Question

A pendulum is placed on a distant planet. The length is one meter, and the measured period is 1.4 seconds, what is the acceleration of gravity on that planet?

Answer 1

Answer:

$a=20.14\ m/s^2$

Explanation:

The time period of the simple pendulum is given by :

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

l is the length of the pendulum

g is the acceleration due to gravity

We have,

T = 1.4 s, l = 1 m

So,

$T^2=\dfrac{4\pi^2 l}{g}\\\\g= \dfrac{4\pi^2 l}{T^2}\\\\g= \dfrac{4\pi^2 \times 1}{(1.4)^2}\\\\g=20.14\ m/s^2$

So, the acceleration due to gravity of that planet is $20.14\ m/s^2$.