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Wave intensity question

  1. Jun 5, 2010 #1
    Statement: 1. The intensity of wave A at point P is I. Intensity of wave B at point P is (4/9)I. The phase difference between these two waves 180 degrees. Wavelength of A and B is 3x10^-4 and 2x10^-4 cm respectively. Frequency of both waves is "f". What is the resultant intensity at P in terms of I?



    2. Formulae: I= (Amp.)^2 x (Freq.)^2



    3. An attempt:
    Resultant intensity = I - (4/9)I = (5/9)I


    My attempt is incorrect, apparently. The correct method is to first find the resultant amplitude of both the waves superposed, and then finding the intensity using a ratio. Why I can't understand is why my method is incorrect. I'll be very glad if someone clears this up for me. Thanks.
     
    Last edited: Jun 5, 2010
  2. jcsd
  3. Jun 5, 2010 #2
    Formula 2 looks weird. Intensity is just amplitued times its conjugate. Same to say module of amplitude squared.

    Solution hint

    [tex]
    I = |A_1 e^{\phi_1} + A_2 e^{\phi_2}|^2 = A_1^2 + A_2^2 + 2|A_1||A_2| \cos(\phi_1 - \phi_2)
    [/tex]

    the rest try to figure out yourself
     
  4. Jun 5, 2010 #3

    Doc Al

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    Staff: Mentor

    You subtracted the intensities, not the amplitudes. But as your formula suggests, intensity is proportional to the square of the amplitude. So if the intensity is I, what would the amplitude be proportional to?
     
  5. Jun 5, 2010 #4
    You nearly knocked me into a coma with the part in bold. I haven't read this formula yet. I can't tell what the "e" stands for.
    Here's the original question:

    http://tinyurl.com/34gkevc

    Q.5, part (c).

    Yes, and that is my question.
    If we are to find the resultant intensity of two waves undergoing complete destructive interference upon superposition, why can't we find it by subtracting the intensities of the two waves, like we find the resultant amplitude by subtracting the amplitudes?

    Square root of I. I can' tell why I'd be needing it, though. I did solve the question using the ratio (Actually, I didn't. I looked it up). I just don't understand why simply subtracting them doesn't work.
     
  6. Jun 5, 2010 #5
    sorry if it is too advanced

    Then follow Doc Al, he gave you basically the same hint. Since there is 180 phase shift you defintely have to subtract smth, because it is the condition for destructive interference. Doc Al gave you hint how to calculate what you need to subtract.
     
  7. Jun 5, 2010 #6
    I forgot an important detail in the question statement.
    Amplitude of the waves are 3x10^-4 and 2x10^-4 cm.
    I know how to solve the question by finding the amplitude of the resultant wave and using a ratio:

    Amp. (Amp') of resultant wave I' = (3-2)x10^-4 cm.

    I'/I= [(Amp'.)^2 x (Freq^2)] / [(Amp.^2) x (Freq^2)]

    Then punching in the values to get I' in terms of I.
     
  8. Jun 5, 2010 #7

    Doc Al

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    Staff: Mentor

    Because intensities of coherent waves do not simply add, but amplitudes do. You cannot simply add intensities. It's the amplitude that carries the phase, not the intensity.

    Example. Say the waves had intensity of I but were exactly in phase. Adding the intensities gives you 2I, but adding the amplitudes gives you 2√I and thus an intensity of 4I. Only one is correct.
     
  9. Jun 5, 2010 #8
    Understood. Thanks a lot :) .
    One more question, please. Is there a difference between, "phase angle" and "phase difference"?
    There's a question I came across (let's call it Q.1) in which we were to find the 'phase difference' between two points on a stationary wave. All we needed to do there was to see in which direction the points were moving. If they were in the same direction of motion, then the "phase difference" was 0. If they were moving in the opposite directions, the "phase difference" was 180 degrees.
    In another question (Q.2), however, we were supposed to calculate the "phase angle" between two points. Their positions were similar to the points in Q.1. But in this question, we had to use this formula to calculate the "phase angle".

    Phase difference = (x/λ) x 360

    I was under the impression that "phase difference" and "phase angle" were the same thing. That's not the case apparently, as if we use the aforementioned formula to calculate the "phase difference" between the points in Q.1, it gives us the wrong answer.

    Is there something wrong with my understanding?
    EDIT: I can't tell whether the wave in Q.2 was stationary or not. Could this have something to do with the calculations?
    EDIT 2: In the formula above, "x" is the horizontal distance between the points and λ is the wavelength.
     
    Last edited: Jun 5, 2010
  10. Jun 5, 2010 #9

    Doc Al

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    Staff: Mentor

    You would specify the phase angle between 0 and 360. (Adding multiples of 360 doesn't change the phase.) So if the distance was 1.5 wavelengths, then the phase difference would be .5 X 360.

    A phase difference of 360 degrees is the same as 0 degrees.

    Could that be the problem with your answer to Q.1?
     
  11. Jun 5, 2010 #10
    No, I know that the phase angle is between 0 and 360, but that's not the problem. According to my teacher, in Q.1, the "phase difference" is either 0 or 180, depending upon the direction of motion of the points. No other value can take place, since two points can either be moving in the same or the opposite directions.

    Here's Q.1:
    http://tinyurl.com/2uyy85l

    [In this document, it's Question #5, part (b)]

    I can't link you to Q.2 because I only have it in my notebook. But if we were to apply that formula on Q.1:

    x = (6/8)λ
    "Phase angle" (which I think might be different from "Phase difference", which is what we were supposed to calculate here) = [ (6/8)λ / λ ] x 360
    // // // = 270.

    The correct answer is 180 degrees.
     
  12. Jun 5, 2010 #11

    Doc Al

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    Staff: Mentor

    Ah... OK. Now I see what you're talking about. Yes, things are different for a standing wave. The spatial portion is stationary, but the sign changes on each side of the node. So, yes, two points on either side of the node are exactly out of phase, wherever they are. (Sorry if I confused you!)
     
  13. Jun 5, 2010 #12
    Thanks a lot for clarifying it! And no you didn't confuse me at all.
    Again, I appreciate your help. I have a major exam in 3 days.
     
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