Question

What is the free-fall acceleration on a planet where the period of a 1.07 m long pendulum is 2.02 s?

Answer 1

Answer:

10.34 m/s^2

Explanation:

length, L = 1.07 m

Time period, T = 2.02 s

The formula of the time period of the pendulum is

$T = 2\pi \sqrt{\frac{L}{g}}$

where, g is the free fall acceleration on that planet.

$g=4\pi ^{2}\frac{L}{T^{2}}$

$g=4\pi ^{2}\frac{1.07}{2.02^{2}}$

g = 10.34 m/s^2