Rearranging angular velocity equation to make $T$ the subject

Question

I want to rearrange the formula for angular velocity $\omega = \dfrac{2\pi}{T}$, to make $T$ the subject as I wish to find the period.

Would the correct answer be $T = \frac{\omega}{2\pi}$ or would it be $T = \frac{2\pi}{\omega}$?

And is there a certain rule you should follow when rearranging ?

Answer 1

Starting with $$\omega = \dfrac{2\pi}{T}$$ First multiply both sides by $T$:

$$\omega\color{red}{\cdot T} = \dfrac{2\pi}{T}\color{red}{\cdot T} = 2\pi$$

Divide both sides by $\omega$ to isolate $T$ on the left:

$$\color{red}{\frac{\color{black}{\omega T}}{\omega}} = \color{red}{\frac{\color{black}{2\pi}}{\omega}}$$

Which, after cancelling, leaves $$T = \frac{2\pi}{\omega}$$

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You could also check your answer for dimensional correctness.

Consider the equation $T = \frac{\omega}{2\pi}$. Assuming everything's in SI units, on the left the units are seconds and on the right the units are inverse seconds. That couldn't be right.

Consider the equation $T = \frac{2\pi}{\omega}$. On the left the units are seconds. On the right, the units are 1 over inverse seconds. I.e. seconds. So this formula is at least correct dimensionally.

Answer 2

From $$ \frac{a}{b}=\frac{c}{d} \tag1 $$ by multiplying out $(1)$ by $bd$ one gets $$ ad=bc\tag2 $$ by dividing $(2)$ by $cd$ one gets $$ \frac{a}{c}=\frac{b}{d}. \tag3 $$ Applying it to $$ w=\frac{2\pi}{T} $$ one gets $$ T=\frac{2\pi}{w}. $$

Answer 3

It is quite simple. You have $\omega = \dfrac{2\pi}{T}$ since you want to make T the subject, multiply the whole equation by T and you will get $$\omega{\cdot T} = \dfrac{2\pi}{T}{\cdot T} = 2\pi$$

On bringing ${\omega}$ on right you will have $T = \frac{2\pi}{\omega}$