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Homework Help: Superposition homework problem

  1. Jun 10, 2006 #1
    Hi guys,

    Couldn't figure this question out even after some thinking through. Hope someone can help me here.

    2 loudspeakers are placed 1.2m vetrically apart. The point A is the horizontal distance of 1.6m from the top speaker. Both speakers are operating in phase with a steady freq of 400Hz and the speed of sound can be assumed to be at 320ms^-1. Does point A have maximum/minimum/somewhere between max and min intensity?

    The answer I was given is minimum intensity.

    I used the formula y = (lambda)D/a where D = 1.6 and a = 1.2 and y = fringe spacing. I worked y out to be 1.067m which I interpret this result as a bright fringe 1.067m above the central maximum. I then used inverse tangent of 0.6/1.6 and sin the resultant angle multiplied by 1.2, to find the path difference. Am I correct till here? How should I continue to find whether if the 2 waves are in phase to point A?
  2. jcsd
  3. Jun 10, 2006 #2
    All of a sudden I managed to come out with the solution with the phase diff of .52 which implies that destructive interference is at point A. I have another question, that is using the calculation for fringe spacing. If I take the inverse tangent of 1.067/1.6 I would get approx .0588 radians.

    Using the sin function on this angle and multiply it by 1.2, I'd get the path difference. When I divide it by 0.8, I'd get a phase diff of .07, which doesn't completely give a minimum, but somewhere between max/min intensity. Why is there this contradictory answer?

  4. Jun 10, 2006 #3


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    There are two ways to solve this type of problem (they are all equivalent but the language is a bit different). One can work in terms of phase difference or in terms of difference of distance (the term "path difference" is sometimes used).

    In any case, one way is to calculate the difference [itex] y_2 - y_1 [/itex] where y_2 and y_1 are the distances between the point and each of the source. Given that the sources are in phase, if this difference of distance is a multiple of the wavelength, you will have constructive interference. If it is an odd multiple of half the wavelength, you have destructive interference

    To summarize, for two sources in phase we have
    [tex] y_2 - y_1 = n \lambda \rightarrow maximum[/tex]
    [tex] y_2 - y_1 = (2 n -1) {\lambda \over 2} \rightarrow minimum [/tex]
    where n =1,2,3,4....

  5. Jun 10, 2006 #4
    This is not right . Only phase differences which are odd multiples of [itex]\pi like \pi,3\pi,5\pi ...[/itex] lead to destructive interferences .Please post how you have obtained this answer so that we may correct you .
    I am assuming that you converted your obtained path difference to phase difference which seems to give your answer, is that right ?
    So that would mean you have made some error in calculating the path difference in the first place .
    Starting from the basics, you seem to know that
    [tex]path diff. = asin\theta[/tex]
    Now as an approximation for small angles we may write,
    [tex]path diff. = atan\theta[/tex]

    From the setup, we know geometrically that

    [tex]tan\theta = \frac{y}{D}[/tex]

    [tex]or path diff. = \frac{ay}{D}[/tex]

    Now in your question you know the values of all parameters in the above equation . See whether the path diff. matches with any one of the conditions for minimum .
    I didn't understand what you did with inverse tangent and the angle .
    Can u follow what I have said ?

    Last edited: Jun 10, 2006
  6. Jun 11, 2006 #5
    Thanks for the response. With regards to your first post, I managed and understood the solutions, I didn't thought of it to be that simple.

    However, one question regarding the phase difference method. I think this method is based on Young's double slit experiment. When determining the angle, why is it required to divide the distance between the 2 speakers by 2? i.e, tangent inverse(0.6/1.6) in this case? Why is it not right by using inverse tangent(1.2/1.6)? I'd get different answers for phase and path diff. for the 2 different angles obtained.

    Edit: I understand that for a single slit experiment, the centre would always be a maxima. What about the centre between 2 sources in a double slit experiment?
    Last edited: Jun 11, 2006
  7. Jun 11, 2006 #6
    In a double slit experiment too, the centre (origin) is usually taken as the central maximum .
    The central maximum is formed where path difference between the two interfering waves is 0 ( usually ) . This is possible only when the point is half way between the two slits , is it not ?
    So this is chosen as the origin . The tangent of the angle and the relation with path difference is just plain geometry as I have shown in my ealier post .
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