Question

A pendulum is taken to another planet where it swings back and forth exactly 17 times every 35.66 seconds. The pendulum's arm is 0.74 m long. What is the acceleration due to gravity on this planet in units of m/s?

Answer 1

Answer: g = 6.65 m/s²

Explanation: The frequency of a simple pendulum is given by the formulae below

f = 1/2π *(√g/l)

Where f = frequency = number of oscillation /time =17/35.66 = 0.477Hz

g = acceleration due gravity in another planet

l = length of simple pendulum = 0.74m

0.477 = 1/2π * (√g/0.74)

0.477 = 1/2 * 3.142 * (√g/0.74)

0.477 = 1/ 6.284 * (√g/0.74)

0.477 = 0.1591 * (√g/0.74)

0.477/0.1591 = (√g/0.74)

2.998 = (√g/0.74)

By taking the square of both sides

8.988 = g/ 0.74

g = 6.65 m/s²