# 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 𝜔 and begin by recalling the relationship between 𝜔 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 𝑘 divided by the mass amount 𝑚. 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 𝜔 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 𝑚 sub 𝑏; or, rearranging to solve for 𝑘, 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 𝜔, will be equal to the square root of 𝑘 over the mass of the system: mass of the board plus the mass of the diver. Substituting in 𝑚 sub 𝑏 times 𝜔 sub 𝑖 squared for 𝑘, we now have an expression we’re ready to plug in for to solve for 𝜔. When we enter these values on our calculator, we find that 𝜔, to three significant figures, is 3.03 hertz. That’s the oscillation frequency of the board plus the diver.