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

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

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

## Answers and Replies

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.

TyErd

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

Mentor

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

TyErd

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

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.

TyErd

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

Mentor

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

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.