What is the period of a simple pendulum 1.00 m long at a place where g is 9.80 m/s squared?

1. The period of a simple pendulum is T = 2 pi sqrt(L/g). 2. Substitute: L / g = 1.00 m / 9.80 m/s squared = 0.102 s squared. 3. The square root of 0.102 s squared is 0.319 s. 4. T = 2 pi x 0.319 s = 2.01 s. 5. Stopping at 0.319 s forgets the factor of 2 pi, which is the trap behind the 0.32 s option. _Source: OpenStax College Physics (CC BY 4.0), Ch 16 "Oscillatory Motion and Waves", section 16.4 The Simple Pendulum_