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What does phase of the motion in terms of cosine displacement mean?

  1. Nov 11, 2012 #1
    What does "phase of the motion in terms of cosine displacement" mean?

    I'm getting tripped up on the wording of this homework question.

    1. The problem statement, all variables and given/known data
    Measured acceleration: An accelerometer has measured the simple harmonic motion shown in the image below.

    2. Relevant equations
    You want to describe the position as a function of time as a cosine: [itex]x(t)=A\cos(\omega t-\phi)[/itex].
    What is the phase of the motion shown in the graph (in terms of a cosine displacement)?
    droHh.jpg

    3. The attempt at a solution
    I've tried reading my textbook for "phase of the motion". There's a section that talks about a rotating vector phasor that references the unit circle. Is this the right path? Are they talking about what the cosine is calculating (i.e. what's inside the brackets of the cosine)?
     
    Last edited: Nov 11, 2012
  2. jcsd
  3. Nov 11, 2012 #2
    Re: What does "phase of the motion in terms of cosine displacement" mean?

    The phase of a periodic function is the displacement the function has from some predefined 0 point. For example, sin(x) is 90° out of phase from cos(x). The question is just asking for [itex]\phi[/itex] in [itex]x(t) = A\cos(\omega t+\phi)[/itex].
     
  4. Nov 11, 2012 #3
    Re: What does "phase of the motion in terms of cosine displacement" mean?

    Thank you.
     
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