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?

Respuesta :

Answer:

0.331 m

Explanation:

We are given that

Acceleration on the moon=[tex]1.62 m/s^2[/tex]

Acceleration due to gravity on earth=[tex]9.8 m/s^2[/tex]

Length of pendulum on the earth=2 m

We know that time period

T=[tex]2\pi\sqrt{\frac{l}{g}}[/tex]

Substitute the values then we get

T=[tex]2\pi\sqrt{\frac{2}{9.8}}[/tex]

Time period on moon=[tex]2\pi\sqrt{\frac{l}{1.62}}[/tex]

Time period on the moon=Time period on the earth

[tex]2\pi\sqrt{\frac{2}{9.8}}=2\pi\sqrt{\frac{l}{1.62}}[/tex]

[tex]\frac{1}{4.9}=\frac{l}{1.62}[/tex]

[tex]l=\frac{1.62}{4.9}[/tex]

l=0.331 m

Hence, the length of pendulum on moon=0.331 m