located at x = +0.5 m and x = -0.5 m. A 680 Hz signal is sent to both speakers. You then walk around the origin, x = 0, in a circle of radius 5.0 m. v_sound = 340 m/s

If you walk once around the complete circle, how many intensity maxima do you hear?


This is what I did: m*wavelength = path difference
Since the maximum path difference is 1.0 m and the wavelength is 0.5 metres, m can only be 0 or 1 or 2. This corresponds to 6 points on the circle (2 points per value of m).
Is this the correct way of doing this question?

The correct answer is 8, not 6, and I don't know where the extra 2 points come from!

Please help me. Thank you very much!
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SystemTheory
#2
Dec13-09, 09:02 PM
P: 279
If you are walking around a circle at 5m doesn't the path to each source depend on sine and cosine relationships, or some other variable geometry? In other words, I am asking, is the path difference to be considered as the distance of the listener to each source, and not the distance between the two sources?
lafalfa
#3
Dec13-09, 09:44 PM
P: 15
The path difference is supposed to be the difference in distance that sound waves from each source must travel to get to you. So, for example, if you are 3 metres from a certain source and there is another source 1 metre directly behind the first source, the path difference would be 4-3=1 metre. I hope that makes sense.
I know for sure that the maximum path difference is 1 m because I had to determine that in another part of the question and the answer key agrees with me on that part.

lafalfa
#4
Dec13-09, 09:45 PM
P: 15
Constructive Interference of Sound

Perhaps I have the entire concept wrong and there's some other mysterious formula that I am supposed to use...?
Bob S
#5
Dec13-09, 10:51 PM
P: 4,663
Are the two speakers being driven in-phase or out-of-phase (polarity reversed)?
Bob S
lafalfa
#6
Dec13-09, 11:05 PM
P: 15
I think we're supposed to assume that they're emitting sound waves in phase and that the only phase difference arises from the path difference.
atyy
#7
Dec13-09, 11:09 PM
Sci Advisor
P: 8,378
Maybe there are 4 points corresponding to m=1? By circular symmetry?
lafalfa
#8
Dec13-09, 11:30 PM
P: 15
How do I find the four points?
atyy
#9
Dec13-09, 11:40 PM
Sci Advisor
P: 8,378
Draw the speakers and the circle. Draw the points where m=0 (no path difference), and the points where m=2 (maximal path difference). Presumably the m=1 points are between those.

To start, let's use x and y axes. I would guess that the two m=0 points are at (x=0, y=5) and (x=0, y=-5). Do you agree?
lafalfa
#10
Dec14-09, 12:22 AM
P: 15
Yes, and the points where m=2 are at (5,0) and (-5,0)?
atyy
#11
Dec14-09, 12:27 AM
Sci Advisor
P: 8,378
Yes. So if you try to place only two m=1 points between your m=0 and m=2 points, will you be able to respect the symmetry of the situation?
atyy
#12
Dec14-09, 12:34 AM
Sci Advisor
P: 8,378
A bit more explicit, assume that an m=1 point lies on the circle between an m=0 and an m=2 point. Show by symmetry that there are 3 other points with the same path difference.
lafalfa
#13
Dec14-09, 02:37 AM
P: 15
I see that by fitting two m=1 points between each m=0 and m=2 points, I would get a total of 8 intensity maxima positions, but I don't understand how we can assume that there are TWO instead of one m=1 points between an m=0 and an m=2 point.
denverdoc
#14
Dec14-09, 03:42 AM
P: 1,351
Maybe I'm missing something, but I don't think that is necessary.

Start the circle on the positive x axis, speakers are situated at -0.5 and +0.5 also on the x axis so you are in an m=2 mode. As I walk around the circle clockwise there is an m=0 maxima at 90 degrees. Presumably I have walked thruough an m=1 max on the way,. If you continue the circle tour there will be 7 maxima before retrurning to the starting point which will be eight.
rl.bhat
#15
Dec14-09, 04:00 AM
HW Helper
P: 4,435
If O is the center of AB, OY the axis and P is the position of the observer, four position of P for m = 1 makes angles pi/4, 3pi/4, 5pi/4 and 7pi/4 with respect to Y-axis. So you get four points for maximum.
atyy
#16
Dec14-09, 09:15 AM
Sci Advisor
P: 8,378
Quote Quote by lafalfa View Post
I see that by fitting two m=1 points between each m=0 and m=2 points, I would get a total of 8 intensity maxima positions, but I don't understand how we can assume that there are TWO instead of one m=1 points between an m=0 and an m=2 point.
It's a good question how many m=1 points lie between each pair of m=0 and m=2 points. But start by assuming that there is only one m=1 point between each pair, and ask, "how many quadrants to a circle are there?"
lafalfa
#17
Dec14-09, 01:26 PM
P: 15
There are four quadrants in a circle, so I would say there should only be four m=1 points, that is, one m=1 point between each m=0 and m=2 points. I don't understand why there should be TWO m=1 points in each of the four quadrants of the circle.
atyy
#18
Dec14-09, 01:51 PM
Sci Advisor
P: 8,378
Quote Quote by atyy View Post
draw the speakers and the circle. Draw the points where m=0 (no path difference), and the points where m=2 (maximal path difference). Presumably the m=1 points are between those.

To start, let's use x and y axes. I would guess that the two m=0 points are at (x=0, y=5) and (x=0, y=-5). Do you agree?
Quote Quote by lafalfa View Post
yes, and the points where m=2 are at (5,0) and (-5,0)?
Quote Quote by lafalfa View Post
there are four quadrants in a circle, so i would say there should only be four m=1 points, that is, one m=1 point between each m=0 and m=2 points. I don't understand why there should be two m=1 points in each of the four quadrants of the circle.
2+2+4=?


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