0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
1
$\begingroup$

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 $G$: $$G \cdot \left(\frac{p}{2\pi}\right)^2 = L$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.