Question

Astronauts on a distant planet set up a simple pendulum of length 1.2 m. The pendulum executes simple harmonic motion and makes 100 complete vibrations in 280 s. What is the magnitude of the acceleration due to gravity on this planet

Answer 1

6.0426 m/s^2

Step-by-step explanation:

The period of a simple pendulum is equal to 2pi sqroot L/g

T = 280s/100 rev = 2.8s

Plug in 1.2 m for L and 2.8s for T, then solve for g to get 6.04 m/s^2 (dependent on how you round)

Answer 2

If you use P = 2 * pi * (L / g)^1/2 for the period of the simple pendulum

g = 4 * pi^2 * 1.2 / 2.8^2 = 6.04 m/s2

Note: omega = 2 pi * f = 2 pi / P and omega = (g / L)^1/2