Question #4b882

1 Answer
Oct 7, 2016

Answer:

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#