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T = sqrt(m/k)^(1/2pi), solve for k

  1. Mar 17, 2010 #1
    how do you make k the subject in the time period of oscillation formula:

    T=sqrt(m/k)^(1/2pi)
     
  2. jcsd
  3. Mar 17, 2010 #2

    Mark44

    Staff: Mentor

    Re: Transposition

    Your formula is ambiguous. This is what it looks like to me.
    [tex]T = \left(\sqrt{\frac{m}{k}}\right)^{\frac{1}{2\pi}[/tex]

    But that doesn't look like anything I've seen.
     
  4. Mar 17, 2010 #3
    Re: Transposition

    oh sorry theres no sqrt, my bad its suppose to be m/k^1/2pi
     
  5. Mar 17, 2010 #4

    Mark44

    Staff: Mentor

    Re: Transposition

    And is the exponent 1/(2pi) or (1/2)pi? IOW, is pi in the numerator or the denominator?
     
  6. Mar 17, 2010 #5
    Re: Transposition

    denominator it is 1/(2pi), like the formula you wrote without the sqrt sign.
     
  7. Mar 17, 2010 #6

    Mark44

    Staff: Mentor

    Re: Transposition

    [tex]T = \left(\frac{m}{k}\right)^{\frac{1}{2\pi}[/tex]
    If it's this one, raise each side to the power 2pi, then take the reciprocal of both sides. That should get you close to being able to solve for k.
     
  8. Mar 17, 2010 #7
    Re: Transposition

    so it is k=m/(T^2pi)???
     
  9. Mar 17, 2010 #8

    Mark44

    Staff: Mentor

    Re: Transposition

    Right
     
  10. Mar 17, 2010 #9

    gabbagabbahey

    User Avatar
    Homework Helper
    Gold Member

    Re: Transposition

    Is this equation supposed to represent the period of oscillation of a Harmonic oscillator with spring constant [itex]k[/itex] and mass [itex]m[/itex]? If so, it is incorrect.

    The actual period is [tex]T=\frac{1}{2\pi}\sqrt{\frac{m}{k}}[/tex], which is quite different from the formula you've written.
     
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