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Suppose a diving board with no one on it bounces up and down in a simple harmonic motion with a frequency of 4.00 Hz. The board has an effective mass of 10.0 kg. What is the frequency of the simple harmonic motion of a 75.0-kg diver on the board?
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Final Answer

$1.37 \textrm{ Hz}$

Solution Video

OpenStax College Physics Solution, Chapter 16, Problem 19 (Problems & Exercises) (1:27)

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Video Transcript

This is College Physics Answers with Shaun Dychko. The frequency of oscillation of the diving board on its own is <i>f1</i> and it's going to be one over two Pi times the square root of the spring constant, divided by the diving board's mass <i>m1</i>. And then, when a person goes on the diving board we'll call that new frequency <i>f2</i> and that equals one over two Pi times the same spring constant because it's the same diving board but divided by different mass, the mass of diving board plus the mass of the person that goes on the diving board. Now, <i>f2</i> divided by <i>f1</i> is going to be <i>f2</i> one over two Pi times square root <i>k</i> over <i>m2</i>, multiplied by the reciprocal of <i>f1</i>. I like to multiply by the reciprocals instead of dividing by fractions. So, multiplying by two Pi times square root <i>m1</i> over <i>k</i>. So each of the factors here was flipped. And this equals, square root <i>m1</i> over <i>m2</i>, because the two <i>k</i> is cancel and the square root <i>k</i> is cancel. So I should say, the two Pi is canceled. And we have <i>f2</i> over <i>f1</i> is square root <i>m1</i> over <i>m2</i>. And then multiply both sides by <i>f1</i>, and you get frequency two is frequency one times square root <i>m1</i> over <i>m2</i>. So that's four Hertz of frequency originally of the diving board on its own times square root of mass of the diving board on its own at ten kilograms divided by <i>m2</i> which is the diving board and the person's mass combined. And this works out to 1.37 Hertz.