This question has been difficult for me to answer for awhile. If anyone could help that would be great.

The period (the time for one complete swing back and forth) $p$, in seconds of a pendulum is related to its length, $L$, in metres, by the formula $$p= 2\pi \sqrt{\frac{L}{g}}$$ where $g= 9.8$ m/s. Rearrange the formula for $L$, and find the length needed for the pendulum to have a period of $1$ second.

I know that $L$ must be smaller than $1$, but I'm not sure how to rearrange the formula.

  • $\begingroup$ Do you mean $$p = 2\pi \sqrt{\frac{L}{g}}$$? In that case, square both sides and multiply by $g/(4\pi^2)$. $\endgroup$ – user61527 Nov 15 '13 at 22:09

Take $$p=2\pi\sqrt{\frac{L}{g}}$$ Divide both sides by $2\pi$: $$\frac{p}{2\pi} = \sqrt{\frac{L}{g}}$$ Square both sides:$$\left(\frac{p}{2\pi}\right)^2 = \frac{L}{G}$$ Multiply both sides by $G$: $$G \cdot \left(\frac{p}{2\pi}\right)^2 = L$$


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