OpenStax College Physics for AP® Courses, Chapter 16, Problem 26 (Problems & Exercises)

This is College Physics Answers with Shaun Dychko. A simple pendulum has a length of 5.00 centimeters which is 5.00 times 10 to the minus 2 meters since the prefix 'centa' means multiply by 10 to the minus 2 and we want to find the frequency of this pendulum. Well we know a formula for its period is 2π times the square root of the length divided by acceleration due to gravity and frequency is the reciprocal of the period—it's 1 over period— and so if we take the reciprocal of this then that will give us the answer for frequency. So that's 1 over 2π times square root of g over L, where I flipped each of the fractions; this 2π is a fraction of 2π over 1 and then taking its reciprocal makes 1 over 2π. So then we have 1 over 2π times square root 9.80 meters per second squared divided by 5.00 times 10 to the minus 2 meters and that is 2.23 hertz. So this pendulum will swing a complete swing starting from its beginning point and back to its beginning point 2.23 times every second.