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Given two waves, find phase angle

  1. Dec 29, 2015 #1
    1. The problem statement, all variables and given/known data
    Given two waves characterized by y1(t) = 3 cos wt; y2(t) = 3 sin(wt + 60deg); does y2(t) lead or lag y1(t) and by what phase angle?

    2. Relevant equations
    y1(t) = 3 cos wt; y2(t) = 3 sin(wt + 60deg)
    The formula I'm given in my book Fundamentals of Applied Electromagnetics is:
    y(x,t) = A cos (2pit/T - 2pi/lamda + reference phase)

    3. The attempt at a solution
    This question seemed really easy. I figured you just shift the sin wave 60 degrees to the left and I can see by this that y2(t) lags y1(t). I searched everywhere of phase difference and I get the peak:
    (distance/wavelength)*2*pi.
    The wavelength I assume is arbitrarily 2pi because I'm given w. I easily thought the answer would be 60 degrees but the back of my book says 54 degrees.

    I'm completely lost and I've tried multiple things. This seems too easy and I'm frustrated (I need to vent a little). Any help would be greatly appreciated. First time posting a homework problem. I'm on winter break and seems like I should know this in the back of my head. My book is for juniors in college and I'm a little embarrassed for asking this. But I guess for Ch.1 they assume I should easily know this. Ch.1 is very vague and literally goes through my physics class in 1 chapter.
     
  2. jcsd
  3. Dec 29, 2015 #2
    Well, trasform cosine in sine or sine in cosine, i suggest using: cosx= sin (x + π/2)

    y1 (t) = 3 sin (ωt + π/2)

    Then you can compare the two functions to find the phase angle
     
  4. Dec 29, 2015 #3
    If I do what you said I get 30 degrees. Am I missing something?
     
  5. Dec 31, 2015 #4


    I got this formula from watching this video. Φ= (2π/λ)*Δx +ΔΦ0 where Δx = distance traveled from origin of both sources to some point. ΔΦ = inherent phase difference. I changed the first y1 equation in to a sine equation. I now have:

    y1(t) = 3*sin(ωt +π/2 )
    y2(t) = 3*sin(ωt + π/3)

    The inherent phase is ΔΦ = Φ2 - Φ1 = π/3-π/2= -π/6 or cause (y1 leads π/6) which is 30°.
    λ=2π so Δx = the length it takes for the wave to arrive at a source.
    For Δx = I get the same thing = π/3-π/2= -π/6 which adding together gives me 60° phase difference.

    In the back of my book I get 54°. I still don't understand what I'm doing wrong. Is there anyone who can help me?
     

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  6. Dec 31, 2015 #5

    ehild

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    The answers in books are wrong quite often.
     
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