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Wave Equations - Superposition

  1. Nov 24, 2013 #1
    1. The problem statement, all variables and given/known data

    Two plane waves are given by

    [itex]E_{1} = \frac{5E_{0}}{((3 \frac{1}{m})x - (4 \frac{1}{s})t)^{2} + 2}[/itex] and
    [itex]E_{2} = -\frac{5E_{0}}{((3 \frac{1}{m})x + (4 \frac{1}{s})t)^{2} - 6}[/itex]

    a) Describe the motion of the two waves.
    b) At what instant is their superposition everywhere zero?
    c) At what point is their superposition always zero?
    2. Relevant equations



    3. The attempt at a solution
    For part b)
    [itex]E_{R} = E_{1} + E_{2} = 5E_{0}(\frac{1}{((3 \frac{1}{m})x - (4 \frac{1}{s})t)^{2} + 2} - \frac{1}{((3 \frac{1}{m})x + (4 \frac{1}{s})t)^{2} - 6}) = 0[/itex]

    Since I'm looking for the instant in which the superposition is zero I set x = 0 and solve for t

    [itex]\frac{1}{16t^{2} + 2} - \frac{1}{16t^{2} - 48t + 36 + 2} = 0[/itex]
    [itex]\frac{1}{16t^{2} + 2} - \frac{1}{16t^{2} - 48t + 38} = 0[/itex]
    [itex]\frac{1}{16t^{2} + 2} = \frac{1}{16t^{2} - 48t + 38}[/itex]
    [itex]\frac{16t^{2} - 48t + 38}{16t^{2} + 2} = 1[/itex]
    [itex]16t^{2} - 48t + 38 = 0[/itex]
    This provides me with an imaginary time. I'm not exactly sure what it is I'm doing wrong here. Thanks for any help you can provide me.
     
  2. jcsd
  3. Nov 24, 2013 #2
    x is not the superposition of the waves here. It is just the spatial coordinate or position.
     
  4. Nov 24, 2013 #3
    So how would I go about solving the problem then?
     
  5. Nov 24, 2013 #4
    Keep the x in your expression for the resultant amplitude and see if you can find some sort of simplified expression for t.
    Why did the s term disappear in your working?
     
  6. Nov 24, 2013 #5
    I thought that the [itex]\frac{1}{s}[/itex] and [itex]\frac{1}{m}[/itex] where units? So I removed them to make my expressions less messy. When I leave x and t in the expression and solve I get

    [itex]\frac{16t^{2} - 24tx + 48t + 9x^{2} - 36x + 38}{16t^{2} - 24tx + 9x^{2} + 2} = 1[/itex]

    I'm not sure how that really helps or how I'm supposed to simplify this further. Thanks for the help.
     
  7. Nov 24, 2013 #6
    I'm not sure how to solve this. Someone else may have to help. In your original solution you can multiply both sides by the denominator of the LHS to get a real time but I still don't think setting x=0 is the right thing to do.

     
  8. Nov 24, 2013 #7
    When I leave x in I do get a neat answer (although I'm not sure if it answers the question). Try the algebra again and if you reach a point similar to this again - multiply by the denominator.
     
  9. Nov 24, 2013 #8
    I start off with this
    [itex]E = E_{1} + E_{2} = \frac{5E_{0}}{(3x - 4t)^{2} + 2} - \frac{5E_{0}}{(3x + 4t - 6)^{2} + 2} = 0[/itex]
    I factor
    [itex]E = E_{1} + E_{2} = 5E_{0}(\frac{1}{(3x - 4t)^{2} + 2} - \frac{1}{(3x + 4t - 6)^{2} + 2} = 0[/itex]
    I expand
    [itex]5E_{0}(\frac{1}{16t^{2} - 24tx + 9x^{2} + 2} - \frac{1}{16t^{2} + 24tx - 48t + 9x^{2} - 36x + 38}) = 0[/itex]
    simplify
    [itex]\frac{1}{16t^{2} - 24tx + 9x^{2} + 2} - \frac{1}{16t^{2} + 24tx - 48t + 9x^{2} - 36x + 38} = 0[/itex]
    move second term to right hand side
    [itex]\frac{1}{16t^{2} - 24tx + 9x^{2} + 2} = \frac{1}{16t^{2} + 24tx - 48t + 9x^{2} - 36x + 38}[/itex]
    multiply both sides by the denominator of the RHS term
    [itex]\frac{16t^{2} + 24tx - 48t + 9x^{2} - 36x + 38}{16t^{2} - 24tx + 9x^{2} + 2} = 1[/itex]
    I set the numerator equal to zero
    [itex]16t^{2} + 24tx - 48t + 9x^{2} - 36x + 38 = 0[/itex]

    And from here it doesn't look like I can solve for a real answer. I don't see what it is I'm doing wrong here.

    The back of my book says that the answer is
    [itex]t = \frac{3}{4} s[/itex]

    I find it interesting that

    [itex]|\frac{∂(16t^{2} + 24tx - 48t + 9x^{2} - 36x + 38)}{∂t}| = \frac{3}{4}[/itex]

    I'm not sure if this is just a coincidence or not though.
     
  10. Nov 25, 2013 #9
    Looking at your first line you have a -6 inside the squared brackets. This is not what you showed originally. Did you make a mistake in your first post?
     
  11. Nov 25, 2013 #10
    You must have written it wrong originally as it now works.

    Setting the numerator to zero is wrong. You have already set the displacement to zero at the start so now you just need to solve it as it appears. Remember what I said to do earlier when you reached a point like this?
     
  12. Nov 25, 2013 #11
    You don't need to do all this common denominator thing.

    As the two fractions have the same numerator (Eo) it is sufficient to equate their denominators in order to have zero amplitude.
     
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