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Time difference between two sine waves

  1. Nov 20, 2004 #1
    "There are two sine waves having a phase difference of 20 degrees. After one reaches its maximum value, how much time will pass until the other reaches its maximum, assuming a frequency of 60 Hz."

    Should I go about this by assuming...
    sin(120pi*t) = sin(120pi(t + x) - 20)

    Any hints appreciated
     
  2. jcsd
  3. Nov 20, 2004 #2

    Galileo

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    You have two sines which have are functions of position and time. (actually, considering them function of time alone is sufficient)
    They both have the form:
    [tex]A\sin(kx-\omega t + \phi)[/tex]

    Assume the first wave reaches its maximum A at time t=0 and position x=0.
    Then you have to find t when the second wave reaches its max A:

    [tex]A\sin(-\omega t + \phi)=A[/tex]
    Where [itex]\phi[/itex] is 20 degrees expressed in radians.
     
    Last edited: Nov 20, 2004
  4. Nov 20, 2004 #3
    I get it. Since 20 degrees = pi/9, moving LHS A to RHS gives
    sin (-120pi*t + pi/9) = 1
    so, -120pi*t + pi/9 = pi/2
    and finally t = 7/2160 seconds

    Many thanks
     
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