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Is this calculation correct for v = 2πfA ?

  1. Aug 25, 2011 #1
    Is this calculation correct for v = 2πfA ??

    1. The problem statement, all variables and given/known data

    Prove that maximum speed of a mass on a spring is given by 2πfA

    Ac = Centripetal acceleration
    r = radius
    v = velocity
    f = frequency
    π = pi

    2. Relevant equations

    Ac = (v^2)/r
    Ac = 4(π^2)r(f^2)


    3. The attempt at a solution

    If Ac = (v^2)/r
    and Ac = 4(π^2)r(f^2)

    then (v^2)/r = 4(π^2)r(f^2)

    so v^2 = 4(π^2)(r^2)(f^2)
    and v = 2πrf


    Just wondering if there is something wrong with this calculation?
     
  2. jcsd
  3. Aug 25, 2011 #2

    PeterO

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    Homework Helper

    Re: Is this calculation correct for v = 2πfA ??

    Is this mass being rotated in a "horizontal" circle on the end of a spring, or bouncing up and down on the end of a spring [where A might be the Amplitude?]
     
  4. Aug 25, 2011 #3
    Re: Is this calculation correct for v = 2πfA ??

    I messed up bad, this is a waste how do i remove the thread?
     
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