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Wave Interference: Minimum Intensity/Hyperbolic

  1. Jul 17, 2017 #1
    1. The problem statement, all variables and given/known data

    Two identical speakers, 10.0 m apart from each other, are stimulated by the same oscillator, with a frequency f, of 21.5 Hz, at a place where the speed of sound is 344 m/s.

    a) Show that a receiver at A will receive the minimum intensity of sound (Amin) due to the inerference of the waves produced by the two speakers.

    b) Show that in order for the sound intensity to remain the same, the receiver, being at the same level as the speakers, has to follow the hyperbolic: 9x2 -16y2 = 144

    c) Can the receiver distance itself from the speakers and still keep receiving the minimum sound intensity? If yes, write down the route it must take. If not, write down how far it can go.

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    PS: I wrote down the right speaker as (1), and the left speaker as (2).

    2. Relevant equations

    Δr = | r2 - r1 |
    v = λf

    3. The attempt at a solution

    a)
    v = λf <=> λ = 344 m/s / 21.5 Hz = 16.0 m

    r1 = (10.0 - 9.00) m = 1.00 m
    r2 = 9.00 m

    Δr = |9.00 - 1.00|m = 8.00 m = 16.0/2 m = λ/2 = 1*λ/2 = n*λ/2, with n = 1

    Therefore, the two waves have a phase difference of π rads or 180 degrees, and cancel each other out. So the receiver... receives, sound with the minimum intensity (0).

    b) And this is where I get stuck. I looked around for some examples, but no luck. The basic principle is essentially that Δr must always result in n*λ/2. I did create an imaginary triangle at the (x,y) point, so that I can have new versions of r1 & r2, but I can't figure out how to connect them with x & y and end up with the hyperbolic.

    Any help is appreciated!
     
  2. jcsd
  3. Jul 17, 2017 #2

    gneill

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    Staff: Mentor

    Look for the ways that a hyberbola can be constructed mathematically. Hint: One involves the relationship between the two "radii" from the focii. (It is analogous to the way that an ellipse can be constructed using two pins to mark the focii, a loop of string, and a pencil :wink:)
     
  4. Jul 17, 2017 #3

    scottdave

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    Homework Helper
    Gold Member

    You are going to need to create two triangles, to represent the distance from each transmitter.
     
  5. Jul 17, 2017 #4
    Is there any link where I can read more about hyperbolas? I looked around and I could only find examples of simple "put numbers in the formula" exercises. My books only covered the basic version (centre is 0) and I don't have them here as well, so I don't know where to look.

    Kinda like this ?

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    I looked around about hyperbolas, and so far I've got:

    >The basic version of a horizontal hyperbola is: (x-h)2/a2 - (y-v)2/b2 = 1

    >For it to be a hyperbola, the difference between the distances of a point from the focii, must be constant.

    >b is how far up from the centre of the hyperbola the point is, and a is how far to the left/right from the centre of the hyperbola it is.

    >h is the x variable of the centre, and v the y variable of the centre.

    In my case, I need the hyperbola to be constructed in such a way, that Δr = n*λ/2. But since each hyperbola must have a constant Δr, and I found that Δr = 8.00 m in (a), that's how it should be. Right? Apart from that, I'm kinda lost. I'm not sure if the axis is constructed in such a way that where y & x connect I have (0,0), for one.
     
  6. Jul 17, 2017 #5
    Update on (b):
    I hypothesized that where the two axis meet, is where (0,0) is located. So, we have each focii being 5.00 m away from the centre, and the place where the hyperbole starts from (according to the book's drawing), being 4.00 m away. So:

    >a = 4.00 m

    >c = 5.00 m

    >c2 = a2 + b2 <=> b = 3.00 m

    >h = 0 = v

    (x-h)2/a2 - (y-v)2/b2 = 1 <=> x2/16 - y2/9 = 1 <=> 9x2 - 16y2 = 144

    So, I have shown that with the current construction, a hyperbola exists. The receiver A (from the book's drawing) is part of the hyperbola. Thus for every point, the difference between the distances from each focii, Δr, is a constant. Δr, for the receiver A, as shown in (a) was 8.00 m, which verifies the Δr = n*λ/2 formula, which is needed for a receiver/point to receive the minimum sound intensity (A = 0).

    Am I done with (b)? Any help on (c) because I don't know where to even start?
     
  7. Jul 17, 2017 #6

    gneill

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    Staff: Mentor

    It looks to me like you have an acceptable demonstration for part (b).

    I'm not sure what part (c) is getting at. Since the hyperbola is not a closed curve, so long as you stick to a given branch you can end up arbitrarily far from the speakers. I suppose you might mention the asymptotes of the hyperbola?
     
  8. Jul 17, 2017 #7
    Whew, thanks for the confirmation.

    Yeah, the book's answer is that the point/receiver can keep on going, and that at the "final ocassion" it follows the path of the asymptotes, y=+- 0.75x
     
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