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Sinusoid question

  1. Apr 27, 2009 #1
    1. The problem statement, all variables and given/known data
    Find the amplitude and phase of the sinusoid (4√3-3)cos(2t + 30°) + (3√3 – 4)cos(2t + 60°)

    2. Relevant equations
    Acos wt + Bsin wt

    3. The attempt at a solution
    Answer should be amplitude 5 and phase 36.9° but I get 4.11 and 5.7
  2. jcsd
  3. Apr 27, 2009 #2


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    Staff: Mentor

    Welcome to the PF. Please show us your work, so that we can try to help you. That's how it works here on the PF.
  4. Apr 27, 2009 #3
    The square root of (4√3-3)^2 + (3√3 – 4)^ 2 = 4.11 and θ = tan^-1 (4√3-3)/(3√3 – 4) = 16.9 degrees
  5. Apr 27, 2009 #4


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    Staff: Mentor

    Not sure what that's meant to represent. You have two sinusoidal functions of time, which can be added to make a new sinusiodal function, with a combined amplitude, and a combined resultant phase. It should look something like this:

    f(t) = A cos(2t + phase)

    What would be a way that you can add two cosine functions that have different phases?
  6. Apr 27, 2009 #5


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    Staff: Mentor

  7. Apr 28, 2009 #6
    I tried changing the second function from cos to sin to match the identity but this does not seem to affect the magnitude in any way. I still do not know how to arrive at A=5.
  8. Apr 28, 2009 #7

    The Electrician

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    Gold Member

    Does this have anything to do with the study of electronics? If so, have you studied phasor arithmetic yet?
  9. Apr 29, 2009 #8
    Actually the problem comes before the section on phasors. Supposedly it is supposed to be solved with the trig identities we are given.
  10. May 1, 2009 #9


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    See the link berkeman gave in post #5. You eventually want to get things in the form

    A cos(2t) + B sin(2t)​

    To do that you'll need to use the addition identity for cosine,

    cos(x+y) = ???​
  11. May 2, 2009 #10
    Ok I am getting 5.1 which sounds right. Thanks everyone.
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