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Phase constant sign (quick question)?

  1. Sep 1, 2016 #1
    1. The problem statement, all variables and given/known data
    A = 20 cm
    14.EX05.jpg
    Given the graph, what is the phase constant(in degrees)?
    (I have already solved for amplitude and frequency)
    2. Relevant equations
    x(t) = A cos (wt + ϕ)
    3. The attempt at a solution
    x(t) = A cos (wt + ϕ)
    Taking from t = 0, we find:
    10 = 20cos(w*0 + ϕ)
    10 = 20cos(ϕ)
    1/2 = cos(ϕ)
    cos-1(1/2) = ϕ
    ϕ = 60 degrees.

    I tried submitting the answer and it said check signs, so I entered -60 degrees.
    Why is this answer negative?
     
  2. jcsd
  3. Sep 1, 2016 #2

    gneill

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    Staff: Mentor

    Hi tyrostoken, Welcome to Physics Forums!

    The cosine trig function is many-to-one: Note that cos(Φ) = cos(-Φ). So when you solved for Φ the answer might have been either +60° or -60°. It's up to you to check which one fulfills the given conditions.
     
  4. Sep 1, 2016 #3

    haruspex

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    2016 Award

    To add to gneill's answer...
    Imagine shifting the curve left until the value at t=0 is once again 0.5. This solves your equation, but the graph is visibly different. So you need another fact about the graph to fix the phase. What obvious difference is there in the graphs?
     
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