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Two parakeets sit on a swing with their combined center of mass 10.0 cm below the pivot. At what frequency do they swing?
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Final Answer

$1.58 \textrm{ Hz}$

Solution Video

OpenStax College Physics Solution, Chapter 16, Problem 27 (Problems & Exercises) (0:38)

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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 <i>g</i> over <i>L</i>. 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.