Answer:
[tex]g=3.76\ m/s^2[/tex]
Explanation:
Given that,
The length of a simple pendulum, l = 2.2 m
The time period of oscillations, T = 4.8 s
We need to find the surface gravity of the planet. The time period of the planet is given by the relation as follows :
[tex]T=2\pi \sqrt{\dfrac{l}{g}} \\\\T^2=4\pi ^2\times \dfrac{l}{g}\\\\g=\dfrac{4\pi ^2 l}{T^2}[/tex]
Put all the values,
[tex]g=\dfrac{4\pi ^2 \times 2.2}{(4.8)^2}\\\\=3.76\ m/s^2[/tex]
So, the value of the surface gravity of the planet is equal to [tex]3.76\ m/s^2[/tex].
