Video: Calculating the Frequency of a Diving Board

An unloaded diving board oscillates at a frequency of 8.06 Hz. The board has an effective mass of 14.2 kg. What is the frequency of the oscillations of the board if a diver of mass 86.5 kg jumps up and down on it?

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Video Transcript

An unloaded diving board oscillates at a frequency of 8.06 hertz. The board has an effective mass of 14.2 kilograms. What is the frequency of the oscillations of the board if a diver of mass 86.5 kilograms jumps up and down on it?

We can call this oscillation frequency we’re solving for πœ” and begin by recalling the relationship between πœ” and a spring system. We can recall that the natural frequency of a mass oscillating on a spring is equal to the square root of the spring constant π‘˜ divided by the mass amount π‘š. In our case, we’re told that the board, when there’s no diver on it, has a natural frequency of 8.06 hertz and that the mass of the board is effectively 14.2 kilograms. We can write that πœ” sub 𝑖, the natural frequency of the board by itself, is equal to the square root of the spring constant of the board divided by π‘š sub 𝑏; or, rearranging to solve for π‘˜, it’s equal to the mass of the board times the board’s natural frequency squared.

We want to solve for the frequency of the oscillations of the board when a diver of mass 86.5 kilograms is on it. This frequency, which we’ve called πœ”, will be equal to the square root of π‘˜ over the mass of the system: mass of the board plus the mass of the diver. Substituting in π‘š sub 𝑏 times πœ” sub 𝑖 squared for π‘˜, we now have an expression we’re ready to plug in for to solve for πœ”. When we enter these values on our calculator, we find that πœ”, to three significant figures, is 3.03 hertz. That’s the oscillation frequency of the board plus the diver.

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