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Homework Help: Simplify the equation - question about the angle

  1. May 4, 2017 #1
    1. The problem statement, all variables and given/known data
    Please, take a look at the following equation and help me to understand where the authors got the value of φ = π/3. I don't see where it is derived from as no additional conditions are given.

    2. Relevant equations
    x(t) = 5e(-t/5) cos(t) + 5e(-t/5) √3 sin(t)

    3. The attempt at a solution

    Here is how I proceed:
    x(t) = 5e(-t/5) (cos(t) + √3 sin(t))

    Now, given the formula:
    f(x) = a sin(wx) + b cos(wx) + B (w > 0) is the same as
    f(x) = √a2+b2 sin(wx + φ) + B

    In my case, B = 0, so I rewrite the expression as:
    x(t) = 5e(-t/5) ( √3 sin(t) + cos(t) )

    x(t) = 5e(-t/5) √√32+12 sin(t + φ)

    x(t) = 10e(-t/5) sin(t + φ)

    But in the book they have:
    x(t) = 10e(-t/5) sin(t + π/3)

    Where did they get φ = π/3 from?

    Thank you!
  2. jcsd
  3. May 4, 2017 #2
    x(t) = 5e(-t/5) ( √3 sin(t) + cos(t) )=##10e^{-t/5}\left(\sin {t }\left(\frac{\sqrt{3}}{2}\right)+\cos{t}\left(\frac{1}{2}\right)\right)##

    ##\sin{(t+\phi)}=\sin t \cos{\phi}+\cos{t}\sin{\phi}##

    What is the value of ##\phi##?
  4. May 4, 2017 #3
    Ah! I see. I should have computed the value of φ based on the coefficients' values:

    coefficient1 x cos(φ) = √3
    coefficient2 x sin(φ) = 1
    (coefficient1 x cos(φ))2 + (coefficient2 x sin(φ))2 = coefficient2
    hence, coefficient = 2, and thus cos(φ) = √3/2 and sin(φ) = ½, and of course φ = π/3
    Now I see how the value appeared.
  5. May 4, 2017 #4

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    You have been shown this more than once already in other threads. It is something you need to learn and commit to memory, especially if you are going into Phhsics or Electrical Engineering.
  6. May 4, 2017 #5
    It takes practice and time to remember these notions. I am doing my best, and as I am learning on my own I don't have enough practice. Happy to be able to ask here. :-)
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