The acceleration due to gravity on the surface of the moon is `1.7ms^(-2)`. What is the time perioid of a simple pendulum on the surface of the moon, if its time period on the surface of earth is `3.5s ?` Take `g=9.8ms^(-2)` on the surface of the earth.

Explanation

To find the time period of a simple pendulum on the surface of the moon, we can use the relationship between the time period of a pendulum and the acceleration due to gravity. The formula for the time period \( T \) of a simple pendulum is given by:

\[ T = 2\pi \sqrt{\frac{L}{g}} \]

where:
- \( T \) is the time period,
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