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Questopn about phase shift SHM

  1. Oct 3, 2007 #1
    Question about phase shift SHM

    The displacement of an object is given by
    [tex]x(t)=x_mcos(\omega t+\phi)[/tex]
    If the inital displacement is 0 and the initial v is in the negative x direction, then the phase constant must be ___rads

    I know that if x=o then the cosine of the phase must be 0
    [tex]\cos(\omega t+\phi)=0[/tex]

    so[tex]\omega t+\phi=\frac{\pi}{2}[/tex]

    and I am stuck from here...hints?
     
    Last edited: Oct 3, 2007
  2. jcsd
  3. Oct 3, 2007 #2

    Dick

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    Initial means t=0. So x(0)=xm*cos(phi). phi=pi/2 works for that. Is v(0) negative? If so then you have a solution.
     
  4. Oct 3, 2007 #3
    v is in -x direction....that is where I am most confused, does that just mean that phi has to be negative?
     
  5. Oct 3, 2007 #4

    Dick

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    It means v=dx/dt is negative at t=0.
     
  6. Oct 3, 2007 #5
    Okay...
    so if [tex]v=-\omega x_m\sin(\omega t+\phi)=-[/tex]
    and x_m=+
    t=0
    then [tex]-\omega x_m\sin(\phi)[/tex] is negative
    so sin(phi) is positive
    Thus, phi=+pi/2

    Does that check out?
     
  7. Oct 3, 2007 #6

    Dick

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    Looks ok to me. Draw a graph of cos(x) if you don't believe me.
     
  8. Oct 3, 2007 #7
    I will......cause I don't.
























    No I won't...cause I do.
    but I probably should so I can see the relationships...

    Thanks again Dick.
     
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