A 900 kg car rests on a suspension of force constant 6.53 x 10^4 N/m and bounces freely on it. What are the frequency and period of the bounce?

1. The frequency of a simple harmonic oscillator is f = (1 / 2 pi) sqrt(k/m). 2. Substitute: k / m = 6.53 x 10^4 N/m divided by 900 kg = 72.6 per second squared. 3. The square root of 72.6 per second squared is 8.52 per second. 4. f = 8.52 / (2 pi) = 1.36 Hz. 5. Period is the reciprocal, T = 1 / 1.36 Hz = 0.738 s. 6. Quoting 8.52 Hz treats the angular frequency in radians per second as if it were a frequency in hertz. 7. Pairing 1.36 Hz with 1.36 s ignores that period and frequency are reciprocals, not equal numbers. _Source: OpenStax College Physics (CC BY 4.0), Ch 16 "Oscillatory Motion and Waves", section 16.3 Simple Harmonic Motion: A Special Periodic Motion_