A 0.500 kg block on a spring of force constant 200 N/m oscillates with an amplitude of 0.100 m. What is its maximum speed?

1. Maximum speed for a mass on a spring is v_max = sqrt(k/m) X. 2. Substitute the ratio: k / m = 200 N/m divided by 0.500 kg = 400 per second squared. 3. Take the square root: sqrt(400 per second squared) = 20.0 per second. 4. Multiply by the amplitude: v_max = (20.0 per second)(0.100 m) = 2.00 m/s. 5. Quoting 20.0 m/s stops before multiplying by the amplitude and leaves an angular frequency behind. 6. Quoting 40.0 m/s skips the square root and then multiplies, which compounds two errors. _Source: OpenStax College Physics (CC BY 4.0), Ch 16 "Oscillatory Motion and Waves", section 16.5 Energy and the Simple Harmonic Oscillator_