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

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

T=sqrt(m/k)^(1/2pi)
 
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TyErd said:
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.
 


oh sorry there's no sqrt, my bad its suppose to be m/k^1/2pi
 


denominator it is 1/(2pi), like the formula you wrote without the sqrt sign.
 


[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.
 


TyErd said:
oh sorry there's 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.