Question

The free-fall acceleration on the moon is 1.62 m/s2 . What is the length of a pendulum whose period on the moon matches the period of a 2.00-m-long pendulum on the earth?

Answer 1

Answer:

0.331 m

Explanation:

We are given that

Acceleration on the moon=$1.62 m/s^2$

Acceleration due to gravity on earth=$9.8 m/s^2$

Length of pendulum on the earth=2 m

We know that time period

T=$2\pi\sqrt{\frac{l}{g}}$

Substitute the values then we get

T=$2\pi\sqrt{\frac{2}{9.8}}$

Time period on moon=$2\pi\sqrt{\frac{l}{1.62}}$

Time period on the moon=Time period on the earth

$2\pi\sqrt{\frac{2}{9.8}}=2\pi\sqrt{\frac{l}{1.62}}$

$\frac{1}{4.9}=\frac{l}{1.62}$

$l=\frac{1.62}{4.9}$

l=0.331 m

Hence, the length of pendulum on moon=0.331 m