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

  • Thread starter TyErd
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  • #1
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how do you make k the subject in the time period of oscillation formula:

T=sqrt(m/k)^(1/2pi)
 

Answers and Replies

  • #2
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how do you make k the subject in the time period of oscillation formula:

T=sqrt(m/k)^(1/2pi)
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.
 
  • #3
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oh sorry theres no sqrt, my bad its suppose to be m/k^1/2pi
 
  • #4
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And is the exponent 1/(2pi) or (1/2)pi? IOW, is pi in the numerator or the denominator?
 
  • #5
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denominator it is 1/(2pi), like the formula you wrote without the sqrt sign.
 
  • #6
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[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.
 
  • #7
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so it is k=m/(T^2pi)???
 
  • #8
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  • #9
gabbagabbahey
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oh sorry theres no sqrt, my bad its suppose to be m/k^1/2pi
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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