Question

An astronaut has landed on planet n-40 and conducts an experiment to determine the acceleration due to gravity on that planet. She uses a simple pendulum that is 0.640 m long and measures that 10 complete oscillations 26.0 s. What is the acceleration of gravity on planet n-40?

Answer 1

3.73 m/s^2

Step-by-step explanation:

The period of a simple pendulum is given by

`T=2\pi \sqrt{(L)/(g)}` (1)

where

L is the length of the pendulum

g is the gravitational acceleration on the planet

The pendulum in the problem has length

L = 0.640 m

and makes 10 oscillations in 26.0 s; it means that its frequency is

`f=(10)/(26 s)=0.385 Hz`

And so its period is

`T=(1)/(f)=(1)/(0.385 Hz)=2.6 s`

So now we can solve equation (1) using L=0.640 m and T=2.6 s, so we can find the value of g on the planet:

`g=((2\pi)/(T))^2L=((2\pi)/(2.6 s))^2 (0.640 m)=3.73 m/s^2`