Question
Two parakeets sit on a swing with their combined center of mass 10.0 cm below the pivot. At what frequency do they swing?
Question by OpenStax is licensed under CC BY 4.0
Final Answer

1.58 Hz1.58 \textrm{ Hz}

Solution video

OpenStax College Physics, Chapter 16, Problem 27 (Problems & Exercises)

OpenStax College Physics, Chapter 16, Problem 27 (PE) video thumbnail

In order to watch this solution you need to have a subscription.

Start free trial Log in
vote with a rating of votes with an average rating of .

Calculator Screenshots

  • OpenStax College Physics, Chapter 16, Problem 27 (PE) calculator screenshot 1
Video Transcript
This is College Physics Answers with Shaun Dychko. The period of oscillation of these swinging parakeets will be two pi times the square root of the distance from the pivot to their center of mass, divided by acceleration due to gravity. Now, frequency is the reciprocal of period so if we take the reciprocal of this which is the same as raising to the exponent negative one, or flipping each factor. We get one over two pi times square root g over L. So it's one over two pi times square root 9.81 meters per second squared, divided by ten centimeters written as ten times ten to the minus two meters. This gives 1.58 Hertz.