Science

Math

Humanities

... and beyond

Socratic Meta
Featured Answers

Topics

Question #4b882

1 Answer

Oct 7, 2016

I found:
$l_("Earth")=1m$
$l_("Moon")=20cm$

Explanation:

Here you could use the fact that the period $T$ of a pendulum depends upon its length $l$ and the acceleration of gravity $g$ as:
$T=2pisqrt(l/g)$

and rearranging:

$l=(T/(2pi))^2*g$

1) on Earth $g=9.8m/s^2$
so you get:
$l_("Earth")=(2/(2pi))^2*9.8=0.99~~1m$

2) On the Moon we have $g_("Moon")=1.6 m/s^2$ (from literature) and so:
$l_("Moon")=(2/(2pi))^2*1.6=0.16~~0.2m=20cm$

Related questions

See all questions in Simple Harmonic Motion - Pendulums

Impact of this question

2261 views around the world

You can reuse this answer
Creative Commons License