Question

After landing on an unfamiliar planet, a space explorer constructs a simple pendulum of length 52.0 cm . The explorer finds that the pendulum completes 104 full swing cycles in a time of 141 s .

What is the magnitude of the gravitational acceleration on this planet?

g_planet=        m/s²

Answer 1

Solution)

We know, g = (4π^2*L)/P^2

Given, pendulum completes 104 full swings in 141 s, so

P = 104/141s

Given that L = 52 cm

g = (4π^2*L)/P^2

= [4π^2*(0.52 m)]/(141/104 s)^2

= 16.37 m/s^2 (Ans)

=========

Good luck!:)