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

1. Mar 17, 2010

### TyErd

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

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

2. Mar 17, 2010

### Staff: Mentor

Re: Transposition

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.

3. Mar 17, 2010

### TyErd

Re: Transposition

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

4. Mar 17, 2010

### Staff: Mentor

Re: Transposition

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

5. Mar 17, 2010

### TyErd

Re: Transposition

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

6. Mar 17, 2010

### Staff: Mentor

Re: Transposition

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

7. Mar 17, 2010

### TyErd

Re: Transposition

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

8. Mar 17, 2010

### Staff: Mentor

Re: Transposition

Right

9. Mar 17, 2010

### gabbagabbahey

Re: Transposition

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.