Question

On earth, you swing a simple pendulum in simple harmonic motion with a period of 1.6 seconds. what is the period of this same pendulum on the moon? assume the moon's acceleration due to gravity is 1/6 of the acceleration due to gravity on earth.

Answer 1

The period of a simple pendulum is given by
$T= 2 \pi \sqrt{ \frac{L}{g} }$
where
L is the pendulum length
g is the acceleration of gravity

If we move the same pendulum from Earth to the Moon, its length L remains the same, while the acceleration of gravity g changes. So we can write the period of the pendulum on Earth as:
$T_e= 2 \pi \sqrt{ \frac{L}{g_e} }$
where $g_e$ is the acceleration of gravity on Earth, while the period of the pendulum on the Moon is
$T_m= 2 \pi \sqrt{ \frac{L}{g_m} }$
where $g_m$ is the acceleration of gravity on the Moon. 

If we do the ratio of the two periods, we get
$\frac{T_m}{T_e} = \sqrt{ \frac{g_e}{g_m} }$
but the gravity acceleration on the Moon is 1/6 of the gravity acceleration on Earth, so we can write $g_e = 6 g_m$ and we can rewrite the previous ratio as
$\frac{T_m}{T_e} = \sqrt{ \frac{6 g_m}{g_m} }= \sqrt{6}$

so the period of the pendulum on the Moon is
$T_m = \sqrt{6} T_e = \sqrt{6} (1.6 s)=3.9 s$