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How to calculate phi

  1. Jan 22, 2012 #1
    A question I have is how is one supposed to calculate phi when simply looking at a sin wave. The equation is of the form x(t) = A*cos(wt + phi). Once again, the only thing i am given is the graph. I appreciate any help you can give!
     
  2. jcsd
  3. Jan 22, 2012 #2

    I like Serena

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    Welcome to PF, jkossis! :smile:

    Find the duration of a complete cycle, which is for instance the time between 2 peaks.
    Let's call this T.

    Now read off the time of the first peak and let's call this t0.

    The first peak corresponds to an angle of zero, so you get: ωt0 + φ=0

    Since ω=2pi/T, this yields:
    (2pi/T)t0 + φ=0
    φ = -(2pi/T)t0
     
  4. Jan 22, 2012 #3

    Ouabache

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    Welcome jcossis !
    Are you sure you need to calculate [itex] \phi [/itex]? If you just wish to determine [itex] \phi [/itex], depending on what your graph looks like, there may be an easier method. You may be able to read [itex] \phi [/itex] directly from your graph. Can you post your graph?
    Have you figured out what each of the elements in your equation [itex]A cos(\omega + \phi) [/itex] means? how it relates to the graphical representation? (hint: google is your friend).
     
  5. Jan 23, 2012 #4
    Hello Bassalisk.

    You should be discussing a cosine wave, not a sine wave.

    go well
     
  6. Jan 23, 2012 #5
    Oh darn :D Curse my fast reading ways
     
  7. Jan 23, 2012 #6

    FOIWATER

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    Cosine leads sine by 90 degrees... and sin begins at the origin, so cosine peaks at the origin.

    where ever the peak on your cos signal is, is the measure of phi (phase shift)
     
  8. Jan 24, 2012 #7

    jim hardy

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    if you're doing it graphically it's way easier and generally more accurate to measure zero crossings where slope is max instead of peaks where slope is zero...
     
  9. Jan 24, 2012 #8

    Ouabache

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    my typo, equation missing time variable t. It should read [itex]A cos(\omega t + \phi) [/itex]
     
    Last edited: Jan 24, 2012
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