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

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

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

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- 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)

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

- #2

Mark44

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Your formula is ambiguous. This is what it looks like to me.how do you make k the subject in the time period of oscillation formula:

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

[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

Mark44

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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

Mark44

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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

Mark44

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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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