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Tricky Trig Function Find the eqn Help

  1. Aug 31, 2005 #1
    Find the equation of a sine function that has a vertical displacement 2 units down, a horizontal phase shift pi/3 to the right, a period of pi/6, reflection in the y-axis and the amplitude of 3.

    my answer [tex] y=3\sin [-12(x+\frac {\pi} {3})] -2 [/tex]

    is this what you get?
     
  2. jcsd
  3. Aug 31, 2005 #2

    mezarashi

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    Homework Helper

    A sine wave function can be expressed generally as:

    [tex] f(x)=A\sin (\frac{2\pi x}{T}+ \theta}) + C [/tex]

    where A is peak to peak amplitude, T is period (or 1/T = frequency f), theta is phase shift (horizontal), and C is the vertical displacement.

    P.S. Your original question. Having the function be both even (reflected about the y-axis), and also forcing a phase-shift generally doesn't make sense (given you want a continuous function). Having both conditions satisfied would be purely coincidental.

    Edit: I forgot to note, when your theta is positive, the phase shift wil be to the left. A negative theta will result in a shift to the right.
     
    Last edited: Aug 31, 2005
  4. Aug 31, 2005 #3
    I think it should be pi/36 instead of pi/3 although I could be wrong.
     
  5. Aug 31, 2005 #4
    Anyone else get the same answer as me?
     
  6. Sep 2, 2005 #5
    At first glance it looks OK. The only thing I found is that for a phase shift to the right, it would be x-pi/3 rather than x+pi/3.
     
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