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Wave interference and resolving resultants

  1. May 8, 2006 #1

    kel

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    Me again ! :surprised

    I have the following question and this time I really have no idea!

    2 sinusoidal waves, identical except for their phase, travel in the same direction along a stretched string and interfere to produce a resultant wave given by:

    x(z,t) = 3Cos (20z-4t+0.82)

    Where x is in mm, z in metres and t in seconds

    a- What is the wavelength of the 2 original waves
    b- What is the phase difference between them
    c- What is their amplitude

    At the present I'm stuck on 'a' and may/may not be able to do b and c once I know how to work out 'a'.

    I can see how to add original waves, but how do I go about resolving the resultant wave? I guess it's similar to vector algebra, but mmmmm.... really don't know where to begin on this one.

    Any help would be very much appreciated.
     
  2. jcsd
  3. May 8, 2006 #2

    Päällikkö

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    This should get you started:
    [tex]\sin \alpha + \sin \beta = 2 \cos \left( \frac{\alpha - \beta}{2} \right) \sin \left( \frac{\alpha + \beta}{2} \right)[/tex]
     
  4. May 8, 2006 #3

    kel

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    You'll have to excuse me for being a bit clueless here, but how do I relate that to the resultant wave?

    Sorry, it's late and my brain needs caffine!:zzz:
    Cheers
     
  5. May 8, 2006 #4

    Päällikkö

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    Adding the two waves gives you the resultant wave.

    Use the fact that the waves are identical, except for phase, to write equations for the two waves.
     
  6. May 9, 2006 #5

    kel

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    So far, I've got the following

    y(z,t) = 3cos(20z-4t) + 3cos(20z-4t+0.82)

    but I don't think this is right
     
  7. May 9, 2006 #6

    Päällikkö

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    The two waves, identical except for phase:
    [tex]
    y(z,t) = A\cos(kz - \omega t + \phi_1) + A\cos(kz - \omega t + \phi_2)
    [/tex]


    The earlier equation for cosines:
    [tex]\cos \alpha + \cos \beta = 2 \cos \left( \frac{\alpha - \beta}{2} \right) \cos \left( \frac{\alpha + \beta}{2} \right)[/tex]

    Can you now see where we're going?
     
  8. May 9, 2006 #7

    kel

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    So, I'd get

    [tex]\cos \alpha + \cos \beta = 2 \cos \left( \frac{\0 - \0.82}{2} \right) \cos \left( \frac{\0.82}{2} \right)[/tex]
    Which leaves
    [tex]\cos \alpha + \cos \beta = 2 \cos \left(-0.41\right) \cos \left(0.41\right)[/tex]
     
    Last edited: May 9, 2006
  9. May 9, 2006 #8

    Päällikkö

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    I get:
    [tex]y(z,t) = A\cos(kz - \omega t + \phi_1) + A\cos(kz - \omega t + \phi_2)[/tex]
    [tex]y(z,t) = 2A \cos\left(\frac{\phi_1-\phi_2}{2}\right) \cos\left(\frac{2kz - 2\omega t + \phi_1 + \phi_2}{2}\right)[/tex]


    [tex]y(z,t) = A' \cos \left( kz - \omega t + \frac{\phi_1 + \phi_2}{2} \right)[/tex] , where
    [tex]A' = 2A \cos \left(\frac{\phi_1-\phi_2}{2}\right)[/tex]
    Now this looks quite a bit like the equation given in the problem.
     
    Last edited: May 9, 2006
  10. May 9, 2006 #9

    kel

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    I tried it like this (adding two waves together):

    a*sin(kx-wt+PI/2) + a*sin(kx-wt+PHI+PI/2) ---> a*cos(kx-wt) +
    a*cos(kx-wt+PHI).

    adding these using the trig identities you get:
    2a*cos(kx-wt+PHI/2)*cos(-PHI/2).

    comparing this with the resultant wave, then:
    k = 20
    w = 4 and
    PHI/2 = 0.82.

    and using k = 2PI/Lambda and w=2P

    wave 2a = 3 this would suggest that the amplitude is 3/2.
     
  11. May 9, 2006 #10

    Päällikkö

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    Good :smile:.
    The thing I'll have to disagree about is the amplitude. The latter cosine term, cos(-PHI/2) = cos(PHI/2), contributes into the amplitude (see A' in my previous post).
     
    Last edited: May 9, 2006
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