Question

An astronaut is doing thee simple pendulum experiment on a different planet to measure the acceleration due to gravity. If the length of the pendulum is 45 cm and the period of oscillations is equal to 1.428 s, what is the acceleration due to the gravity of the planet

Answer 1

The acceleration due to gravity on the given planet is 8.71 m/s².

The given parameters;

• Length of the pendulum, L = 45 cm
• Period of the oscillation, T = 1.428 s

The acceleration due to gravity on the planet is calculated by applying following formula as follows;

$T = 2\pi \sqrt{\frac{l}{g} } \\\\\frac{T}{2 \pi } = \sqrt{\frac{l}{g}} \\\\\frac{T^2}{4\pi ^2} = \frac{l}{g} \\\\g = \frac{4\pi^2 l}{T^2} \\\\g = \frac{4 \times (\pi)^2 \times 0.45}{1.428^2} \\\\g = 8.71 \ m/s^2$

Thus, the acceleration due to gravity on the given planet is 8.71 m/s².

Learn more about period of oscillation here: https://brainly.com/question/20070798