A diving board oscillates with simple harmonic motion of frequency \({\bf{2}}{\bf{.8}}\;{\bf{cycles per second}}\). What is the maximum amplitude with which the end of the board can oscillate in order that a pebble placed there (Fig. 11–56) does not lose contact with the board during the oscillation?

Maximum acceleration applied on any object in vertical motion must be lesser than or equal to the acceleration due to gravity applied on that object.

The frequency of diving board is \(f = 2.8\;{\rm{Hz}}\).

According to the law of equilibrium, all forces must be balanced by the counter forces acting in opposite direction. Now, as shown in the figure below:

Here as shown in the figure, pebble is placed on the diving board without any constraints attached or any other forces acting upon it. Only force pebble is dealing with is gravity. So, the maximum acceleration that can be applied on the pebble is acceleration due to gravity\(g\). Now, if pebble oscillates on the board due to the oscillations made by the diving board then it will come back down due to gravity. But, if diving board also starts accelerating downwards then pebble will lose contact with the board.

Now, the maximum acceleration and the amplitude of the diving board varies as:

\(\begin{aligned}{c}{a_{{\rm{max}}}} &= \frac{{kA}}{m}\\ &= {\omega ^2}A\\ &= 4{{\rm{\pi }}^2}{f^2}A\end{aligned}\)

But, considering the forces acting on the pebble, maximum acceleration will act as:

\(\begin{aligned}{c}{a_{{\rm{max}}}} \le g\\4{{\rm{\pi }}^2}{f^2}A \le g\\A \le \frac{g}{{4{{\rm{\pi }}^2}{f^2}}}\\ \le \frac{{9.8\;{\rm{m/}}{{\rm{s}}^2}}}{{4{{\rm{\pi }}^2}{{\left( {2.8\;{\rm{Hz}}} \right)}^2}}}\\ \le 3.2 \times {10^{ - 2}}\;{\rm{m}}\end{aligned}\)

Hence, the maximum amplitude of the oscillations is \(3.2 \times {10^{ - 2}}\;{\rm{m}}\).