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Phase constant explanation needed!

  1. Dec 7, 2011 #1
    Can anyone explain to me what the phase constant (phi) is regarding the equation for displacement in simple harmonic motion and tell me if there is any equation to get it because I have looked at some answers to simple questions and in the answers they have just neglected the phase constant altogether so they must have let it equal to zero or something.
    Urgent assistance needed as physics test is on Monday!
     
  2. jcsd
  3. Dec 7, 2011 #2

    NascentOxygen

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    Staff: Mentor

    Why not provide the equation you are concerned about?
     
  4. Dec 7, 2011 #3
    It's not so much the equation just the phase constant itself.
    x(t)=Acos[wt+(phi)]
     
  5. Dec 7, 2011 #4
    The phase constant allows for different starting positions or velocities, if your equation is
    [tex] x(t) = A\cos (\omega t + \phi ) [/tex]
    and you know that
    [tex] x(0) = 0 [/tex]
    Then you can solve for the phase angle
    [tex] 0 = A\cos (\omega (0) + \phi ) \Rightarrow \phi = \frac{\pi}{2} [/tex]
     
  6. Dec 7, 2011 #5

    ehild

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    The phase constant depends on the initial conditions. Those mean the initial displacement and velocity. If the displacement x=Acos(ωt+φ) was maximum at t=0, the phase constant has to be zero, φ=0, as the the cosine function has its maximum at zero argument, cos0=1.

    If the SHM started with a push, at zero displacement but maximum velocity, the derivative of x(t), v=-Aωsin(ωt+φ) has to be maximum at t=0. That happens when sin(φ)=-1, that is φ=-pi/2. In his case, the displacement is x=Acos(ωt-π/2), and it is equivalent to x=Asin(ωt).

    In general case, x(0)=x0 and v(0)=v0. You can determine both A and φ in

    x(t)=Acos(ωt+φ), v(t)=-Aωsin(ωt+φ)

    which correspond to these initial conditions:

    At t=0

    x0=Acosφ, v0=-Aωsinφ,----->

    A=sqrt(x02+(v0/ω)^2) and

    tanφ= -v0/(ωx0).

    ehild
     
  7. Dec 7, 2011 #6
    Thanks a million lads!
     
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