A 0.400 kg block on a spring of force constant 100 N/m oscillates freely on a frictionless surface. What is the period of the motion?

1. The period of a mass on a spring is T = 2 pi sqrt(m/k). 2. Substitute the values: m / k = 0.400 kg / 100 N/m = 4.00 x 10^-3 s squared. 3. The square root of 4.00 x 10^-3 is 6.32 x 10^-2 s. 4. T = 2 pi x 6.32 x 10^-2 s = 0.397 s. 5. Leaving out the square root gives 0.0251 s, and inverting the ratio gives 2.52 s, which is the frequency route taken by mistake. _Source: OpenStax College Physics (CC BY 4.0), Ch 16 "Oscillatory Motion and Waves", section 16.3 Simple Harmonic Motion: A Special Periodic Motion_