### 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 weβre solving for π and begin by recalling the relationship between π
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. In our case, weβre told that the
board, when thereβs 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 sub π, the
natural frequency of the board by itself, is equal to the square root of the spring
constant of the board divided by π sub π; or, rearranging to solve for π, itβs
equal to the mass of the board times the boardβs 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 weβve called
π, 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 sub π times π
sub π squared for π, we now have an expression weβre ready to plug in for to solve
for π. When we enter these values on our
calculator, we find that π, to three significant figures, is 3.03 hertz. Thatβs the oscillation frequency of
the board plus the diver.