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

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

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

Mark44
Mentor

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.
$$T = \left(\sqrt{\frac{m}{k}}\right)^{\frac{1}{2\pi}$$

But that doesn't look like anything I've seen.

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

Mark44
Mentor

And is the exponent 1/(2pi) or (1/2)pi? IOW, is pi in the numerator or the denominator?

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

Mark44
Mentor

$$T = \left(\frac{m}{k}\right)^{\frac{1}{2\pi}$$
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.

so it is k=m/(T^2pi)???

Mark44
Mentor

so it is k=m/(T^2pi)???
Right

gabbagabbahey
Homework Helper
Gold Member

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 $k$ and mass $m$? If so, it is incorrect.

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