To solve this problem it is necessary to apply the concepts related to the Period based on gravity and length.
Mathematically this concept can be expressed as
[tex]T= 2\pi \sqrt{\frac{l}{g}}[/tex]
Where,
l = Length
g = Gravitational acceleration
First we will find the period that with the characteristics presented can be given on Mars and then we can find the length of the pendulum at the desired time.
The period on Mars with the given length of 0.99396m and the gravity of the moon (approximately [tex]1.62m / s ^ 2)[/tex] will be
[tex]T= 2\pi \sqrt{\frac{l}{g}}[/tex]
[tex]T= 2\pi \sqrt{\frac{0.99396}{1.62}}[/tex]
[tex]T = 4.921seg[/tex]
For the second question posed, it would be to find the length so that the period is 2 seconds, that is:
[tex]T= 2\pi \sqrt{\frac{l}{g}}[/tex]
[tex]2= 2\pi \sqrt{\frac{l}{1.62}}[/tex]
[tex]l = 0.16414m[/tex]
Therefore, we can observe also that the shorter distance would be the period compared to the first result given.
