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Solving for Phase Constant: Simple Harmonic Motion

  1. Apr 5, 2010 #1
    1. The problem statement, all variables and given/known data
    Another SHM question. Thanks for the help on my first one.

    The block is at x0 = +5 cm with a positive velocity V0 at time t = 0. Its motion is SHM with amplitude 10 cm and period 2 seconds. If x(t) = Acos(wt + [tex]\varphi[/tex]), the phase constant [tex]\varphi[/tex] should be:

    2. Relevant equations
    x(t) = Acos(wt + [tex]\varphi[/tex])


    3. The attempt at a solution
    0.05 = 0.10cos([tex]\pi[/tex] + [tex]\varphi[/tex])
    [tex]\varphi[/tex] = 60 - [tex]\pi[/tex] (since the 60 is in degrees...do I just use pi radians = 3.14 radians?)
    = 57 degrees

    I got the pi for wt this way btw:
    wT = 2pi
    w = 2pi/T = pi (since T = 2)

    Not sure if I screwed up that or the radians/degrees thing at the end. Any help would be appreciated.
     
  2. jcsd
  3. Apr 5, 2010 #2

    rl.bhat

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    Homework Helper

    The relevant equation is
    x(t) = Acos(wt + φ)
    If you put t = 0, then
    x(0) = Acos(φ)
     
  4. Apr 5, 2010 #3
    Okay. So it is 60 degrees then. Thanks!
     
  5. Apr 5, 2010 #4
    Just one more thing about this actually...my first inclination upon reading the 5cm was to use that as the phase constant. It describes it as starting out from there...I thought it normally started at 0 so in order to 'get it to 5cm' I would need to use the phase constant to represent that. Obviously that is wrong...I'm just not quite sure why...

    EDIT: Ah...so we are starting 5cm from 0...but 5cm is not a degree or radian measure...what we're doing is finding how much of a shift 5cm constitutes in terms of a phase constant. Is that right?
     
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