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Revision question on waves

  1. Apr 29, 2014 #1
    Two waves are identical apart from a phase difference, they create a resultant wave of 25% increase of the original amplitude, what is the phase difference.



    sin(alpha)+sin(beta)=2sin(1/2)(alpha+beta)cos(1/2)(alpha-beta)



    Asin(kx+wt)=2sin(1/2)(alpha+beta)cos(1/2)alpha-beta)
    Asin(kx+wt)=2sin(1/2)(alpha+beta)cos(1/2)alpha-beta)
    =1.2Asin(kx+wt)

    I assume that's right then you rearrange but I'm highly doubting myself
     
    Last edited by a moderator: Apr 29, 2014
  2. jcsd
  3. Apr 29, 2014 #2

    Simon Bridge

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    It is a LOT easier if you use the phasor representation of the waves.
    Then you just add the phasor arrows head-to-tail like vectors.
     
  4. Apr 29, 2014 #3

    So like.. e^i(wt-kx) ? I never really understood this method so I should probably look it up, how would you apply that to this question? If I may ask

    Thanks for the help!
     
  5. Apr 29, 2014 #4

    Simon Bridge

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    The two waves vary at the same rate - so they will always maintain the same phase difference between them.

    Executive summary:

    Waves can be represented by vectors that rotate. The length of the vector is the amplitude of the wave.

    Two waves that are identical but for a phase difference are two vectors with an constant angle between them.

    The result of the two waves combining is the same as adding the two vectors.

    So two waves with a phase difference of ##\phi## but identical otherwise, will form a iscoseles triangle where the external angle at the apex is the phase difference. Thus the internal angle is ##\theta=\pi-\phi##

    The resultant is the third side - use the cosine rule.
    Sketch it and you'll see.

    i.e. if the phase difference is ##\pi/2##, and the amplitudes are A, then the resultant wave will have amplitude ##\sqrt{2}A## (pythagoras).

    What you are trying to do is:
    ##\sin(\omega t-kx)+\sin(\omega t-kx+\phi) = A\sin(\omega t-kx+\delta)##
    ... you are given ##A## and you need to find ##\phi##
     
  6. May 3, 2014 #5
    So to ask simply, apologies if I'm wrong. As we know that both waves have the same amplitude and their resultant is 25% greater, could I draw a line joined to another line which as at an angle. Label the angle outside of the two lines (apex) as theta. After this I could use Pythagoras to determine the inside angles and then knowing the angle is 180 determine the unknown angle which is he phase difference?

    I could label lines with length 4, 4 and 5
    So sorry for the lateness!!
     
  7. May 5, 2014 #6

    Simon Bridge

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    I'd draw arrows, not lines. In this case they are the same length.
    As vectors: ##\vec r = (r,\theta)## ... i.e. a length and an angle measured anticlockwise from some reference direction (i.e. the x axis).

    If the first wave is ##\vec r_1 = (1,0)##
    Then the second wave is ##\vec r_2 = (1,\theta)##
    Sketch these head-to-tail - careful: ##\theta\neq\pi/2## so don't draw a right-angle.

    Since there is no right angle, you cannot use pythagoras directly.
    I'd use the cosine rule instead.
     
  8. May 5, 2014 #7

    Simon Bridge

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    I'd draw arrows, not lines. In this case they are the same length.
    As vectors: ##\vec r = (r,\theta)## ... i.e. a length and an angle measured anticlockwise from some reference direction (i.e. the x axis).

    If the first wave is ##\vec r_1 = (1,0)##
    Then the second wave is ##\vec r_2 = (1,\theta)##
    Sketch these head-to-tail - careful: ##\theta\neq\pi/2## so don't draw a right-angle.

    Since there is no right angle, you cannot use pythagoras directly.
    I'd use the cosine rule instead.
     
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